Inner product/K/Norm/Cauchy-Schwarz/Section

If an inner product is given, then we can define the length of a vector, and then also the distance between two vectors.

Definition

Let ${}V$ denote a vector space over ${}{\mathbb {K} }$, endowed with an inner product ${}\left\langle -,-\right\rangle$. For a vector ${}v\in V$, we call the real number

${}\Vert {v}\Vert ={\sqrt {\left\langle v,v\right\rangle }}\,$

the norm of ${}v$.

The inner product ${}\left\langle v,v\right\rangle$ is always real and not negative; therefore, its square root is a uniquely determined real number. For a complex vector space with an inner product, it does not make any difference whether we determine the norm directly, or via the underlying real vector space, see exercise.

Theorem

Let ${}V$ denote a vector space over ${}{\mathbb {K} }$, endowed with an inner product ${}\left\langle -,-\right\rangle$. Let ${}\Vert {-}\Vert$ denote the associated norm. Then the Cauchy-Schwarz estimate holds, that is,

${}\vert {\left\langle v,w\right\rangle }\vert \leq \Vert {v}\Vert \cdot \Vert {w}\Vert \,$

for all

${}v,w\in V$.

Proof

For ${}w=0$, the statement holds. So suppose ${}w\neq 0$, hence, also ${}\Vert {w}\Vert \neq 0$ holds. Therefore, we have the estimates

${}{\begin{aligned}0&\leq \left\langle v-{\frac {\left\langle v,w\right\rangle }{\Vert {w}\Vert ^{2}}}w,v-{\frac {\left\langle v,w\right\rangle }{\Vert {w}\Vert ^{2}}}w\right\rangle \\&=\left\langle v,v\right\rangle -{\frac {\left\langle v,w\right\rangle }{\Vert {w}\Vert ^{2}}}\left\langle w,v\right\rangle -{\frac {\overline {\left\langle v,w\right\rangle }}{\Vert {w}\Vert ^{2}}}\left\langle v,w\right\rangle +{\frac {\left\langle v,w\right\rangle {\overline {\left\langle v,w\right\rangle }}}{\Vert {w}\Vert ^{4}}}\left\langle w,w\right\rangle \\&=\left\langle v,v\right\rangle -{\frac {\left\langle v,w\right\rangle }{\Vert {w}\Vert ^{2}}}{\overline {\left\langle v,w\right\rangle }}-{\frac {\overline {\left\langle v,w\right\rangle }}{\Vert {w}\Vert ^{2}}}\left\langle v,w\right\rangle +{\frac {\left\langle v,w\right\rangle {\overline {\left\langle v,w\right\rangle }}}{\Vert {w}\Vert ^{2}}}\\&=\left\langle v,v\right\rangle -{\frac {\left\langle v,w\right\rangle {\overline {\left\langle v,w\right\rangle }}}{\Vert {w}\Vert ^{2}}}\\&=\left\langle v,v\right\rangle -{\frac {\vert {\left\langle v,w\right\rangle }\vert ^{2}}{\Vert {w}\Vert ^{2}}}.\end{aligned}}$

Multiplication with ${}\Vert {w}\Vert ^{2}$ and taking the square root yields the result.

$\Box$

Lemma

Let ${}V$ denote a vector space over ${}{\mathbb {K} }$, endowed with an inner product ${}\left\langle -,-\right\rangle$. Then the corresponding norm

satisfies the following properties.

1. We have ${}\Vert {v}\Vert \geq 0$.
2. We have ${}\Vert {v}\Vert =0$ if and only if ${}v=0$.
3. For ${}\lambda \in {\mathbb {K} }$ and ${}v\in V$, we have

   ${}\Vert {\lambda v}\Vert =\vert {\lambda }\vert \cdot \Vert {v}\Vert \,.$

4. For ${}v,w\in V$, we have

   ${}\Vert {v+w}\Vert \leq \Vert {v}\Vert +\Vert {w}\Vert \,.$

Proof

The first two properties follow directly from the definition of an inner product.
The compatibility with multiplication follows from

${}\Vert {\lambda v}\Vert ^{2}=\left\langle \lambda v,\lambda v\right\rangle =\lambda \left\langle v,\lambda v\right\rangle =\lambda {\overline {\lambda }}\left\langle v,v\right\rangle =\vert {\lambda }\vert ^{2}\Vert {v}\Vert ^{2}\,.$

In order to prove the triangle estimate, we write

${}{\begin{aligned}\Vert {v+w}\Vert ^{2}&=\left\langle v+w,v+w\right\rangle \\&=\Vert {v}\Vert ^{2}+\Vert {w}\Vert ^{2}+\left\langle v,w\right\rangle +{\overline {\left\langle v,w\right\rangle }}\\&=\Vert {v}\Vert ^{2}+\Vert {w}\Vert ^{2}+2\operatorname {Re} \,{\left(\left\langle v,w\right\rangle \right)}\\&\leq \Vert {v}\Vert ^{2}+\Vert {w}\Vert ^{2}+2\vert {\left\langle v,w\right\rangle }\vert .\end{aligned}}$

Due to fact, this is ${}\leq {\left(\Vert {v}\Vert +\Vert {w}\Vert \right)}^{2}$. This estimate transfers to the square roots.

$\Box$

With the following statement, the polarization identity, we can reconstruct the inner product from its associated norm.

Lemma

Let ${}V$ denote a vector space over ${}{\mathbb {K} }$, endowed with an inner product ${}\left\langle -,-\right\rangle$. Let ${}\Vert {-}\Vert$ denote the associated norm. Then, in case ${}{\mathbb {K} }=\mathbb {R}$, the relation

${}\left\langle v,w\right\rangle ={\frac {1}{2}}{\left(\Vert {v+w}\Vert ^{2}-\Vert {v}\Vert ^{2}-\Vert {w}\Vert ^{2}\right)}\,$

holds, and, in case ${}{\mathbb {K} }=\mathbb {C}$, the relation

$\left\langle v,w\right\rangle ={\frac {1}{4}}{\left(\Vert {v+w}\Vert ^{2}-\Vert {v-w}\Vert ^{2}+{\mathrm {i} }\Vert {v+{\mathrm {i} }w}\Vert ^{2}-{\mathrm {i} }\Vert {v-{\mathrm {i} }w}\Vert ^{2}\right)}\,$

holds.

Proof

See exercise.

$\Box$