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This quesiton has already been asked for the simple $\mathbb{R}^n$ vector space here. I however would like to have a general proof for the following:

Let $V$ be a vector space over the field $F$, i.e. we have:

  1. $\langle\cdot,\cdot \rangle: V\times V \rightarrow F$
  2. $\langle x,y\rangle =\overline{\langle y,x\rangle}$
  3. $\langle \alpha x,y\rangle =\alpha \langle x,y\rangle$
  4. $\langle x+y,z\rangle = \langle x,z\rangle +\langle y,z\rangle$
  5. $\langle x,x\rangle \geq 0$, $\langle x,x\rangle=0\Leftrightarrow x = 0$

Let $B_1(V)$ be the unit sphere of $V$ (with respect to $\langle \cdot ,\cdot \rangle$)

Now to the actual question:

Let $v\in B_1(V)$. Show that $$\sup_{w\in B_1(V)} |\langle v,w\rangle| = 1.$$

Optimally the proof would be built on the axioms itself, i.e. without the use of advanced theorems.

uniquesolution
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Jürg W. Spaak
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1 Answers1

2

If you take $\mathbb{F}$ as the real or complex fields, then it can be shown that any inner product satisfies the Cauchy-Scwartz inequality, namely:

$$|\langle u,v\rangle|\leq \|u\| \|v\|$$ for every $u,v$, where $\|u\|=\sqrt{\langle u,u\rangle}$

Now it follows immediately from the C-S inequality that given $v\in B_1(V)$, for every $w\in B_1(V)$ you have $|\langle w,v\rangle|\leq 1$, and if $v\neq 0$, you can take $w=\frac{v}{\|v\|}$, so that the supremum is actually a maximum.

uniquesolution
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