Cauchy–Schwarz inequality

Let ${\displaystyle u}$ and ${\displaystyle v}$ be arbitrary vectors in a vector space over ${\displaystyle \mathbb {F} }$ with an inner product, where ${\displaystyle \mathbb {F} }$ is the field of real or complex numbers. We prove the inequality

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

and that equality holds if and only if either ${\displaystyle u}$ or ${\displaystyle v}$ is a multiple of the other (which includes the special case that either is the zero vector).

If ${\displaystyle v=0}$, it is clear that there is equality, and in this case ${\displaystyle u}$ and ${\displaystyle v}$ are also linearly dependent, regardless of ${\displaystyle u}$, so the theorem is true. Similarly if ${\displaystyle u=0}$. One henceforth assumes that ${\displaystyle v}$ is nonzero.

Let

${\displaystyle z=u-u_{v}=u-{\frac {\langle u,v\rangle }{\langle v,v\rangle }}v.}$

Then, by linearity of the inner product in its first argument, one has

${\displaystyle \langle z,v\rangle =\left\langle u-{\frac {\langle u,v\rangle }{\langle v,v\rangle }}v,v\right\rangle =\langle u,v\rangle -{\frac {\langle u,v\rangle }{\langle v,v\rangle }}\langle v,v\rangle =0.}$

Therefore, ${\displaystyle z}$ is a vector orthogonal to the vector ${\displaystyle v}$ (Indeed, ${\displaystyle z}$ is the projection of ${\displaystyle u}$ onto the plane orthogonal to ${\displaystyle v}$ .) We can thus apply the Pythagorean theorem to

${\displaystyle u={\frac {\langle u,v\rangle }{\langle v,v\rangle }}v+z}$

which gives

${\displaystyle \|u\|^{2}=\left|{\frac {\langle u,v\rangle }{\langle v,v\rangle }}\right|^{2}\|v\|^{2}+\|z\|^{2}={\frac {|\langle u,v\rangle |^{2}}{(\|v\|^{2})^{2}}}\,\|v\|^{2}+\|z\|^{2}={\frac {|\langle u,v\rangle |^{2}}{\|v\|^{2}}}+\|z\|^{2}\geq {\frac {|\langle u,v\rangle |^{2}}{\|v\|^{2}}}}$

and, after multiplication by ${\displaystyle \|v\|^{2}}$ and taking square root, we get the Cauchy–Schwarz inequality. Moreover, if the relation ${\displaystyle \geq }$ in the above expression is actually an equality, then ${\displaystyle \|z\|^{2}=0}$ and hence ${\displaystyle z=0}$; the definition of ${\displaystyle z}$ then establishes a relation of linear dependence between ${\displaystyle u}$ and ${\displaystyle v}$. On the other hand, if ${\displaystyle u}$ and ${\displaystyle v}$ are linearly dependent, then there exists ${\displaystyle c\in \mathbb {F} }$ such that ${\displaystyle u=c\cdot v}$ (since ${\displaystyle v\neq 0}$). Then

${\displaystyle |\langle u,v\rangle |=|\langle c\cdot v,v\rangle |=\left|c\|v\|^{2}\right|=|c|\|v\|^{2}=\|c\cdot v\|\|v\|=\|u\|\|v\|.}$

This establishes the theorem.

Let ${\displaystyle u}$ and ${\displaystyle v}$ be arbitrary vectors in an inner product space over ${\displaystyle \mathbb {C} }$.

In the special case ${\displaystyle v=0}$ the theorem is trivially true. Now assume that ${\displaystyle v\neq 0}$. Let ${\displaystyle \lambda \in \mathbb {C} }$ be given by ${\displaystyle \lambda =\langle u,v\rangle /\|v\|^{2}}$, then

${\displaystyle {\begin{aligned}0&\leq \|u-\lambda \cdot v\|^{2}\\&=\langle u,u\rangle -\langle \lambda \cdot v,u\rangle -\langle u,\lambda \cdot v\rangle +\langle \lambda \cdot v,\lambda \cdot v\rangle \\&=\langle u,u\rangle -\lambda \langle v,u\rangle -{\overline {\lambda }}\langle u,v\rangle +\lambda {\overline {\lambda }}\langle v,v\rangle \\&=\|u\|^{2}-\lambda {\overline {\langle u,v\rangle }}-{\overline {\lambda }}\langle u,v\rangle +\lambda {\overline {\lambda }}\|v\|^{2}\\&=\|u\|^{2}-{\frac {|\langle u,v\rangle |^{2}}{\|v\|^{2}}}-{\frac {|\langle u,v\rangle |^{2}}{\|v\|^{2}}}+{\frac {|\langle u,v\rangle |^{2}}{\|v\|^{2}}}\\&=\|u\|^{2}-{\frac {|\langle u,v\rangle |^{2}}{\|v\|^{2}}}.\end{aligned}}}$

Therefore, ${\displaystyle 0\leq \|u\|^{2}-{\frac {|\langle u,v\rangle |^{2}}{\|v\|^{2}}}}$, or ${\displaystyle |\langle u,v\rangle |\leq \|u\|\|v\|}$.

If the inequality holds as an equality, then ${\displaystyle \|u-\lambda \cdot v\|=0}$, and so ${\displaystyle u-\lambda \cdot v=0}$, thus ${\displaystyle u}$ and ${\displaystyle v}$ are linearly dependent. On the other hand, if ${\displaystyle u}$ and ${\displaystyle v}$ are linearly dependent, then ${\displaystyle |\langle u,v\rangle |=\|u\|\|v\|}$, as shown in the first proof.

There are many different proofs[6] of the Cauchy–Schwarz inequality other than the above two examples.[1][3] 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 ${\displaystyle \mathbb {R} }$ and not ${\displaystyle \mathbb {C} }$.[7]

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.

In the usual 2-dimensional space with the dot product, let ${\displaystyle v=(v_{1},v_{2})}$ and ${\displaystyle u=(u_{1},u_{2})}$. The Cauchy–Schwarz inequality is that

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

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

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

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

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

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

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

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

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,

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

which yields the Cauchy–Schwarz inequality.

For the inner product space of square-integrable complex-valued functions, one has

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

A generalization of this is the Hölder inequality.

The triangle inequality for the standard norm is often shown as a consequence of the Cauchy–Schwarz inequality, as follows: given vectors x and y:

${\displaystyle {\begin{aligned}\|x+y\|^{2}&=\langle x+y,x+y\rangle \\&=\|x\|^{2}+\langle x,y\rangle +\langle y,x\rangle +\|y\|^{2}\\&=\|x\|^{2}+2\operatorname {Re} \langle x,y\rangle +\|y\|^{2}\\&\leq \|x\|^{2}+2|\langle x,y\rangle |+\|y\|^{2}\\&\leq \|x\|^{2}+2\|x\|\|y\|+\|y\|^{2}\\&=(\|x\|+\|y\|)^{2}\end{aligned}}}$

Taking square roots gives the triangle inequality.

${\displaystyle \|x+y\|\leq \|x\|+\|y\|}$

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.[8][9]

The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any real inner-product space by defining:[10][11]

${\displaystyle \cos \theta _{xy}={\frac {\langle x,y\rangle }{\|x\|\|y\|}}.}$

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,[12][13] as is done when extracting a metric from quantum fidelity.

Let X, Y be random variables, then the covariance inequality[14][15] is given by

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

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

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

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