Cauchy–Schwarz inequality

The Cauchy–Schwarz inequality (also called Cauchy–Bunyakovsky-Schwarz inequality)^[1][2][3][4] is considered one of the most important and widely used inequalities in mathematics.^[5]

The inequality for sums was published by Augustin-Louis Cauchy (1821). The corresponding inequality for integrals was published by Viktor Bunyakovsky (1859)^[2] and Hermann Schwarz (1888). Schwarz gave the modern proof of the integral version.^[5]

Statement of the inequality[]

The Cauchy–Schwarz inequality states that for all vectors ${\displaystyle u}$ and ${\displaystyle v}$ of an inner product space it is true that

${\displaystyle \left|\left\langle \mathbf {u} ,\mathbf {v} \right\rangle \right|^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle ,}$

(Cauchy-Schwarz inequality [written using only the inner product])

where ${\displaystyle \langle \cdot ,\cdot \rangle }$ 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 ${\displaystyle \mathbf {u} }$ is denoted and defined by:

${\displaystyle \left\|\mathbf {u} \right\|:={\sqrt {\left\langle \mathbf {u} ,\mathbf {u} \right\rangle }}}$

so that this norm and the inner product are related by the defining condition ${\displaystyle \left\|\mathbf {u} \right\|^{2}=\left\langle \mathbf {u} ,\mathbf {u} \right\rangle ,}$ where ${\displaystyle \left\langle \mathbf {u} ,\mathbf {u} \right\rangle }$ 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:^[6][7]

${\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |\leq \|\mathbf {u} \|\|\mathbf {v} \|.}$

(Cauchy-Schwarz inequality [written using norm and inner product])

Moreover, the two sides are equal if and only if ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$ are linearly dependent.^[8][9][10]

Special cases[]

Titu's lemma - Positive real numbers[]

Titu's lemma (named after Titu Andreescu, also known as T2 lemma, Engel's form, or Sedrakyan's inequality) states that for positive reals, one has

${\displaystyle {\frac {\left(\sum _{i=1}^{n}u_{i}\right)^{2}}{\sum _{i=1}^{n}v_{i}}}\leq \sum _{i=1}^{n}{\frac {u_{i}^{2}}{v_{i}}}.}$

It is a direct consequence of the Cauchy–Schwarz inequality, obtained upon substituting ${\displaystyle u_{i}'={\frac {u_{i}}{\sqrt {v_{i}}}}}$ and ${\displaystyle v_{i}'={\sqrt {v_{i}}}.}$ This form is especially helpful when the inequality involves fractions where the numerator is a perfect square.

ℝ² - The plane[]

The real vector space ${\displaystyle \mathbb {R} ^{2}}$ denotes the 2-dimensional plane. It is also the 2-dimensional Euclidean space where the inner product is the dot product. If ${\displaystyle \mathbf {v} =\left(v_{1},v_{2}\right)}$ and ${\displaystyle \mathbf {u} =\left(u_{1},u_{2}\right)}$ then the Cauchy–Schwarz inequality becomes:

${\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle ^{2}=(\|\mathbf {u} \|\|\mathbf {v} \|\cos \theta )^{2}\leq \|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2},}$

where ${\displaystyle \theta }$ is the angle between ${\displaystyle u}$ and ${\displaystyle v}$.

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 ${\displaystyle v_{1},v_{2},u_{1}}$ and ${\displaystyle u_{2}}$ as

${\displaystyle \left(u_{1}v_{1}+u_{2}v_{2}\right)^{2}\leq \left(u_{1}^{2}+u_{2}^{2}\right)\left(v_{1}^{2}+v_{2}^{2}\right),}$

where equality holds if and only if the vector ${\displaystyle (u_{1},u_{2})}$ is in the same or opposite direction as the vector ${\displaystyle \left(v_{1},v_{2}\right)}$, or if one of them is the zero vector.

ℝⁿ - n-dimensional Euclidean space[]

In Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ with the standard inner product, which is the dot product, the Cauchy–Schwarz inequality becomes:

${\displaystyle \left(\sum _{i=1}^{n}u_{i}v_{i}\right)^{2}\leq \left(\sum _{i=1}^{n}u_{i}^{2}\right)\left(\sum _{i=1}^{n}v_{i}^{2}\right)}$

The Cauchy–Schwarz inequality can be proved using only ideas from elementary algebra in this case. Consider the following quadratic polynomial in ${\displaystyle x}$

${\displaystyle 0\leq \left(u_{1}x+v_{1}\right)^{2}+\cdots +\left(u_{n}x+v_{n}\right)^{2}=\left(\sum _{i}u_{i}^{2}\right)x^{2}+2\left(\sum _{i}u_{i}v_{i}\right)x+\sum _{i}v_{i}^{2}.}$

Since it is nonnegative, it has at most one real root for ${\displaystyle x,}$ hence its discriminant is less than or equal to zero. That is,

${\displaystyle \left(\sum _{i}u_{i}v_{i}\right)^{2}-\left(\sum _{i}{u_{i}^{2}}\right)\left(\sum _{i}{v_{i}^{2}}\right)\leq 0,}$

which yields the Cauchy–Schwarz inequality.

ℂⁿ - n-dimensional Complex space[]

If ${\displaystyle \mathbf {u} ,\mathbf {v} \in \mathbb {C} ^{n}}$ with ${\displaystyle \mathbf {u} =\left(u_{1},\ldots ,u_{n}\right)}$ and ${\displaystyle \mathbf {v} =\left(v_{1},\ldots ,v_{n}\right)}$ (where ${\displaystyle u_{1},\ldots ,u_{n}\in \mathbb {C} }$ and ${\displaystyle v_{1},\ldots ,v_{n}\in \mathbb {C} }$) and if the inner product on the vector space ${\displaystyle \mathbb {C} ^{n}}$ is the canonical complex inner product (defined by ${\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle :=u_{1}{\overline {v_{1}}}+\cdots +u_{n}{\overline {v_{n}}}}$), then the inequality may be restated more explicitly as follows (where the bar notation is used for complex conjugation):

${\displaystyle \left|\langle \mathbf {u} ,\mathbf {v} \rangle \right|^{2}=\left|\sum _{k=1}^{n}u_{k}{\bar {v}}_{k}\right|^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \langle \mathbf {v} ,\mathbf {v} \rangle =\left(\sum _{k=1}^{n}u_{k}{\bar {u}}_{k}\right)\left(\sum _{k=1}^{n}v_{k}{\bar {v}}_{k}\right)=\sum _{j=1}^{n}\left|u_{j}\right|^{2}\sum _{k=1}^{n}\left|v_{k}\right|^{2}.}$

That is,

${\displaystyle \left|u_{1}{\bar {v}}_{1}+\cdots +u_{n}{\bar {v}}_{n}\right|^{2}\leq \left(\left|u_{1}\right|^{2}+\cdots +\left|u_{n}\right|^{2}\right)\left(\left|v_{1}\right|^{2}+\cdots +\left|v_{n}\right|^{2}\right).}$

L²[]

For the inner product space of square-integrable complex-valued functions, the following inequality:

${\displaystyle \left|\int _{\mathbb {R} ^{n}}f(x){\overline {g(x)}}\,dx\right|^{2}\leq \int _{\mathbb {R} ^{n}}|f(x)|^{2}\,dx\int _{\mathbb {R} ^{n}}|g(x)|^{2}\,dx.}$

The Hölder inequality is a generalization of this.

Proof[]

For real vector spaces[]

Let ${\displaystyle (V,\langle \cdot ,\cdot \rangle )}$ be an inner product space. Consider an arbitary pair ${\displaystyle u,v\in V}$ and the function ${\displaystyle p\colon \mathbb {R} \to \mathbb {R} }$ defined by ${\displaystyle p(t)=\langle tu+v,tu+v\rangle }$. Since the inner product is positive-definite, ${\displaystyle p(t)}$ only takes non-negative values. On the other hand, ${\displaystyle p(t)}$ can be expanded using the bilinearity of the inner product:

${\displaystyle p(t)=\Vert u\Vert ^{2}t^{2}+2t\langle u,v\rangle +\Vert v\Vert ^{2}}$.

Thus, ${\displaystyle p}$ is a polynomial of degree ${\displaystyle 2}$ (unless ${\displaystyle u=0}$, in which case there is nothing to show). Since the sign of ${\displaystyle p}$ does not change, the discriminant of this polynomial must be non-positive:

${\displaystyle \Delta =4\left(\langle u,v\rangle ^{2}-\Vert u\Vert ^{2}\Vert v\Vert ^{2}\right)\leqslant 0}$.

The conclusion follows.

For the equality case, notice that ${\displaystyle \Delta =0}$ happens if and only if ${\displaystyle p(t)=(t\Vert u\Vert +\Vert v\Vert )^{2}.}$ If ${\displaystyle t_{0}=-\Vert v\Vert /\Vert u\Vert ,}$ then ${\displaystyle p(t_{0})=\langle t_{0}u+v,t_{0}u+v\rangle =0,}$ and hence ${\displaystyle v=-t_{0}u.}$

For arbitrary vector spaces[]

Let ${\displaystyle u,v\in V.}$ Ignoring the trivial case where ${\displaystyle v=0,}$ define ${\displaystyle \textstyle w=u-{\frac {\langle u,v\rangle }{\Vert v\Vert ^{2}}}v.}$ Since ${\displaystyle \langle w,v\rangle =0,}$ by the Pythagorean theorem,

${\displaystyle \Vert u\Vert ^{2}={\frac {|\langle u,v\rangle |^{2}}{\Vert v\Vert ^{2}}}+\Vert w\Vert ^{2}\geq {\frac {|\langle u,v\rangle |^{2}}{\Vert v\Vert ^{2}}},}$

with the equality holding if and only if ${\displaystyle u={\frac {\langle u,v\rangle }{\Vert v\Vert ^{2}}}v.}$

Applications[]

Analysis[]

In any inner product space, the triangle inequality is a consequence of the Cauchy–Schwarz inequality, as is now shown:

${\displaystyle {\begin{alignedat}{4}\|\mathbf {u} +\mathbf {v} \|^{2}&=\langle \mathbf {u} +\mathbf {v} ,\mathbf {u} +\mathbf {v} \rangle &&\\&=\|\mathbf {u} \|^{2}+\langle \mathbf {u} ,\mathbf {v} \rangle +\langle \mathbf {v} ,\mathbf {u} \rangle +\|\mathbf {v} \|^{2}~&&~{\text{ where }}\langle \mathbf {v} ,\mathbf {u} \rangle ={\overline {\langle \mathbf {u} ,\mathbf {v} \rangle }}\\&=\|\mathbf {u} \|^{2}+2\operatorname {Re} \langle \mathbf {u} ,\mathbf {v} \rangle +\|\mathbf {v} \|^{2}&&\\&\leq \|\mathbf {u} \|^{2}+2|\langle \mathbf {u} ,\mathbf {v} \rangle |+\|\mathbf {v} \|^{2}&&\\&\leq \|\mathbf {u} \|^{2}+2\|\mathbf {u} \|\|\mathbf {v} \|+\|\mathbf {v} \|^{2}&&\\&=(\|\mathbf {u} \|+\|\mathbf {v} \|)^{2}.&&\end{alignedat}}}$

Taking square roots gives the triangle inequality:

${\displaystyle \|\mathbf {u} +\mathbf {v} \|\leq \|\mathbf {u} \|+\|\mathbf {v} \|.}$

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.^[11][12]

Geometry[]

The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any real inner-product space by defining:^[13][14]

${\displaystyle \cos \theta _{\mathbf {u} \mathbf {v} }={\frac {\langle \mathbf {u} ,\mathbf {v} \rangle }{\|\mathbf {u} \|\|\mathbf {v} \|}}.}$

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,^[15][16] as is done when extracting a metric from quantum fidelity.

Probability theory[]

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be random variables, then the covariance inequality:^[17][18] is given by

${\displaystyle \operatorname {Var} (Y)\geq {\frac {\operatorname {Cov} (Y,X)^{2}}{\operatorname {Var} (X)}}.}$

After defining an inner product on the set of random variables using the expectation of their product,

${\displaystyle \langle X,Y\rangle :=\operatorname {E} (XY),}$

the Cauchy–Schwarz inequality becomes

${\displaystyle |\operatorname {E} (XY)|^{2}\leq \operatorname {E} (X^{2})\operatorname {E} (Y^{2}).}$

To prove the covariance inequality using the Cauchy–Schwarz inequality, let ${\displaystyle \mu =\operatorname {E} (X)}$ and ${\displaystyle \nu =\operatorname {E} (Y),}$ then

${\displaystyle {\begin{aligned}|\operatorname {Cov} (X,Y)|^{2}&=|\operatorname {E} \left((X-\mu )(Y-\nu )\right)|^{2}\\&=|\langle X-\mu ,Y-\nu \rangle |^{2}\\&\leq \langle X-\mu ,X-\mu \rangle \langle Y-\nu ,Y-\nu \rangle \\&=\operatorname {E} \left((X-\mu )^{2}\right)\operatorname {E} \left((Y-\nu )^{2}\right)\\&=\operatorname {Var} (X)\operatorname {Var} (Y),\end{aligned}}}$

where ${\displaystyle \operatorname {Var} }$ denotes variance and ${\displaystyle \operatorname {Cov} }$ denotes covariance.

Generalizations[]

Various generalizations of the Cauchy–Schwarz inequality exist. Hölder's inequality generalizes it to ${\displaystyle L^{p}}$ 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 ${\displaystyle L^{2}(m),m}$ being a finite measure, the standard inner product gives rise to a positive functional ${\displaystyle \varphi }$ by ${\displaystyle \varphi (g)=\langle g,1\rangle .}$ Conversely, every positive linear functional ${\displaystyle \varphi }$ on ${\displaystyle L^{2}(m)}$ can be used to define an inner product ${\displaystyle \langle f,g\rangle _{\varphi }:=\varphi \left(g^{*}f\right),}$ where ${\displaystyle g^{*}}$ is the pointwise complex conjugate of ${\displaystyle g.}$ In this language, the Cauchy–Schwarz inequality becomes^[19]

${\displaystyle \left|\varphi \left(g^{*}f\right)\right|^{2}\leq \varphi \left(f^{*}f\right)\varphi \left(g^{*}g\right),}$

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

The next two theorems are further examples in operator algebra.

This extends the fact ${\displaystyle \varphi \left(a^{*}a\right)\cdot 1\geq \varphi (a)^{*}\varphi (a)=|\varphi (a)|^{2},}$ when ${\displaystyle \varphi }$ is a linear functional. The case when ${\displaystyle a}$ is self-adjoint, i.e. ${\displaystyle a=a^{*},}$ is sometimes known as Kadison's inequality.

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

This theorem can be deduced from Hölder's inequality.^[26] There are also non commutative versions for operators and tensor products of matrices.^[27]

See also[]

• Bessel's inequality
• Hölder's inequality
• Jensen's inequality
• Kunita–Watanabe inequality
• Minkowski inequality

Notes[]

Citations[]

References[]

External links[]

• Earliest Uses: The entry on the Cauchy–Schwarz inequality has some historical information.
• Example of application of Cauchy–Schwarz inequality to determine Linearly Independent Vectors Tutorial and Interactive program.