Difference between revisions of "Cauchy-Schwarz Formula"

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=== Sedrakyan's lemma - Positive real numbers ===
 
=== Sedrakyan's lemma - Positive real numbers ===
−
[[Sedrakyan's inequality]], also called [[Arthur Engel (mathematician)|Engel]]'s form, the T2 lemma, or [[Titu Andreescu|Titu]]'s lemma, states that for positive reals:
+
Sedrakyan's inequality, also called Engel's form, the T2 lemma, or Titu's lemma, states that for positive reals:
−
<math display=block>\frac{\left(\sum_{i=1}^n u_i\right)^2 }{\sum_{i=1}^n v_i} \leq \sum_{i=1}^n \frac{u_i^2}{v_i} \quad \text{ or equivalently, } \quad \frac{u^2_1}{v_1} + \frac{u^2_2}{v_2} + \cdots + \frac{u^2_n}{v_n} \geq \frac{\left(u_1 + u_2 + \cdots + u_n\right)^2}{v_1 + v_2 + \cdots + v_n}.</math>
+
<math>\frac{\left(\sum_{i=1}^n u_i\right)^2 }{\sum_{i=1}^n v_i} \leq \sum_{i=1}^n \frac{u_i^2}{v_i} \quad \text{ or equivalently, } \quad \frac{u^2_1}{v_1} + \frac{u^2_2}{v_2} + \cdots + \frac{u^2_n}{v_n} \geq \frac{\left(u_1 + u_2 + \cdots + u_n\right)^2}{v_1 + v_2 + \cdots + v_n}.</math>
  
 
It is a direct consequence of the Cauchy–Schwarz inequality, obtained by using the [[dot product]] on <math>\R^n</math> upon substituting <math>u_i' = \frac{u_i}{\sqrt{v_i}} \text{ and } v_i' = \sqrt{v_i}.</math> This form is especially helpful when the inequality involves fractions where the numerator is a [[Square number|perfect square]].
 
It is a direct consequence of the Cauchy–Schwarz inequality, obtained by using the [[dot product]] on <math>\R^n</math> upon substituting <math>u_i' = \frac{u_i}{\sqrt{v_i}} \text{ and } v_i' = \sqrt{v_i}.</math> This form is especially helpful when the inequality involves fractions where the numerator is a [[Square number|perfect square]].
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The real vector space <math>\R^2</math> denotes the 2-dimensional plane. It is also the 2-dimensional [[Euclidean space]] where the inner product is the [[dot product]].  
 
The real vector space <math>\R^2</math> denotes the 2-dimensional plane. It is also the 2-dimensional [[Euclidean space]] where the inner product is the [[dot product]].  
 
If <math>\mathbf{v} = \left(v_1, v_2\right)</math> and <math>\mathbf{u} = \left(u_1, u_2\right)</math> then the Cauchy–Schwarz inequality becomes:
 
If <math>\mathbf{v} = \left(v_1, v_2\right)</math> and <math>\mathbf{u} = \left(u_1, u_2\right)</math> then the Cauchy–Schwarz inequality becomes:
−
<math display=block>\langle \mathbf{u}, \mathbf{v} \rangle^2 = (\|\mathbf{u}\| \|\mathbf{v}\| \cos \theta)^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2,</math>
+
<math>\langle \mathbf{u}, \mathbf{v} \rangle^2 = (\|\mathbf{u}\| \|\mathbf{v}\| \cos \theta)^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2,</math>
 
where <math>\theta</math> is the [[angle]] between <math>u</math> and <math>v.</math>
 
where <math>\theta</math> is the [[angle]] between <math>u</math> and <math>v.</math>
  
 
The form above is perhaps the easiest in which to understand the inequality, since the square of the cosine can be at most 1, which occurs when the vectors are in the same or opposite directions. It can also be restated in terms of the vector coordinates <math>v_1, v_2, u_1, \text{ and } u_2</math> as
 
The form above is perhaps the easiest in which to understand the inequality, since the square of the cosine can be at most 1, which occurs when the vectors are in the same or opposite directions. It can also be restated in terms of the vector coordinates <math>v_1, v_2, u_1, \text{ and } u_2</math> as
−
<math display=block>\left(u_1 v_1 + u_2 v_2\right)^2 \leq \left(u_1^2 + u_2^2\right) \left(v_1^2 + v_2^2\right),</math>
+
<math>\left(u_1 v_1 + u_2 v_2\right)^2 \leq \left(u_1^2 + u_2^2\right) \left(v_1^2 + v_2^2\right),</math>
 
where equality holds if and only if the vector <math>\left(u_1, u_2\right)</math> is in the same or opposite direction as the vector <math>\left(v_1, v_2\right),</math> or if one of them is the zero vector.
 
where equality holds if and only if the vector <math>\left(u_1, u_2\right)</math> is in the same or opposite direction as the vector <math>\left(v_1, v_2\right),</math> or if one of them is the zero vector.
  
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=== ℝ<sup>''n''</sup> - ''n''-dimensional Euclidean space ===
 
=== ℝ<sup>''n''</sup> - ''n''-dimensional Euclidean space ===
 
{{anchor|real_number_proof}}In [[Euclidean space]] <math>\R^n</math> with the standard inner product, which is the [[dot product]], the Cauchy–Schwarz inequality becomes:
 
{{anchor|real_number_proof}}In [[Euclidean space]] <math>\R^n</math> with the standard inner product, which is the [[dot product]], the Cauchy–Schwarz inequality becomes:
−
<math display=block>\left(\sum_{i=1}^n u_i v_i\right)^2 \leq \left(\sum_{i=1}^n u_i^2\right) \left(\sum_{i=1}^n v_i^2\right).</math>
+
<math>\left(\sum_{i=1}^n u_i v_i\right)^2 \leq \left(\sum_{i=1}^n u_i^2\right) \left(\sum_{i=1}^n v_i^2\right).</math>
  
 
The Cauchy–Schwarz inequality can be proved using only ideas from elementary algebra in this case.  
 
The Cauchy–Schwarz inequality can be proved using only ideas from elementary algebra in this case.  
 
Consider the following [[quadratic polynomial]] in <math>x</math>
 
Consider the following [[quadratic polynomial]] in <math>x</math>
−
<math display=block>0 \leq \left(u_1 x + v_1\right)^2 + \cdots + \left(u_n x + v_n\right)^2 = \left(\sum_i u_i^2\right) x^2 + 2 \left(\sum_i u_i v_i\right) x + \sum_i v_i^2.</math>
+
<math>0 \leq \left(u_1 x + v_1\right)^2 + \cdots + \left(u_n x + v_n\right)^2 = \left(\sum_i u_i^2\right) x^2 + 2 \left(\sum_i u_i v_i\right) x + \sum_i v_i^2.</math>
  
 
Since it is nonnegative, it has at most one real root for <math>x,</math> hence its [[discriminant]] is less than or equal to zero. That is,
 
Since it is nonnegative, it has at most one real root for <math>x,</math> hence its [[discriminant]] is less than or equal to zero. That is,
−
<math display=block>\left(\sum_i u_i v_i\right)^2 - \left(\sum_i {u_i^2}\right)  \left(\sum_i {v_i^2}\right) \leq 0,</math>
+
<math>\left(\sum_i u_i v_i\right)^2 - \left(\sum_i {u_i^2}\right)  \left(\sum_i {v_i^2}\right) \leq 0,</math>
  
 
which yields the Cauchy–Schwarz inequality.
 
which yields the Cauchy–Schwarz inequality.
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=== ℂ<sup>''n''</sup> - ''n''-dimensional Complex space===
 
=== ℂ<sup>''n''</sup> - ''n''-dimensional Complex space===
 
If <math>\mathbf{u}, \mathbf{v} \in \Complex^n</math> with <math>\mathbf{u} = \left(u_1, \ldots, u_n\right)</math> and <math>\mathbf{v} = \left(v_1, \ldots, v_n\right)</math> (where <math>u_1, \ldots, u_n \in \Complex</math> and <math>v_1, \ldots, v_n \in \Complex</math>) and if the inner product on the vector space <math>\Complex^n</math> is the canonical complex inner product (defined by <math>\langle \mathbf{u}, \mathbf{v} \rangle := u_1 \overline{v_1} + \cdots + u_{n} \overline{v_n},</math> where the bar notation is used for [[Complex conjugate|complex conjugation]]), then the inequality may be restated more explicitly as follows:  
 
If <math>\mathbf{u}, \mathbf{v} \in \Complex^n</math> with <math>\mathbf{u} = \left(u_1, \ldots, u_n\right)</math> and <math>\mathbf{v} = \left(v_1, \ldots, v_n\right)</math> (where <math>u_1, \ldots, u_n \in \Complex</math> and <math>v_1, \ldots, v_n \in \Complex</math>) and if the inner product on the vector space <math>\Complex^n</math> is the canonical complex inner product (defined by <math>\langle \mathbf{u}, \mathbf{v} \rangle := u_1 \overline{v_1} + \cdots + u_{n} \overline{v_n},</math> where the bar notation is used for [[Complex conjugate|complex conjugation]]), then the inequality may be restated more explicitly as follows:  
−
<math display=block>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2  
+
<math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2  
 
= \left|\sum_{k=1}^{n}u_k\bar{v}_k\right|^2  
 
= \left|\sum_{k=1}^{n}u_k\bar{v}_k\right|^2  
 
\leq \langle \mathbf{u}, \mathbf{u} \rangle \langle \mathbf{v}, \mathbf{v} \rangle  
 
\leq \langle \mathbf{u}, \mathbf{u} \rangle \langle \mathbf{v}, \mathbf{v} \rangle  
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That is,
 
That is,
−
<math display=block>\left|u_1 \bar{v}_1 + \cdots + u_n \bar{v}_n\right|^2 \leq \left(\left|u_1\right|^2 + \cdots + \left|u_n\right|^2\right) \left(\left|v_1\right|^2 + \cdots + \left|v_n\right|^2\right).</math>
+
<math>\left|u_1 \bar{v}_1 + \cdots + u_n \bar{v}_n\right|^2 \leq \left(\left|u_1\right|^2 + \cdots + \left|u_n\right|^2\right) \left(\left|v_1\right|^2 + \cdots + \left|v_n\right|^2\right).</math>
  
 
=== L<sup>2</sup> ===
 
=== L<sup>2</sup> ===
  
 
For the inner product space of [[square-integrable]] complex-valued [[function (mathematics)|functions]], the following inequality:
 
For the inner product space of [[square-integrable]] complex-valued [[function (mathematics)|functions]], the following inequality:
−
<math display=block>\left|\int_{\R^n} f(x) \overline{g(x)}\,dx\right|^2  \leq  \int_{\R^n} |f(x)|^2\,dx \int_{\R^n} |g(x)|^2 \,dx.</math>
+
<math>\left|\int_{\R^n} f(x) \overline{g(x)}\,dx\right|^2  \leq  \int_{\R^n} |f(x)|^2\,dx \int_{\R^n} |g(x)|^2 \,dx.</math>
  
 
The [[Hölder inequality]] is a generalization of this.
 
The [[Hölder inequality]] is a generalization of this.
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=== Analysis ===
 
=== Analysis ===
 
In any [[inner product space]], the [[triangle inequality]] is a consequence of the Cauchy–Schwarz inequality, as is now shown:  
 
In any [[inner product space]], the [[triangle inequality]] is a consequence of the Cauchy–Schwarz inequality, as is now shown:  
−
<math display=block>\begin{alignat}{4}
+
<math>\begin{alignat}{4}
 
\|\mathbf{u} + \mathbf{v}\|^2  
 
\|\mathbf{u} + \mathbf{v}\|^2  
 
&= \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v} \rangle && \\
 
&= \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v} \rangle && \\
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Taking square roots gives the triangle inequality:
 
Taking square roots gives the triangle inequality:
−
<math display=block>\|\mathbf{u} + \mathbf{v}\| \leq \|\mathbf{u}\| + \|\mathbf{v}\|.</math>
+
<math>\|\mathbf{u} + \mathbf{v}\| \leq \|\mathbf{u}\| + \|\mathbf{v}\|.</math>
  
 
The Cauchy–Schwarz inequality is used to prove that the inner product is a [[continuous function]] with respect to the [[topology]] induced by the inner product itself.<ref>{{cite book|last1=Bachman|first1=George|last2=Narici|first2=Lawrence|date=2012-09-26|title=Functional Analysis|publisher=Courier Corporation|isbn=9780486136554|pages=141|url=https://books.google.com/books?id=_lTDAgAAQBAJ}}</ref><ref>{{cite book|last=Swartz|first=Charles|date=1994-02-21|title=Measure, Integration and Function Spaces|publisher=World Scientific|isbn=9789814502511|pages=236|url=https://books.google.com/books?id=SsbsCgAAQBAJ}}</ref>
 
The Cauchy–Schwarz inequality is used to prove that the inner product is a [[continuous function]] with respect to the [[topology]] induced by the inner product itself.<ref>{{cite book|last1=Bachman|first1=George|last2=Narici|first2=Lawrence|date=2012-09-26|title=Functional Analysis|publisher=Courier Corporation|isbn=9780486136554|pages=141|url=https://books.google.com/books?id=_lTDAgAAQBAJ}}</ref><ref>{{cite book|last=Swartz|first=Charles|date=1994-02-21|title=Measure, Integration and Function Spaces|publisher=World Scientific|isbn=9789814502511|pages=236|url=https://books.google.com/books?id=SsbsCgAAQBAJ}}</ref>
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=== Geometry ===
 
=== Geometry ===
 
The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any [[real numbers|real]] inner-product space by defining:<ref>{{cite book|last=Ricardo|first=Henry|date=2009-10-21|title=A Modern Introduction to Linear Algebra|publisher=CRC Press|isbn=9781439894613|pages=18|url=https://books.google.com/books?id=s7bMBQAAQBAJ}}</ref><ref>{{cite book|last1=Banerjee|first1=Sudipto|last2=Roy|first2=Anindya|date=2014-06-06|title=Linear Algebra and Matrix Analysis for Statistics|publisher=CRC Press|isbn=9781482248241|pages=181|url=https://books.google.com/books?id=WDTcBQAAQBAJ}}</ref>
 
The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any [[real numbers|real]] inner-product space by defining:<ref>{{cite book|last=Ricardo|first=Henry|date=2009-10-21|title=A Modern Introduction to Linear Algebra|publisher=CRC Press|isbn=9781439894613|pages=18|url=https://books.google.com/books?id=s7bMBQAAQBAJ}}</ref><ref>{{cite book|last1=Banerjee|first1=Sudipto|last2=Roy|first2=Anindya|date=2014-06-06|title=Linear Algebra and Matrix Analysis for Statistics|publisher=CRC Press|isbn=9781482248241|pages=181|url=https://books.google.com/books?id=WDTcBQAAQBAJ}}</ref>
−
<math display=block>\cos\theta_{\mathbf{u} \mathbf{v}} = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \|\mathbf{v}\|}.</math>
+
<math>\cos\theta_{\mathbf{u} \mathbf{v}} = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \|\mathbf{v}\|}.</math>
  
 
The Cauchy–Schwarz inequality proves that this definition is sensible, by showing that the right-hand side lies in the interval {{math|[&minus;1,&nbsp;1]}} and justifies the notion that (real) [[Hilbert space]]s are simply generalizations of the [[Euclidean space]]. It can also be used to define an angle in [[complex numbers|complex]] [[inner-product space]]s, by taking the absolute value or the real part of the right-hand side,<ref>{{cite book|last=Valenza|first=Robert J.|date=2012-12-06|title=Linear Algebra: An Introduction to Abstract Mathematics|publisher=Springer Science & Business Media|isbn=9781461209010|pages=146|url=https://books.google.com/books?id=7x8MCAAAQBAJ}}</ref><ref>{{cite book|last=Constantin|first=Adrian|date=2016-05-21|title=Fourier Analysis with Applications|publisher=Cambridge University Press|isbn=9781107044104|pages=74|url=https://books.google.com/books?id=JnMZDAAAQBAJ}}</ref> as is done when extracting a metric from [[Fidelity of quantum states|quantum fidelity]].
 
The Cauchy–Schwarz inequality proves that this definition is sensible, by showing that the right-hand side lies in the interval {{math|[&minus;1,&nbsp;1]}} and justifies the notion that (real) [[Hilbert space]]s are simply generalizations of the [[Euclidean space]]. It can also be used to define an angle in [[complex numbers|complex]] [[inner-product space]]s, by taking the absolute value or the real part of the right-hand side,<ref>{{cite book|last=Valenza|first=Robert J.|date=2012-12-06|title=Linear Algebra: An Introduction to Abstract Mathematics|publisher=Springer Science & Business Media|isbn=9781461209010|pages=146|url=https://books.google.com/books?id=7x8MCAAAQBAJ}}</ref><ref>{{cite book|last=Constantin|first=Adrian|date=2016-05-21|title=Fourier Analysis with Applications|publisher=Cambridge University Press|isbn=9781107044104|pages=74|url=https://books.google.com/books?id=JnMZDAAAQBAJ}}</ref> as is done when extracting a metric from [[Fidelity of quantum states|quantum fidelity]].
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=== Probability theory ===
 
=== Probability theory ===
 
<!-- For the multivariate case,{{clarify|reason=define GE operator here|date=July 2011}}<ref>{{cite journal| last=Gautam|first=Tripathi|title=A matrix extension of the Cauchy-Schwarz inequality|journal=Economics Letters|date=4 December 1998|url=http://web2.uconn.edu/tripathi/published-papers/cs.pdf|doi=10.1016/s0165-1765(99)00014-2}}</ref>
 
<!-- For the multivariate case,{{clarify|reason=define GE operator here|date=July 2011}}<ref>{{cite journal| last=Gautam|first=Tripathi|title=A matrix extension of the Cauchy-Schwarz inequality|journal=Economics Letters|date=4 December 1998|url=http://web2.uconn.edu/tripathi/published-papers/cs.pdf|doi=10.1016/s0165-1765(99)00014-2}}</ref>
−
<math display=block>\operatorname{Var}(Y) \geq \operatorname{Cov} (Y, X) \operatorname{Var}^{-1}(X) \operatorname{Cov}(X, Y)</math>
+
<math>\operatorname{Var}(Y) \geq \operatorname{Cov} (Y, X) \operatorname{Var}^{-1}(X) \operatorname{Cov}(X, Y)</math>
 
This inequality means that the difference is semidefinite positive. -->
 
This inequality means that the difference is semidefinite positive. -->
  
 
Let <math>X</math> and <math>Y</math> be [[random variable]]s, then the covariance inequality:<ref>{{cite book|last=Mukhopadhyay|first=Nitis|date=2000-03-22|title=Probability and Statistical Inference|publisher=CRC Press|isbn=9780824703790|pages=150|url=https://books.google.com/books?id=TMSnGkr_DxwC}}</ref><ref>{{cite book|last=Keener|first=Robert W.|date=2010-09-08|title=Theoretical Statistics: Topics for a Core Course|publisher=Springer Science & Business Media|isbn=9780387938394|pages=71|url=https://books.google.com/books?id=aVJmcega44cC}}</ref> is given by
 
Let <math>X</math> and <math>Y</math> be [[random variable]]s, then the covariance inequality:<ref>{{cite book|last=Mukhopadhyay|first=Nitis|date=2000-03-22|title=Probability and Statistical Inference|publisher=CRC Press|isbn=9780824703790|pages=150|url=https://books.google.com/books?id=TMSnGkr_DxwC}}</ref><ref>{{cite book|last=Keener|first=Robert W.|date=2010-09-08|title=Theoretical Statistics: Topics for a Core Course|publisher=Springer Science & Business Media|isbn=9780387938394|pages=71|url=https://books.google.com/books?id=aVJmcega44cC}}</ref> is given by
−
<math display=block>\operatorname{Var}(Y) \geq \frac{\operatorname{Cov}(Y, X)^2}{\operatorname{Var}(X)}.</math>
+
<math>\operatorname{Var}(Y) \geq \frac{\operatorname{Cov}(Y, X)^2}{\operatorname{Var}(X)}.</math>
  
 
After defining an inner product on the set of random variables using the expectation of their product,
 
After defining an inner product on the set of random variables using the expectation of their product,
−
<math display=block>\langle X, Y \rangle := \operatorname{E}(X Y),</math>
+
<math>\langle X, Y \rangle := \operatorname{E}(X Y),</math>
 
the Cauchy–Schwarz inequality becomes
 
the Cauchy–Schwarz inequality becomes
−
<math display=block>|\operatorname{E}(XY)|^2 \leq \operatorname{E}(X^2) \operatorname{E}(Y^2).</math>
+
<math>|\operatorname{E}(XY)|^2 \leq \operatorname{E}(X^2) \operatorname{E}(Y^2).</math>
  
 
To prove the covariance inequality using the Cauchy–Schwarz inequality, let <math>\mu = \operatorname{E}(X)</math> and <math>\nu = \operatorname{E}(Y),</math> then
 
To prove the covariance inequality using the Cauchy–Schwarz inequality, let <math>\mu = \operatorname{E}(X)</math> and <math>\nu = \operatorname{E}(Y),</math> then
−
<math display=block>\begin{align}
+
<math>\begin{align}
 
|\operatorname{Cov}(X, Y)|^2  
 
|\operatorname{Cov}(X, Y)|^2  
 
&= |\operatorname{E}\left((X - \mu)(Y - \nu)\right)|^2 \\
 
&= |\operatorname{E}\left((X - \mu)(Y - \nu)\right)|^2 \\
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By definition, <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are linearly dependent if and only if one is a scalar multiple of the other.  
 
By definition, <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are linearly dependent if and only if one is a scalar multiple of the other.  
 
If <math>\mathbf{u} = c \mathbf{v}</math> where <math>c</math> is some scalar then  
 
If <math>\mathbf{u} = c \mathbf{v}</math> where <math>c</math> is some scalar then  
−
<math display=block>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|  
+
<math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|  
 
= \left|\langle c \mathbf{v}, \mathbf{v} \rangle\right|  
 
= \left|\langle c \mathbf{v}, \mathbf{v} \rangle\right|  
 
= \left|c \langle \mathbf{v}, \mathbf{v} \rangle\right|  
 
= \left|c \langle \mathbf{v}, \mathbf{v} \rangle\right|  
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which shows that equality holds in the {{EquationNote|Cauchy-Schwarz Inequality}}.  
 
which shows that equality holds in the {{EquationNote|Cauchy-Schwarz Inequality}}.  
 
The case where <math>\mathbf{v} = c \mathbf{u}</math> for some scalar <math>c</math> is very similar, with the main difference between the complex conjugation of {{nowrap|1=<math>c</math>:}}
 
The case where <math>\mathbf{v} = c \mathbf{u}</math> for some scalar <math>c</math> is very similar, with the main difference between the complex conjugation of {{nowrap|1=<math>c</math>:}}
−
<math display=block>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|  
+
<math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|  
 
= \left|\langle \mathbf{u}, c \mathbf{u} \rangle\right|  
 
= \left|\langle \mathbf{u}, c \mathbf{u} \rangle\right|  
 
= \left|\overline{c} \langle \mathbf{u}, \mathbf{u} \rangle\right|  
 
= \left|\overline{c} \langle \mathbf{u}, \mathbf{u} \rangle\right|  
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Let <math>(V, \langle \cdot, \cdot \rangle)</math> be an real inner product space. Consider an arbitrary pair <math>u, v \in V</math> and the function <math>p : \R \to \R</math> defined by <math>p(t) = \langle tu + v, tu + v\rangle.</math>  
 
Let <math>(V, \langle \cdot, \cdot \rangle)</math> be an real inner product space. Consider an arbitrary pair <math>u, v \in V</math> and the function <math>p : \R \to \R</math> defined by <math>p(t) = \langle tu + v, tu + v\rangle.</math>  
 
Since the inner product is positive-definite, <math>p(t)</math> only takes non-negative values. On the other hand, <math>p(t)</math> can be expanded using the bilinearity of the inner product and using the fact that <math>\langle u, v \rangle = \langle v, u \rangle</math> for real inner products:
 
Since the inner product is positive-definite, <math>p(t)</math> only takes non-negative values. On the other hand, <math>p(t)</math> can be expanded using the bilinearity of the inner product and using the fact that <math>\langle u, v \rangle = \langle v, u \rangle</math> for real inner products:
−
<math display=block>p(t) = \Vert u \Vert^2 t^2 + t \left[\langle u, v \rangle + \langle v, u \rangle\right] + \Vert v \Vert^2 = \Vert u \Vert^2 t^2 + 2t \langle u, v \rangle + \Vert v \Vert^2.</math>  
+
<math>p(t) = \Vert u \Vert^2 t^2 + t \left[\langle u, v \rangle + \langle v, u \rangle\right] + \Vert v \Vert^2 = \Vert u \Vert^2 t^2 + 2t \langle u, v \rangle + \Vert v \Vert^2.</math>  
 
Thus, <math>p</math> is a polynomial of degree <math>2</math> (unless <math>u = 0,</math> which is a case that can be independently verified). Since the sign of <math>p</math> does not change, the discriminant of this polynomial must be non-positive:
 
Thus, <math>p</math> is a polynomial of degree <math>2</math> (unless <math>u = 0,</math> which is a case that can be independently verified). Since the sign of <math>p</math> does not change, the discriminant of this polynomial must be non-positive:
−
<math display=block>\Delta = 4 \left(\langle u, v \rangle ^2 - \Vert u \Vert^2 \Vert v \Vert^2\right) \leqslant 0.</math>
+
<math>\Delta = 4 \left(\langle u, v \rangle ^2 - \Vert u \Vert^2 \Vert v \Vert^2\right) \leqslant 0.</math>
 
The conclusion follows.
 
The conclusion follows.
  
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The Cauchy-Schwarz inequality in the case where the inner product is the [[dot product]] on <math>\R^n</math> is now proven.
 
The Cauchy-Schwarz inequality in the case where the inner product is the [[dot product]] on <math>\R^n</math> is now proven.
 
The Cauchy-Schwarz inequality may be rewritten as <math>\left|\vec{a} \cdot \vec{b}\right|^2 \leq \left\|\vec{a}\right\|^2 \, \left\|\vec{b}\right\|^2</math> or equivalently, <math>\left(\vec{a} \cdot \vec{b}\right)^2 \leq \left(\vec{a} \cdot \vec{a}\right) \, \left(\vec{b} \cdot \vec{b}\right)</math> for <math>\vec{a} := \left(a_1, \ldots, a_n\right), \vec{b} := \left(b_1, \ldots, b_n\right) \in \R^n,</math> which expands to:
 
The Cauchy-Schwarz inequality may be rewritten as <math>\left|\vec{a} \cdot \vec{b}\right|^2 \leq \left\|\vec{a}\right\|^2 \, \left\|\vec{b}\right\|^2</math> or equivalently, <math>\left(\vec{a} \cdot \vec{b}\right)^2 \leq \left(\vec{a} \cdot \vec{a}\right) \, \left(\vec{b} \cdot \vec{b}\right)</math> for <math>\vec{a} := \left(a_1, \ldots, a_n\right), \vec{b} := \left(b_1, \ldots, b_n\right) \in \R^n,</math> which expands to:
−
<math display=block>\left(a_1^2 + a_2^2 + \cdots + a_n^2\right) \left(b_1^2 + b_2^2 + \cdots + b_n^2\right) \geq \left(a_1b_1 + a_2b_2 + \cdots + a_nb_n\right)^2.</math>  
+
<math>\left(a_1^2 + a_2^2 + \cdots + a_n^2\right) \left(b_1^2 + b_2^2 + \cdots + b_n^2\right) \geq \left(a_1b_1 + a_2b_2 + \cdots + a_nb_n\right)^2.</math>  
  
 
To simplify, let  
 
To simplify, let  
−
<math display=block>\begin{align} A &= a_1^2 + a_2^2 + \cdots + a_n^2, \\ B &= b_1^2 + b_2^2 + \cdots + b_n^2 \\ D &= a_1 b_1 + a_2 b_2 + \cdots + a_n b_n \\\end{align}</math>
+
<math>\begin{align} A &= a_1^2 + a_2^2 + \cdots + a_n^2, \\ B &= b_1^2 + b_2^2 + \cdots + b_n^2 \\ D &= a_1 b_1 + a_2 b_2 + \cdots + a_n b_n \\\end{align}</math>
 
so that the statement that remains to be to proven can be written as <math>A B \geq D^2,</math> which can be rearranged to <math>D^2 - A B \leq 0.</math> The [[discriminant]] of the [[quadratic equation]] <math>A x^2 + 2 D x + B</math> is <math>4 D^2 - 4 A B.</math>  
 
so that the statement that remains to be to proven can be written as <math>A B \geq D^2,</math> which can be rearranged to <math>D^2 - A B \leq 0.</math> The [[discriminant]] of the [[quadratic equation]] <math>A x^2 + 2 D x + B</math> is <math>4 D^2 - 4 A B.</math>  
  
 
Therefore, to complete the proof it is sufficient to prove that this quadratic either has no real roots or exactly one real root, because this will imply:  
 
Therefore, to complete the proof it is sufficient to prove that this quadratic either has no real roots or exactly one real root, because this will imply:  
−
<math display=block>4 \left(D^2 - A B\right) \leq 0.</math>
+
<math>4 \left(D^2 - A B\right) \leq 0.</math>
  
 
Substituting the values of <math>A, B, D</math> into <math>A x^2 + 2 D x + B</math> gives:
 
Substituting the values of <math>A, B, D</math> into <math>A x^2 + 2 D x + B</math> gives:
−
<math display=block>\begin{alignat}{4}
+
<math>\begin{alignat}{4}
 
A x^2 + 2 D x + B  
 
A x^2 + 2 D x + B  
 
&= \left(a_1^2 + a_2^2 + \cdots + a_n^2\right) x^2 + 2 \left(a_1 b_1 + a_2 b_2 + \cdots + a_n b_n\right) x + \left(b_1^2 + b_2^2 + \cdots + b_n^2\right) \\
 
&= \left(a_1^2 + a_2^2 + \cdots + a_n^2\right) x^2 + 2 \left(a_1 b_1 + a_2 b_2 + \cdots + a_n b_n\right) x + \left(b_1^2 + b_2^2 + \cdots + b_n^2\right) \\
Line 201: Line 201:
 
This proves the inequality and so to finish the proof, it remains to show that equality is achievable.
 
This proves the inequality and so to finish the proof, it remains to show that equality is achievable.
 
The equality <math>a_i x = - b_i</math> is the equality case for Cauchy-Schwarz after inspecting  
 
The equality <math>a_i x = - b_i</math> is the equality case for Cauchy-Schwarz after inspecting  
−
<math display=block>\left(a_1 x + b_1\right)^2 + \left(a_2 x + b_2\right)^2 + \cdots + \left(a_n x + b_n\right)^2 \geq 0,</math>
+
<math>\left(a_1 x + b_1\right)^2 + \left(a_2 x + b_2\right)^2 + \cdots + \left(a_n x + b_n\right)^2 \geq 0,</math>
 
which proves that equality is achievable. <math>\blacksquare</math>
 
which proves that equality is achievable. <math>\blacksquare</math>
  
Line 223: Line 223:
 
Details of <math>\left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2}</math>'s elementary expansion are now given for the interested reader. Let <math>V = \|\mathbf{v}\|^2</math> and <math>c = \langle \mathbf{u}, \mathbf{v} \rangle</math> so that <math>\overline{c} c = \left|c\right|^2 = \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2</math> and <math>\overline{c} = \overline{\langle \mathbf{u}, \mathbf{v} \rangle} = \langle \mathbf{v}, \mathbf{u} \rangle.</math>  
 
Details of <math>\left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2}</math>'s elementary expansion are now given for the interested reader. Let <math>V = \|\mathbf{v}\|^2</math> and <math>c = \langle \mathbf{u}, \mathbf{v} \rangle</math> so that <math>\overline{c} c = \left|c\right|^2 = \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2</math> and <math>\overline{c} = \overline{\langle \mathbf{u}, \mathbf{v} \rangle} = \langle \mathbf{v}, \mathbf{u} \rangle.</math>  
 
Then
 
Then
−
<math display=block>\begin{alignat}{4}
+
<math>\begin{alignat}{4}
 
\left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2}  
 
\left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2}  
 
&= \left\|V \mathbf{u} - c \mathbf{v}\right\|^{2}  
 
&= \left\|V \mathbf{u} - c \mathbf{v}\right\|^{2}  
Line 247: Line 247:
 
This expansion does not require <math>\mathbf{v}</math> to be non-zero; however, <math>\mathbf{v}</math> must be non-zero in order to divide both sides by <math>\|\mathbf{v}\|^2</math> and to deduce the Cauchy-Schwarz inequality from it.  
 
This expansion does not require <math>\mathbf{v}</math> to be non-zero; however, <math>\mathbf{v}</math> must be non-zero in order to divide both sides by <math>\|\mathbf{v}\|^2</math> and to deduce the Cauchy-Schwarz inequality from it.  
 
Swapping <math>\mathbf{u}</math> and <math>\mathbf{v}</math> gives rise to:
 
Swapping <math>\mathbf{u}</math> and <math>\mathbf{v}</math> gives rise to:
−
<math display=block>\left\|\|\mathbf{u}\|^2 \mathbf{v} - \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \mathbf{u}\right\|^{2} =\|\mathbf{u}\|^2 \left[\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2\right]</math>
+
<math>\left\|\|\mathbf{u}\|^2 \mathbf{v} - \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \mathbf{u}\right\|^{2} =\|\mathbf{u}\|^2 \left[\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2\right]</math>
 
and thus
 
and thus
−
<math display=block>\begin{alignat}{4}
+
<math>\begin{alignat}{4}
 
\|\mathbf{u}\|^2\|\mathbf{v}\|^2 \left[\|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2\right]
 
\|\mathbf{u}\|^2\|\mathbf{v}\|^2 \left[\|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2\right]
 
&=\|\mathbf{u}\|^2 \left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2} \\
 
&=\|\mathbf{u}\|^2 \left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2} \\
Line 260: Line 260:
 
The special case of <math>\mathbf{v} = \mathbf{0}</math> was proven above so it is henceforth assumed that <math>\mathbf{v} \neq \mathbf{0}.</math>  
 
The special case of <math>\mathbf{v} = \mathbf{0}</math> was proven above so it is henceforth assumed that <math>\mathbf{v} \neq \mathbf{0}.</math>  
 
Let
 
Let
−
<math display=block>\mathbf{z} := \mathbf{u} - \frac {\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}.</math>
+
<math>\mathbf{z} := \mathbf{u} - \frac {\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}.</math>
  
 
It follows from the linearity of the inner product in its first argument that:
 
It follows from the linearity of the inner product in its first argument that:
−
<math display=block>\langle \mathbf{z}, \mathbf{v} \rangle  
+
<math>\langle \mathbf{z}, \mathbf{v} \rangle  
 
= \left\langle \mathbf{u} - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}, \mathbf{v} \right\rangle  
 
= \left\langle \mathbf{u} - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}, \mathbf{v} \right\rangle  
 
= \langle \mathbf{u}, \mathbf{v} \rangle - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \langle \mathbf{v}, \mathbf{v} \rangle  
 
= \langle \mathbf{u}, \mathbf{v} \rangle - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \langle \mathbf{v}, \mathbf{v} \rangle  
Line 269: Line 269:
  
 
Therefore, <math>\mathbf{z}</math> is a vector orthogonal to the vector <math>\mathbf{v}</math> (Indeed, <math>\mathbf{z}</math> is the [[vector projection|projection]] of <math>\mathbf{u}</math> onto the plane orthogonal to <math>\mathbf{v}.</math>) We can thus apply the [[Pythagorean theorem#Inner product spaces|Pythagorean theorem]] to
 
Therefore, <math>\mathbf{z}</math> is a vector orthogonal to the vector <math>\mathbf{v}</math> (Indeed, <math>\mathbf{z}</math> is the [[vector projection|projection]] of <math>\mathbf{u}</math> onto the plane orthogonal to <math>\mathbf{v}.</math>) We can thus apply the [[Pythagorean theorem#Inner product spaces|Pythagorean theorem]] to
−
<math display=block>\mathbf{u}= \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} + \mathbf{z}</math>
+
<math>\mathbf{u}= \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} + \mathbf{z}</math>
 
which gives
 
which gives
−
<math display=block>\|\mathbf{u}\|^2  
+
<math>\|\mathbf{u}\|^2  
 
= \left|\frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle}\right|^2 \|\mathbf{v}\|^2 + \|\mathbf{z}\|^2  
 
= \left|\frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle}\right|^2 \|\mathbf{v}\|^2 + \|\mathbf{z}\|^2  
 
= \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{(\|\mathbf{v}\|^2 )^2} \,\|\mathbf{v}\|^2 + \|\mathbf{z}\|^2  
 
= \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{(\|\mathbf{v}\|^2 )^2} \,\|\mathbf{v}\|^2 + \|\mathbf{z}\|^2  
Line 283: Line 283:
  
 
An inner product can be used to define a [[positive linear functional]]. For example, given a Hilbert space <math>L^2(m), m</math> being a finite measure, the standard inner product gives rise to a positive functional <math>\varphi</math> by <math>\varphi (g) = \langle g, 1 \rangle.</math>  Conversely, every positive linear functional <math>\varphi</math> on <math>L^2(m)</math> can be used to define an  inner product <math>\langle f, g \rangle _\varphi := \varphi\left(g^* f\right),</math> where <math>g^*</math> is the [[Pointwise product|pointwise]] [[complex conjugate]] of <math>g.</math> In this language, the Cauchy–Schwarz inequality becomes<ref>{{cite book|last1=Faria|first1=Edson de|last2=Melo|first2=Welington de|date=2010-08-12|title=Mathematical Aspects of Quantum Field Theory|publisher=Cambridge University Press|isbn=9781139489805|pages=273|url=https://books.google.com/books?id=u9M9PFLNpMMC}}</ref>
 
An inner product can be used to define a [[positive linear functional]]. For example, given a Hilbert space <math>L^2(m), m</math> being a finite measure, the standard inner product gives rise to a positive functional <math>\varphi</math> by <math>\varphi (g) = \langle g, 1 \rangle.</math>  Conversely, every positive linear functional <math>\varphi</math> on <math>L^2(m)</math> can be used to define an  inner product <math>\langle f, g \rangle _\varphi := \varphi\left(g^* f\right),</math> where <math>g^*</math> is the [[Pointwise product|pointwise]] [[complex conjugate]] of <math>g.</math> In this language, the Cauchy–Schwarz inequality becomes<ref>{{cite book|last1=Faria|first1=Edson de|last2=Melo|first2=Welington de|date=2010-08-12|title=Mathematical Aspects of Quantum Field Theory|publisher=Cambridge University Press|isbn=9781139489805|pages=273|url=https://books.google.com/books?id=u9M9PFLNpMMC}}</ref>
−
<math display=block>\left|\varphi\left(g^* f\right)\right|^2 \leq \varphi\left(f^* f\right) \varphi\left(g^* g\right),</math>
+
<math>\left|\varphi\left(g^* f\right)\right|^2 \leq \varphi\left(f^* f\right) \varphi\left(g^* g\right),</math>
  
 
which extends verbatim to positive functionals on C*-algebras:
 
which extends verbatim to positive functionals on C*-algebras:
Line 301: Line 301:
 
{{math theorem|name=Cauchy-Schwarz inequality|note=Modified Schwarz inequality for 2-positive maps<ref>{{cite book|last=Paulsen|first=Vern|year=2002|title=Completely Bounded Maps and Operator Algebras|series=Cambridge Studies in Advanced Mathematics|volume=78|publisher=Cambridge University Press|isbn=9780521816694|page=40|url=https://books.google.com/books?id=VtSFHDABxMIC&pg=PA40}}</ref>|style=|math_statement=  
 
{{math theorem|name=Cauchy-Schwarz inequality|note=Modified Schwarz inequality for 2-positive maps<ref>{{cite book|last=Paulsen|first=Vern|year=2002|title=Completely Bounded Maps and Operator Algebras|series=Cambridge Studies in Advanced Mathematics|volume=78|publisher=Cambridge University Press|isbn=9780521816694|page=40|url=https://books.google.com/books?id=VtSFHDABxMIC&pg=PA40}}</ref>|style=|math_statement=  
 
For a 2-positive map <math>\varphi</math> between C*-algebras, for all <math>a, b</math> in its domain,
 
For a 2-positive map <math>\varphi</math> between C*-algebras, for all <math>a, b</math> in its domain,
−
<math display=block>\varphi(a)^*\varphi(a) \leq \Vert\varphi(1)\Vert\varphi\left(a^*a\right), \text{ and }</math>
+
<math>\varphi(a)^*\varphi(a) \leq \Vert\varphi(1)\Vert\varphi\left(a^*a\right), \text{ and }</math>
−
<math display=block>\Vert\varphi\left(a^* b\right)\Vert^2 \leq \Vert\varphi\left(a^*a\right)\Vert \cdot \Vert\varphi\left(b^*b\right)\Vert.</math>
+
<math>\Vert\varphi\left(a^* b\right)\Vert^2 \leq \Vert\varphi\left(a^*a\right)\Vert \cdot \Vert\varphi\left(b^*b\right)\Vert.</math>
 
}}
 
}}
  
Line 309: Line 309:
 
{{math theorem|name=Callebaut's Inequality<ref>{{cite journal|last1=Callebaut|first1=D.K.|date=1965|title=Generalization of the Cauchy–Schwarz inequality|journal=J. Math. Anal. Appl.|volume=12|issue=3|pages=491–494|doi=10.1016/0022-247X(65)90016-8|doi-access=free}}</ref>|note=|style=|math_statement=   
 
{{math theorem|name=Callebaut's Inequality<ref>{{cite journal|last1=Callebaut|first1=D.K.|date=1965|title=Generalization of the Cauchy–Schwarz inequality|journal=J. Math. Anal. Appl.|volume=12|issue=3|pages=491–494|doi=10.1016/0022-247X(65)90016-8|doi-access=free}}</ref>|note=|style=|math_statement=   
 
For reals <math>0 \leqslant s \leqslant t \leqslant 1,</math>  
 
For reals <math>0 \leqslant s \leqslant t \leqslant 1,</math>  
−
<math display=block>\left(\sum_{i=1}^n a_i b_i\right)^2  
+
<math>\left(\sum_{i=1}^n a_i b_i\right)^2  
 
~\leqslant~ \left(\sum_{i=1}^n a_i^{1+s} b_i^{1-s}\right) \left(\sum_{i=1}^n a_i^{1-s} b_i^{1+s}\right)
 
~\leqslant~ \left(\sum_{i=1}^n a_i^{1+s} b_i^{1-s}\right) \left(\sum_{i=1}^n a_i^{1-s} b_i^{1+s}\right)
 
~\leqslant~ \left(\sum_{i=1}^n a_i^{1+t} b_i^{1-t}\right) \left(\sum_{i=1}^n a_i^{1-t} b_i^{1+t}\right)
 
~\leqslant~ \left(\sum_{i=1}^n a_i^{1+t} b_i^{1-t}\right) \left(\sum_{i=1}^n a_i^{1-t} b_i^{1+t}\right)

Revision as of 15:04, 28 October 2021

Statement of the inequality

The Cauchy–Schwarz inequality states that for all vectors '"`UNIQ--postMath-00000001-QINU`"' and '"`UNIQ--postMath-00000002-QINU`"' of an inner product space it is true that

'"`UNIQ--postMath-00000003-QINU`"'

where '"`UNIQ--postMath-00000004-QINU`"' is the inner product. Examples of inner products include the real and complex dot product; see the examples in inner product. Every inner product gives rise to a norm, called the canonical or induced norm, where the norm of a vector '"`UNIQ--postMath-00000005-QINU`"' is denoted and defined by: '"`UNIQ--postMath-00000006-QINU`"' so that this norm and the inner product are related by the defining condition '"`UNIQ--postMath-00000007-QINU`"' where '"`UNIQ--postMath-00000008-QINU`"' is always a non-negative real number (even if the inner product is complex-valued). By taking the square root of both sides of the above inequality, the Cauchy–Schwarz inequality can be written in its more familiar form:

'"`UNIQ--postMath-00000009-QINU`"'

Moreover, the two sides are equal if and only if '"`UNIQ--postMath-0000000A-QINU`"' and '"`UNIQ--postMath-0000000B-QINU`"' are linearly dependent.

Special cases

Sedrakyan's lemma - Positive real numbers

Sedrakyan's inequality, also called Engel's form, the T2 lemma, or Titu's lemma, states that for positive reals: '"`UNIQ--postMath-0000000C-QINU`"'

It is a direct consequence of the Cauchy–Schwarz inequality, obtained by using the dot product on '"`UNIQ--postMath-0000000D-QINU`"' upon substituting '"`UNIQ--postMath-0000000E-QINU`"' This form is especially helpful when the inequality involves fractions where the numerator is a perfect square.

ℝ2 - The plane

Cauchy-Schwarz inequality in a unit circle of the Euclidean plane

The real vector space '"`UNIQ--postMath-0000000F-QINU`"' denotes the 2-dimensional plane. It is also the 2-dimensional Euclidean space where the inner product is the dot product. If '"`UNIQ--postMath-00000010-QINU`"' and '"`UNIQ--postMath-00000011-QINU`"' then the Cauchy–Schwarz inequality becomes: '"`UNIQ--postMath-00000012-QINU`"' where '"`UNIQ--postMath-00000013-QINU`"' is the angle between '"`UNIQ--postMath-00000014-QINU`"' and '"`UNIQ--postMath-00000015-QINU`"'

The form above is perhaps the easiest in which to understand the inequality, since the square of the cosine can be at most 1, which occurs when the vectors are in the same or opposite directions. It can also be restated in terms of the vector coordinates '"`UNIQ--postMath-00000016-QINU`"' as '"`UNIQ--postMath-00000017-QINU`"' where equality holds if and only if the vector '"`UNIQ--postMath-00000018-QINU`"' is in the same or opposite direction as the vector '"`UNIQ--postMath-00000019-QINU`"' or if one of them is the zero vector.

Template:Anchor

ℝn - n-dimensional Euclidean space

Template:AnchorIn Euclidean space '"`UNIQ--postMath-0000001A-QINU`"' with the standard inner product, which is the dot product, the Cauchy–Schwarz inequality becomes: '"`UNIQ--postMath-0000001B-QINU`"'

The Cauchy–Schwarz inequality can be proved using only ideas from elementary algebra in this case. Consider the following quadratic polynomial in '"`UNIQ--postMath-0000001C-QINU`"' '"`UNIQ--postMath-0000001D-QINU`"'

Since it is nonnegative, it has at most one real root for '"`UNIQ--postMath-0000001E-QINU`"' hence its discriminant is less than or equal to zero. That is, '"`UNIQ--postMath-0000001F-QINU`"'

which yields the Cauchy–Schwarz inequality.

ℂn - n-dimensional Complex space

If '"`UNIQ--postMath-00000020-QINU`"' with '"`UNIQ--postMath-00000021-QINU`"' and '"`UNIQ--postMath-00000022-QINU`"' (where '"`UNIQ--postMath-00000023-QINU`"' and '"`UNIQ--postMath-00000024-QINU`"') and if the inner product on the vector space '"`UNIQ--postMath-00000025-QINU`"' is the canonical complex inner product (defined by '"`UNIQ--postMath-00000026-QINU`"' where the bar notation is used for complex conjugation), then the inequality may be restated more explicitly as follows: '"`UNIQ--postMath-00000027-QINU`"'

That is, '"`UNIQ--postMath-00000028-QINU`"'

L2

For the inner product space of square-integrable complex-valued functions, the following inequality: '"`UNIQ--postMath-00000029-QINU`"'

The Hölder inequality is a generalization of this.

Applications

Analysis

In any inner product space, the triangle inequality is a consequence of the Cauchy–Schwarz inequality, as is now shown: '"`UNIQ--postMath-0000002A-QINU`"'

Taking square roots gives the triangle inequality: '"`UNIQ--postMath-0000002B-QINU`"'

The Cauchy–Schwarz inequality is used to prove that the inner product is a continuous function with respect to the topology induced by the inner product itself.[1][2]

Geometry

The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any real inner-product space by defining:[3][4] '"`UNIQ--postMath-0000002C-QINU`"'

The Cauchy–Schwarz inequality proves that this definition is sensible, by showing that the right-hand side lies in the interval [−1, 1] and justifies the notion that (real) Hilbert spaces are simply generalizations of the Euclidean space. It can also be used to define an angle in complex inner-product spaces, by taking the absolute value or the real part of the right-hand side,[5][6] as is done when extracting a metric from quantum fidelity.

Probability theory

Let '"`UNIQ--postMath-0000002D-QINU`"' and '"`UNIQ--postMath-0000002E-QINU`"' be random variables, then the covariance inequality:[7][8] is given by '"`UNIQ--postMath-0000002F-QINU`"'

After defining an inner product on the set of random variables using the expectation of their product, '"`UNIQ--postMath-00000030-QINU`"' the Cauchy–Schwarz inequality becomes '"`UNIQ--postMath-00000031-QINU`"'

To prove the covariance inequality using the Cauchy–Schwarz inequality, let '"`UNIQ--postMath-00000032-QINU`"' and '"`UNIQ--postMath-00000033-QINU`"' then '"`UNIQ--postMath-00000034-QINU`"' where '"`UNIQ--postMath-00000035-QINU`"' denotes variance and '"`UNIQ--postMath-00000036-QINU`"' denotes covariance.

Proofs

There are many different proofs[9] of the Cauchy–Schwarz inequality other than those given below.[10][11] When consulting other sources, there are often two sources of confusion. First, some authors define ⟨⋅,⋅⟩ to be linear in the second argument rather than the first. Second, some proofs are only valid when the field is '"`UNIQ--postMath-00000037-QINU`"' and not '"`UNIQ--postMath-00000038-QINU`"'[12]

This section gives proofs of the following theorem:

Template:Math theorem

Moreover, if this equality holds and if '"`UNIQ--postMath-00000039-QINU`"' then '"`UNIQ--postMath-0000003A-QINU`"'

In all of the proofs given below, the proof in the trivial case where at least one of the vectors is zero (or equivalently, in the case where '"`UNIQ--postMath-0000003B-QINU`"') is the same. It is presented immediately below only once to reduce repetition. It also includes the easy part of the proof the above Template:EquationNote; that is, it proves that if '"`UNIQ--postMath-0000003C-QINU`"' and '"`UNIQ--postMath-0000003D-QINU`"' are linearly dependent then '"`UNIQ--postMath-0000003E-QINU`"'

Template:Collapse top By definition, '"`UNIQ--postMath-0000003F-QINU`"' and '"`UNIQ--postMath-00000040-QINU`"' are linearly dependent if and only if one is a scalar multiple of the other. If '"`UNIQ--postMath-00000041-QINU`"' where '"`UNIQ--postMath-00000042-QINU`"' is some scalar then '"`UNIQ--postMath-00000043-QINU`"'

which shows that equality holds in the Template:EquationNote. The case where '"`UNIQ--postMath-00000044-QINU`"' for some scalar '"`UNIQ--postMath-00000045-QINU`"' is very similar, with the main difference between the complex conjugation of Template:Nowrap '"`UNIQ--postMath-00000046-QINU`"'

If at least one of '"`UNIQ--postMath-00000047-QINU`"' and '"`UNIQ--postMath-00000048-QINU`"' is the zero vector then '"`UNIQ--postMath-00000049-QINU`"' and '"`UNIQ--postMath-0000004A-QINU`"' are necessarily linearly dependent (just scalar multiply the non-zero vector by the number '"`UNIQ--postMath-0000004B-QINU`"' to get the zero vector; e.g. if '"`UNIQ--postMath-0000004C-QINU`"' then let '"`UNIQ--postMath-0000004D-QINU`"' so that '"`UNIQ--postMath-0000004E-QINU`"'), which proves the converse of this characterization in this special case; that is, this shows that if at least one of '"`UNIQ--postMath-0000004F-QINU`"' and '"`UNIQ--postMath-00000050-QINU`"' is '"`UNIQ--postMath-00000051-QINU`"' then the Template:EquationNote holds.

If '"`UNIQ--postMath-00000052-QINU`"' which happens if and only if '"`UNIQ--postMath-00000053-QINU`"' then '"`UNIQ--postMath-00000054-QINU`"' and '"`UNIQ--postMath-00000055-QINU`"' so that in particular, the Cauchy-Schwarz inequality holds because both sides of it are '"`UNIQ--postMath-00000056-QINU`"' The proof in the case of '"`UNIQ--postMath-00000057-QINU`"' is identical. Template:Collapse bottom

Consequently, the Cauchy-Schwarz inequality only needs to be proven only for non-zero vectors and also only the non-trivial direction of the Template:EquationNote must be shown.

For real inner products spaces

Let '"`UNIQ--postMath-00000058-QINU`"' be an real inner product space. Consider an arbitrary pair '"`UNIQ--postMath-00000059-QINU`"' and the function '"`UNIQ--postMath-0000005A-QINU`"' defined by '"`UNIQ--postMath-0000005B-QINU`"' Since the inner product is positive-definite, '"`UNIQ--postMath-0000005C-QINU`"' only takes non-negative values. On the other hand, '"`UNIQ--postMath-0000005D-QINU`"' can be expanded using the bilinearity of the inner product and using the fact that '"`UNIQ--postMath-0000005E-QINU`"' for real inner products: '"`UNIQ--postMath-0000005F-QINU`"' Thus, '"`UNIQ--postMath-00000060-QINU`"' is a polynomial of degree '"`UNIQ--postMath-00000061-QINU`"' (unless '"`UNIQ--postMath-00000062-QINU`"' which is a case that can be independently verified). Since the sign of '"`UNIQ--postMath-00000063-QINU`"' does not change, the discriminant of this polynomial must be non-positive: '"`UNIQ--postMath-00000064-QINU`"' The conclusion follows.

For the equality case, notice that '"`UNIQ--postMath-00000065-QINU`"' happens if and only if '"`UNIQ--postMath-00000066-QINU`"' If '"`UNIQ--postMath-00000067-QINU`"' then '"`UNIQ--postMath-00000068-QINU`"' and hence '"`UNIQ--postMath-00000069-QINU`"'

Proof for the dot product

The Cauchy-Schwarz inequality in the case where the inner product is the dot product on '"`UNIQ--postMath-0000006A-QINU`"' is now proven. The Cauchy-Schwarz inequality may be rewritten as '"`UNIQ--postMath-0000006B-QINU`"' or equivalently, '"`UNIQ--postMath-0000006C-QINU`"' for '"`UNIQ--postMath-0000006D-QINU`"' which expands to: '"`UNIQ--postMath-0000006E-QINU`"'

To simplify, let '"`UNIQ--postMath-0000006F-QINU`"' so that the statement that remains to be to proven can be written as '"`UNIQ--postMath-00000070-QINU`"' which can be rearranged to '"`UNIQ--postMath-00000071-QINU`"' The discriminant of the quadratic equation '"`UNIQ--postMath-00000072-QINU`"' is '"`UNIQ--postMath-00000073-QINU`"'

Therefore, to complete the proof it is sufficient to prove that this quadratic either has no real roots or exactly one real root, because this will imply: '"`UNIQ--postMath-00000074-QINU`"'

Substituting the values of '"`UNIQ--postMath-00000075-QINU`"' into '"`UNIQ--postMath-00000076-QINU`"' gives: '"`UNIQ--postMath-00000077-QINU`"' which is a sum of terms that are each '"`UNIQ--postMath-00000078-QINU`"' by the trivial inequality: '"`UNIQ--postMath-00000079-QINU`"' for all '"`UNIQ--postMath-0000007A-QINU`"' This proves the inequality and so to finish the proof, it remains to show that equality is achievable. The equality '"`UNIQ--postMath-0000007B-QINU`"' is the equality case for Cauchy-Schwarz after inspecting '"`UNIQ--postMath-0000007C-QINU`"' which proves that equality is achievable. '"`UNIQ--postMath-0000007D-QINU`"'

For arbitrary vector spaces

Proof 1

The special case of '"`UNIQ--postMath-0000007E-QINU`"' was proven above so it is henceforth assumed that '"`UNIQ--postMath-0000007F-QINU`"' As is now shown, the Cauchy–Schwarz Template:Emequality (and the rest of the theorem) is an almost immediate corollary of the following Template:Em:Lua error in package.lua at line 80: module 'Module:No globals' not found.

Template:NumBlk

which is readily verified by elementarily expanding '"`UNIQ--postMath-00000080-QINU`"' (via the definition of the norm) and then simplifying.

Observing that the left hand side of Template:EquationNote is non-negative (which makes this also true of the right hand side) proves that '"`UNIQ--postMath-00000081-QINU`"' from which the Template:EquationNote follows (by taking the square root of both sides). If '"`UNIQ--postMath-00000082-QINU`"' then the right hand side (and thus also the left hand side) of Template:EquationNote is '"`UNIQ--postMath-00000083-QINU`"' which is only possible if Template:Nowrap[note 1] thus '"`UNIQ--postMath-00000084-QINU`"' which shows that '"`UNIQ--postMath-00000085-QINU`"' and '"`UNIQ--postMath-00000086-QINU`"' are linearly dependent.Lua error in package.lua at line 80: module 'Module:No globals' not found. Since the (trivial) converse was proved above, the proof of the theorem is complete.


Details of 's elementary expansion are now given for the interested reader. Let and so that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{c} c = \left|c\right|^2 = \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{c} = \overline{\langle \mathbf{u}, \mathbf{v} \rangle} = \langle \mathbf{v}, \mathbf{u} \rangle.} Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{4} \left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2} &= \left\|V \mathbf{u} - c \mathbf{v}\right\|^{2} = \left\langle V \mathbf{u} - c \mathbf{v}, V \mathbf{u} - c \mathbf{v} \right\rangle && ~\text{ By definition of the norm } \\ &= \left\langle V \mathbf{u}, V \mathbf{u} \right\rangle - \left\langle V \mathbf{u}, c \mathbf{v} \right\rangle - \left\langle c \mathbf{v}, V \mathbf{u} \right\rangle + \left\langle c \mathbf{v}, c \mathbf{v} \right\rangle && ~\text{ Expand } \\ &= V^2 \left\langle \mathbf{u}, \mathbf{u} \right\rangle - V \overline{c} \left\langle \mathbf{u}, \mathbf{v} \right\rangle - c V \left\langle \mathbf{v}, \mathbf{u} \right\rangle + c \overline{c} \left\langle \mathbf{v}, \mathbf{v} \right\rangle && ~\text{ Pull out scalars (note that } V := \|\mathbf{v}\|^2 \text{ is real) } \\ &= V^2\|\mathbf{u}\|^2 ~~- V \overline{c} c ~~~~~~~~~- c V \overline{c} ~~~~~~~~~+ c \overline{c} V && ~\text{ Use definitions of } c := \langle \mathbf{u}, \mathbf{v} \rangle \text{ and } V \\ &= V^2\|\mathbf{u}\|^2 ~~- V \overline{c} c ~=~ V \left[V\|\mathbf{u}\|^2 - \overline{c} c\right] && ~\text{ Simplify } \\ &=\|\mathbf{v}\|^2 \left[\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2\right] && ~\text{ Rewrite in terms of } \mathbf{u} \text{ and } \mathbf{v}. \\ \end{alignat} }

This expansion does not require Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}} to be non-zero; however, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}} must be non-zero in order to divide both sides by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \|\mathbf{v}\|^2} and to deduce the Cauchy-Schwarz inequality from it. Swapping Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{u}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}} gives rise to: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\|\|\mathbf{u}\|^2 \mathbf{v} - \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \mathbf{u}\right\|^{2} =\|\mathbf{u}\|^2 \left[\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2\right]} and thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{4} \|\mathbf{u}\|^2\|\mathbf{v}\|^2 \left[\|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2\right] &=\|\mathbf{u}\|^2 \left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2} \\ &=\|\mathbf{v}\|^2 \left\|\|\mathbf{u}\|^2 \mathbf{v} - \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \mathbf{u}\right\|^{2}. \blacksquare \\ \end{alignat} }

Proof 2

The special case of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v} = \mathbf{0}} was proven above so it is henceforth assumed that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v} \neq \mathbf{0}.} Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{z} := \mathbf{u} - \frac {\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}.}

It follows from the linearity of the inner product in its first argument that: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle \mathbf{z}, \mathbf{v} \rangle = \left\langle \mathbf{u} - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}, \mathbf{v} \right\rangle = \langle \mathbf{u}, \mathbf{v} \rangle - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \langle \mathbf{v}, \mathbf{v} \rangle = 0.}

Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{z}} is a vector orthogonal to the vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}} (Indeed, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{z}} is the projection of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{u}} onto the plane orthogonal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}.} ) We can thus apply the Pythagorean theorem to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{u}= \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} + \mathbf{z}} which gives Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \|\mathbf{u}\|^2 = \left|\frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle}\right|^2 \|\mathbf{v}\|^2 + \|\mathbf{z}\|^2 = \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{(\|\mathbf{v}\|^2 )^2} \,\|\mathbf{v}\|^2 + \|\mathbf{z}\|^2 = \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{\|\mathbf{v}\|^2} + \|\mathbf{z}\|^2 \geq \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{\|\mathbf{v}\|^2}.}

The Cauchy–Schwarz inequality follows by multiplying by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \|\mathbf{v}\|^2} and then taking the square root. Moreover, if the relation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \geq} in the above expression is actually an equality, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \|\mathbf{z}\|^2 = 0} and hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{z} = \mathbf{0};} the definition of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{z}} then establishes a relation of linear dependence between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{u}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{v}.} The converse was proved at the beginning of this section, so the proof is complete. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

Generalizations

Various generalizations of the Cauchy–Schwarz inequality exist. Hölder's inequality generalizes it to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^p} norms. More generally, it can be interpreted as a special case of the definition of the norm of a linear operator on a Banach space (Namely, when the space is a Hilbert space). Further generalizations are in the context of operator theory, e.g. for operator-convex functions and operator algebras, where the domain and/or range are replaced by a C*-algebra or W*-algebra.

An inner product can be used to define a positive linear functional. For example, given a Hilbert space Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^2(m), m} being a finite measure, the standard inner product gives rise to a positive functional Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi (g) = \langle g, 1 \rangle.} Conversely, every positive linear functional Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^2(m)} can be used to define an inner product Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle f, g \rangle _\varphi := \varphi\left(g^* f\right),} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g^*} is the pointwise complex conjugate of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g.} In this language, the Cauchy–Schwarz inequality becomes[13] Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|\varphi\left(g^* f\right)\right|^2 \leq \varphi\left(f^* f\right) \varphi\left(g^* g\right),}

which extends verbatim to positive functionals on C*-algebras:

Template:Math theorem

The next two theorems are further examples in operator algebra.

Template:Math theorem

This extends the fact Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi\left(a^*a\right) \cdot 1 \geq \varphi(a)^* \varphi(a) = |\varphi(a)|^2,} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} is a linear functional. The case when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} is self-adjoint, that is, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a = a^*,} is sometimes known as Kadison's inequality.

Template:Math theorem

Another generalization is a refinement obtained by interpolating between both sides of the Cauchy-Schwarz inequality:

Template:Math theorem

This theorem can be deduced from Hölder's inequality.

Licensing

Content obtained and/or adapted from:

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