Difference between revisions of "Cauchy-Schwarz Formula"

From Department of Mathematics at UTSA
Jump to navigation Jump to search
 
Line 1: Line 1:
−
== Statement of the inequality ==
+
<p>One of the most important inequalities in mathematics is inarguably the famous Cauchy-Schwarz inequality whose use appears in many important proofs. We will prove this important inequality and prove an analogue of the triangle inequality in higher dimension Euclidean <math>n</math>-space.</p>
−
The Cauchy–Schwarz inequality states that for all vectors <math>u</math> and <math>v</math> of an inner product space it is true that
+
==The Cauchy-Schwarz Inequality==
  
−
<math>\left|\left\langle \mathbf{u}, \mathbf{v} \right\rangle\right|^2 \leq \langle \mathbf{u}, \mathbf{u} \rangle \cdot \langle \mathbf{v}, \mathbf{v} \rangle,</math>
+
<td><strong>Theorem 1 (The Cauchy-Schwarz Inequality):</strong> If <math>\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n</math> then <math>(\mathbf{x} \cdot \mathbf{y})^2 \leq \| \mathbf{x} \|^2 \| \mathbf{y} \|^2</math>.</td>
  
−
where <math>\langle \cdot, \cdot \rangle</math> 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 <math>\mathbf{u}</math> is denoted and defined by:
+
<ul>
−
<math>\left\|\mathbf{u}\right\| := \sqrt{\left\langle \mathbf{u}, \mathbf{u} \right\rangle}</math>
+
<li><strong>Proof:</strong> Let <math>\mathbf{x}, \mathbf{y} \in \mathbb{R}^n</math>. Then we want to prove that:</li>
−
so that this norm and the inner product are related by the defining condition <math>\left\|\mathbf{u}\right\|^2 = \left\langle \mathbf{u}, \mathbf{u} \right\rangle,</math> where <math>\left\langle \mathbf{u}, \mathbf{u} \right\rangle</math> is always a non-negative real number (even if the inner product is complex-valued).
+
</ul>
−
By taking the square root of both sides of the above inequality, the Cauchy–Schwarz inequality can be written in its more familiar form:
+
<math>\begin{align} \quad ( \mathbf{x} \cdot \mathbf{y} )^2 = \left ( \sum_{i=1}^{n} x_iy_i \right)^2 \leq \left ( \sum_{i=1}^{n} x_i^2 \right ) \left ( \sum_{i=1}^{n} y_i^2 \right ) = \| \mathbf{x} \|^2 \| \mathbf{y} \|^2 \quad (*) \end{align}</math>
 +
<ul>
 +
<li>Notice that the sum of squares is always nonnegative, and so for all <math>t \in \mathbb{R}</math> we have that:</li>
 +
</ul>
 +
<math>\begin{align} \quad 0 \leq \sum_{i=1}^{n} (x_it + y_i)^2 = \sum_{i=1}^{n} (x_i^2 t^2 + 2x_iy_it + y_i^2) = \sum_{i=1}^{n} x_i^2 t^2 + \sum_{i=1}^{n} 2x_iy_i t + \sum_{i=1}^{n} y_i^2 \end{align}</math>
 +
<ul>
 +
<li>Let <math>\displaystyle{A = \sum_{i=1}^{n} x_i^2}</math>, <math>\displaystyle{B = \sum_{i=1}^{n} x_iy_i}</math>, and <math>\displaystyle{C = \sum_{i=1}^{n} y_i^2}</math>. Then:</li>
 +
</ul>
 +
<math>\begin{align} \quad 0 \leq At^2 + 2Bt + C \end{align}</math>
 +
<ul>
 +
<li>Suppose that <math>A = 0</math>. Then <math>(*)</math> reduces to <math>0 \leq 0</math> which is true. If <math>A >  0</math>, then let <math>t = -\frac{B}{A}</math>. Then:</li>
 +
</ul>
 +
<math>\begin{align} \quad 0 \leq A \left ( \frac{-B}{A} \right)^2 + 2B \left(\frac{-B}{A} \right ) + C = \frac{B^2}{A} - \frac{2B^2}{A} + C = - \frac{B^2}{A} + C \end{align}</math>
 +
<ul>
 +
<li>Therefore we have that:</li>
 +
</ul>
 +
<math>\begin{align} \quad 0 \leq -\frac{B^2}{A} + C \\ \quad 0 \leq -B^2 + AC \\ \quad B^2 \leq AC \\ \quad \left ( \sum_{i=1}^{n} x_iy_i \right )^2 \leq \left ( \sum_{i=1}^{n} x_i^2 \right ) \left ( \sum_{i=1}^{n} y_i^2 \right ) \\ \quad (\mathbf{x} \cdot \mathbf{y})^2 \leq \| \mathbf{x} \|^2 \| \mathbf{y} \|^2 \quad \blacksquare \end{align}</math>
 +
<p>Often times the Cauchy-Schwarz inequality is stated by squaring both sides of the inequality above:</p>
 +
<math>\begin{align} \quad \mid \mathbf{x} \cdot \mathbf{y} | \leq \| \mathbf{x} \| \| \mathbf{y} \| \end{align}</math>
 +
==The Triangle Inequality==
  
−
<math>\langle \mathbf{u}, \mathbf{v} \rangle| \leq \|\mathbf{u}\| \|\mathbf{v}\|.</math>
+
<td><strong>Theorem 2 (The Triangle Inequality):</strong> If <math>\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n</math> then <math>\| \mathbf{x} + \mathbf{y} \| \leq \| \mathbf{x} \| + \| \mathbf{y} \|</math>.</td>
  
−
Moreover, the two sides are equal if and only if <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are linearly dependent.
+
<ul>
 +
<li><strong>Proof:</strong> Let <math>\mathbf{x}, \mathbf{y} \in \mathbb{R}^n</math>. Then:</li>
 +
</ul>
 +
<math>\begin{align} \quad \| \mathbf{x} + \mathbf{y} \| = \| (x_1 + y_1, x_2 + y_2, ..., x_n + y_n) \| = \left ( \sum_{i=1}^{n} (x_i + y_i)^2 \right )^{1/2} \end{align}</math>
 +
<ul>
 +
<li>Square both sides of the equation and apply the Cauchy-Schwarz inequality at <math>(*)</math> to get:</li>
 +
</ul>
 +
<math>\begin{align} \quad \| \mathbf{x} + \mathbf{y} \|^2 = \sum_{i=1}^{n} (x_i + y_i)^2 \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 = (x_1 + y_1)^2 + (x_2 + y_2)^2 + ... + (x_n + y_n)^2 \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 = (x_1^2 + 2x_1y_1 + y_1^2) + (x_2^2 + 2x_2y_2 + y_2^2) + ... + (x_n^2 + 2x_ny_n + y_n^2) \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 = (x_1^2 + x_2^2 + ... + x_n^2) + 2(x_1y_1 + x_2y_2 + ... + x_ny_n) + (y_1^2 + y_2^2 + ... + y_n^2) \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 = \| \mathbf{x} \|^2 + 2 (\mathbf{x} \cdot \mathbf{y}) + \| \mathbf{y} \|^2 \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 \overset{(*)} \leq \| \mathbf{x} \|^2 + 2 \| \mathbf{x} \| \| \mathbf{y} \| + \| \mathbf{y} \|^2 \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 \leq (\| \mathbf{x} \| + \| \mathbf{y} \|)^2 \end{align}</math>
 +
<ul>
 +
<li>Square rooting both sides of the inequality above yields <math>\| \mathbf{x} + \mathbf{y} \| \leq \| \mathbf{x} \| + \| \mathbf{y} \|</math> as desired. <math>\blacksquare</math></li>
 +
</ul>
  
−
== 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:
 
−
<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]].
 
−
 
−
=== ℝ<sup>2</sup> - The plane ===
 
−
[[File:Cauchy-Schwarz inequation in Euclidean plane.gif|thumb|300px|Cauchy-Schwarz inequality in a unit circle of the Euclidean plane]]
 
−
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:
 
−
<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>
 
−
 
−
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>\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.
 
−
 
−
{{anchor|Real 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:
 
−
<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.
 
−
Consider the following [[quadratic polynomial]] in <math>x</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,
 
−
<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.
 
−
 
−
=== 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:
 
−
<math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\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
 
−
= \left(\sum_{k=1}^{n} u_k \bar{u}_k\right) \left(\sum_{k=1}^{n} v_k \bar{v}_k\right)
 
−
= \sum_{j=1}^n \left|u_j\right|^2 \sum_{k=1}^n \left|v_k\right|^2.</math>
 
−
 
−
That is,
 
−
<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> ===
 
−
 
−
For the inner product space of [[square-integrable]] complex-valued [[function (mathematics)|functions]], the following inequality:
 
−
<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.
 
−
 
−
== Applications ==
 
−
 
−
=== Analysis ===
 
−
In any [[inner product space]], the [[triangle inequality]] is a consequence of the Cauchy–Schwarz inequality, as is now shown:
 
−
<math>\begin{alignat}{4}
 
−
\|\mathbf{u} + \mathbf{v}\|^2
 
−
&= \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v} \rangle && \\
 
−
&= \|\mathbf{u}\|^2 + \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{u} \rangle + \|\mathbf{v}\|^2 ~ && ~ \text{ where } \langle \mathbf{v}, \mathbf{u} \rangle = \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \\
 
−
&= \|\mathbf{u}\|^2 + 2 \operatorname{Re} \langle \mathbf{u}, \mathbf{v} \rangle + \|\mathbf{v}\|^2 && \\
 
−
&\leq \|\mathbf{u}\|^2 + 2|\langle \mathbf{u}, \mathbf{v} \rangle| + \|\mathbf{v}\|^2 && \\
 
−
&\leq \|\mathbf{u}\|^2 + 2\|\mathbf{u}\|\|\mathbf{v}\| + \|\mathbf{v}\|^2 && \\
 
−
&= (\|\mathbf{u}\| + \|\mathbf{v}\|)^2. &&
 
−
\end{alignat}</math>
 
−
 
−
Taking square roots gives the triangle inequality:
 
−
<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>
 
−
 
−
=== 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>
 
−
<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]].
 
−
 
−
=== 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>
 
−
<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. -->
 
−
 
−
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>\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,
 
−
<math>\langle X, Y \rangle := \operatorname{E}(X Y),</math>
 
−
the Cauchy–Schwarz inequality becomes
 
−
<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
 
−
<math>\begin{align}
 
−
|\operatorname{Cov}(X, Y)|^2
 
−
&= |\operatorname{E}\left((X - \mu)(Y - \nu)\right)|^2 \\
 
−
&=|\langle X - \mu, Y - \nu \rangle |^2\\
 
−
&\leq \langle X - \mu, X - \mu \rangle \langle Y - \nu, Y - \nu \rangle \\
 
−
& = \operatorname{E}\left((X - \mu)^2\right) \operatorname{E}\left((Y - \nu)^2\right) \\
 
−
& = \operatorname{Var}(X) \operatorname{Var}(Y),
 
−
\end{align}</math>
 
−
where <math>\operatorname{Var}</math> denotes [[variance]] and <math>\operatorname{Cov}</math> denotes [[covariance]].
 
−
 
−
== Proofs ==
 
−
 
−
There are many different proofs<ref>{{cite journal|last1=Wu|first1=Hui-Hua|last2=Wu|first2=Shanhe|date=April 2009|title=Various proofs of the Cauchy-Schwarz inequality|journal=Octogon Mathematical Magazine|issn=1222-5657|isbn=978-973-88255-5-0|volume=17|issue=1|pages=221–229|url=http://www.uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_19.pdf|access-date=18 May 2016}}</ref> of the Cauchy–Schwarz inequality other than those given below.<ref name="Steele" /><ref name=":0" />
 
−
When consulting other sources, there are often two sources of confusion. First, some authors define {{math|⟨⋅,⋅⟩}} to be linear in the [[Inner product space#Alternative definitions, notations and remarks|second argument]] rather than the first.
 
−
Second, some proofs are only valid when the field is <math>\mathbb R</math> and not <math>\mathbb C.</math><ref>{{cite book|last1=Aliprantis|first1=Charalambos D.|last2=Border|first2=Kim C.|date=2007-05-02|title=Infinite Dimensional Analysis: A Hitchhiker's Guide|publisher=Springer Science & Business Media|isbn=9783540326960|url=https://books.google.com/books?id=4hIq6ExH7NoC}}</ref>
 
−
 
−
This section gives proofs of the following theorem:
 
−
 
−
{{math theorem|name=Cauchy-Schwarz inequality|note=|style=|math_statement=
 
−
Let <math>\mathbf{u}</math> and <math>\mathbf{v}</math> be arbitrary vectors in an [[inner product space]] over the scalar field <math>\mathbb{F},</math> where <math>\mathbb F</math> is the field of real numbers <math>\R</math> or complex numbers <math>\Complex.</math> Then
 
−
 
−
{{NumBlk|:|<math>\big| \langle \mathbf{u}, \mathbf{v} \rangle \big| \leq\|\mathbf{u}\|\|\mathbf{v}\|</math>|{{EquationRef|Cauchy-Schwarz Inequality}}}}
 
−
 
−
where in addition, equality holds in the {{EquationNote|Cauchy-Schwarz Inequality}} [[if and only if]] <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are [[Linear independence|linearly dependent]].|{{EquationRef|Characterization of Equality in Cauchy-Schwarz}}}}
 
−
 
−
Moreover, if this equality holds and if <math>\mathbf{v} \neq \mathbf{0}</math> then <math>\mathbf{u} = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{v}\|^2} \mathbf{v}.</math>
 
−
 
−
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 <math>\|\mathbf{u}\|\|\mathbf{v}\|= 0</math>) is the same. It is presented immediately below only once to reduce repetition. It also includes the easy part of the proof the above {{EquationNote|Characterization of Equality in Cauchy-Schwarz|2=Equality Characterization}}; that is, it proves that if <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are linearly dependent then <math>\big| \langle \mathbf{u}, \mathbf{v} \rangle \big| =\|\mathbf{u}\|\|\mathbf{v}\|.</math>
 
−
 
−
{{collapse top|title=Proof of the trivial parts: Case where a vector is <math>\mathbf{0}</math> and also one direction of the {{EquationNote|Characterization of Equality in Cauchy-Schwarz|2=Equality Characterization}}|left=true}}
 
−
<!-- Note to Editors:
 
−
Anyone reading this part of the proof is very likely to be new to this subject, which is why this proof is much more detailed than the proofs given elsewhere in this article.
 
−
-->
 
−
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
 
−
<math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|
 
−
= \left|\langle c \mathbf{v}, \mathbf{v} \rangle\right|
 
−
= \left|c \langle \mathbf{v}, \mathbf{v} \rangle\right|
 
−
= \left|c\right|\|\mathbf{v}\|\|\mathbf{v}\|
 
−
=\|c \mathbf{v}\|\|\mathbf{v}\|
 
−
=\|\mathbf{u}\|\|\mathbf{v}\|</math>
 
−
 
−
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>:}}
 
−
<math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|
 
−
= \left|\langle \mathbf{u}, c \mathbf{u} \rangle\right|
 
−
= \left|\overline{c} \langle \mathbf{u}, \mathbf{u} \rangle\right|
 
−
= \left|\overline{c}\right| \|\mathbf{u}\|\|\mathbf{u}\|
 
−
= \left|c\right| \|\mathbf{u}\|\|\mathbf{u}\|
 
−
=\|\mathbf{u}\| \|c \mathbf{u}\|
 
−
=\|\mathbf{u}\| \|\mathbf{v}\|.</math>
 
−
 
−
If at least one of <math>\mathbf{u}</math> and <math>\mathbf{v}</math> is the zero vector then <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are necessarily linearly dependent (just scalar multiply the non-zero vector by the number <math>0</math> to get the zero vector; e.g. if <math>\mathbf{u} = \mathbf{0}</math> then let <math>c = 0</math> so that <math>\mathbf{u} = c \mathbf{v}</math>), which proves the converse of this characterization in this special case; that is, this shows that if at least one of <math>\mathbf{u}</math> and <math>\mathbf{v}</math> is <math>\mathbf{0}</math> then the {{EquationNote|Characterization of Equality in Cauchy-Schwarz|2=Equality Characterization}} holds.
 
−
 
−
If <math>\mathbf{u} = \mathbf{0},</math> which happens if and only if <math>\|\mathbf{u}\|= 0,</math> then <math>\|\mathbf{u}\| \|\mathbf{v}\| = 0</math> and <math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|
 
−
= \left|\langle \mathbf{0}, \mathbf{v} \rangle\right|
 
−
= |0| = 0</math> so that in particular, the Cauchy-Schwarz inequality holds because both sides of it are <math>0.</math>
 
−
The proof in the case of <math>\mathbf{v} = \mathbf{0}</math> is identical.
 
−
{{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 {{EquationNote|Characterization of Equality in Cauchy-Schwarz|2=Equality Characterization}} must be shown.
 
−
 
−
=== For real inner products spaces ===
 
−
 
−
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:
 
−
<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:
 
−
<math>\Delta = 4 \left(\langle u, v \rangle ^2 - \Vert u \Vert^2 \Vert v \Vert^2\right) \leqslant 0.</math>
 
−
The conclusion follows.
 
−
 
−
For the equality case, notice that <math>\Delta = 0</math> happens if and only if <math>p(t) = (t\Vert u \Vert + \Vert v \Vert)^2.</math> If <math>t_0 = -\Vert v \Vert / \Vert u \Vert,</math> then <math>p(t_0) = \langle t_0u + v,t_0u + v\rangle = 0,</math> and hence <math>v = -t_0u.</math>
 
−
 
−
==== Proof for the dot product ====
 
−
 
−
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:
 
−
<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
 
−
<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>
 
−
 
−
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>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:
 
−
<math>\begin{alignat}{4}
 
−
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 x^2 + 2a_1 b_1 x + b_1^2\right) + \left(a_2^2 x^2 + 2a_2 b_2 x + b_2^2\right) + \cdots + \left(a_n^2 x^2 + 2a_n b_n x + b_n^2\right) \\
 
−
&= \left(a_1x + b_1\right)^2 + \left(a_2 x + b_2\right)^2 + \cdots + \left(a_n x + b_n\right)^2 \\
 
−
&\geq 0
 
−
\end{alignat}</math>
 
−
which is a sum of terms that are each <math>\,\geq 0\,</math> by the [https://artofproblemsolving.com/wiki/index.php/Trivial_Inequality trivial inequality:] <math>r^2 \geq 0</math> for all <math>r \in \R.</math>
 
−
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
 
−
<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>
 
−
 
−
=== For arbitrary vector spaces ===
 
−
 
−
==== Proof 1 ====
 
−
 
−
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>
 
−
As is now shown, the Cauchy–Schwarz {{em|in}}equality (and the rest of the theorem) is an almost immediate corollary of the following {{em|equality}}:{{sfn|Halmos|1982|pp=2, 167}}
 
−
 
−
{{NumBlk|:|<math>\frac{1}{\|\mathbf{v}\|^2} \left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2} =\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2</math>|{{EquationRef|Eq. 1}}}}
 
−
 
−
which is readily verified by elementarily expanding <math>\left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^{2}</math> (via the definition of the norm) and then simplifying.
 
−
 
−
Observing that the left hand side of {{EquationNote|Eq. 1}} is non-negative (which makes this also true of the right hand side) proves that <math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right|^2 \leq \|\mathbf{u}\|^2\|\mathbf{v}\|^2,</math> from which the {{EquationNote|Cauchy-Schwarz Inequality}} follows (by taking the square root of both sides).
 
−
If <math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right| =\|\mathbf{u}\|\|\mathbf{v}\|</math> then the right hand side (and thus also the left hand side) of {{EquationNote|Eq. 1}} is <math>0,</math> which is only possible if {{nowrap|<math>\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v} = \mathbf{0}</math>;}}<ref group=note>In fact, it follows immediately from {{EquationRef|Eq. 1}} that <math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right| = \|\mathbf{u}\|\|\mathbf{v}\|</math> if and only if <math>\|\mathbf{v}\|^2 \mathbf{u} = \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}.</math></ref>
 
−
thus <math>\mathbf{u} = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{v}\|^2} \mathbf{v},</math> which shows that <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are linearly dependent.{{sfn|Halmos|1982|pp=2, 167}}
 
−
Since the (trivial) converse was proved above, the proof of the theorem is complete. <math>\blacksquare</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
 
−
<math>\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}
 
−
</math>
 
−
 
−
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:
 
−
<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
 
−
<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 \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}
 
−
</math>
 
−
 
−
==== Proof 2 ====
 
−
 
−
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
 
−
<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:
 
−
<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
 
−
= \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.</math>
 
−
 
−
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>\mathbf{u}= \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} + \mathbf{z}</math>
 
−
which gives
 
−
<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
 
−
= \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}.</math>
 
−
 
−
The Cauchy–Schwarz inequality follows by multiplying by <math>\|\mathbf{v}\|^2</math> and then taking the square root.
 
−
Moreover, if the relation <math>\geq</math> in the above expression is actually an equality, then <math>\|\mathbf{z}\|^2 = 0</math> and hence <math>\mathbf{z} = \mathbf{0};</math> the definition of <math>\mathbf{z}</math> then establishes a relation of linear dependence between <math>\mathbf{u}</math> and <math>\mathbf{v}.</math> The converse was proved at the beginning of this section, so the proof is complete. <math>\blacksquare</math>
 
−
 
−
== Generalizations ==
 
−
Various generalizations of the Cauchy–Schwarz inequality exist. [[Hölder's inequality]] generalizes it to <math>L^p</math> 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 algebra]]s, 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 <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>\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:
 
−
 
−
{{math theorem|name=Cauchy–Schwarz inequality for  positive functionals on C*-algebras<ref>{{cite book|last=Lin|first=Huaxin|date=2001-01-01|title=An Introduction to the Classification of Amenable C*-algebras|publisher=World Scientific|isbn=9789812799883|pages=27|url=https://books.google.com/books?id=2qru8d7BCAAC}}</ref><ref>{{cite book|last=Arveson|first=W.|date=2012-12-06|title=An Invitation to C*-Algebras|publisher=Springer Science & Business Media|isbn=9781461263715|pages=28|url=https://books.google.com/books?id=d5TqBwAAQBAJ}}</ref>|note=|style=|math_statement=
 
−
If <math>\varphi</math> is a positive linear functional on a C*-algebra <math>A,</math> then for all <math>a, b \in A,</math> <math>\left|\varphi\left(b^*a\right)\right|^2 \leq \varphi\left(b^*b\right)\varphi\left(a^*a\right).</math>
 
−
}}
 
−
 
−
The next two theorems are further examples in operator algebra.
 
−
 
−
{{math theorem|name=Kadison–Schwarz inequality<ref>{{cite book|last=Størmer|first=Erling|date=2012-12-13|title=Positive Linear Maps of Operator Algebras|series=Springer Monographs in Mathematics|publisher=Springer Science & Business Media|isbn=9783642343698|url=https://books.google.com/books?id=lQtKAIONqwIC}}</ref><ref>{{cite journal|last=Kadison|first=Richard V.|date=1952-01-01|title=A Generalized Schwarz Inequality and Algebraic Invariants for Operator Algebras|jstor=1969657|journal=Annals of Mathematics|doi=10.2307/1969657|volume=56|number=3|pages=494–503}}</ref>|note=Named after [[Richard Kadison]]|style=|math_statement=
 
−
If <math>\varphi</math> is a unital positive map, then for every [[normal operator|normal element]] <math>a</math> in its domain, we have <math>\varphi(a^*a) \geq \varphi\left(a^*\right) \varphi(a)</math> and <math>\varphi\left(a^*a\right) \geq \varphi(a) \varphi\left(a^*\right).</math>
 
−
}}
 
−
 
−
This extends the fact <math>\varphi\left(a^*a\right) \cdot 1 \geq \varphi(a)^* \varphi(a) = |\varphi(a)|^2,</math> when <math>\varphi</math> is a linear functional. The case when <math>a</math> is self-adjoint, that is, <math>a = a^*,</math> is sometimes known as '''Kadison's inequality'''.
 
−
 
−
{{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,
 
−
<math>\varphi(a)^*\varphi(a) \leq \Vert\varphi(1)\Vert\varphi\left(a^*a\right), \text{ and }</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>
 
−
}}
 
−
 
−
Another generalization is a refinement obtained by interpolating between both sides of the Cauchy-Schwarz inequality: 
 
−
 
−
{{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>
 
−
<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+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^2\right) \left(\sum_{i=1}^n b_i^2\right).</math>
 
−
}}
 
−
 
−
This theorem can be deduced from [[Hölder's inequality]].
 
  
 
==Licensing==
 
==Licensing==
 
Content obtained and/or adapted from:
 
Content obtained and/or adapted from:
−
* [https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality Cauchy-Schwarz inequality, Wikipedia] under a CC BY-SA license
+
* [http://mathonline.wikidot.com/the-cauchy-schwarz-and-triangle-inequalities The Cauchy-Schwarz and Triangle Inequalities, mathonline.wikidot.com] under a CC BY-SA license

Latest revision as of 15:19, 28 October 2021

One of the most important inequalities in mathematics is inarguably the famous Cauchy-Schwarz inequality whose use appears in many important proofs. We will prove this important inequality and prove an analogue of the triangle inequality in higher dimension Euclidean -space.

The Cauchy-Schwarz Inequality

Theorem 1 (The Cauchy-Schwarz Inequality): If then .

  • Proof: Let . Then we want to prove that:

  • Notice that the sum of squares is always nonnegative, and so for all we have that:

  • Let , , and . Then:

  • Suppose that . Then reduces to which is true. If , then let . Then:

  • Therefore we have that:

Often times the Cauchy-Schwarz inequality is stated by squaring both sides of the inequality above:

The Triangle Inequality

Theorem 2 (The Triangle Inequality): If then .

  • Proof: Let . Then:

  • Square both sides of the equation and apply the Cauchy-Schwarz inequality at to get:

  • Square rooting both sides of the inequality above yields as desired.


Licensing

Content obtained and/or adapted from:

Retrieved from "https://mathresearch.utsa.edu/wiki/index.php?title=Cauchy-Schwarz_Formula&oldid=3162"