Does $\frac{x}{||x||}$ maximize inner product with $x$ among vectors of norm $1$?

Question

Let $x \in \mathbb{R}^{n}$ (with a Euclidean norm and inner product) be fixed. Intuitively inner product between two vector combines two pieces of information: how large the two vectors are in norm and how much they point in the same "direction". This intuition suggests the following result: $$ \sup_{||y|| = 1} \langle x, y \rangle\ = \langle x, \frac{x}{||x||} \rangle $$

In other words $\frac{x}{||x||}$ is the argmax of $y \mapsto \langle x, y \rangle$ defined for all norm $1$ vectors. Is this true? How to show it? Is it true for all inner products on $\mathbb{R}^{n}$ with their associated norms?

Answer 1

Note that$$\left\langle x,\frac x{\|x\|}\right\rangle=\|x\|.\tag1$$So, your question is: is $\sup_{\|y\|=1}\langle x,y\rangle=\|x\|$? Yes. From $(1)$, you know that there is some $y$ with norm $1$ such that $\langle x,y\rangle=1$. On the other hand, by the Cauchy-Schwarz inequality, if $\|y\|=1$, you have$$\langle x,y\rangle\leqslant|\langle x,y\rangle|\leqslant\|x\|.\|y\|=\|x\|.$$

Answer 2

This is a special case of the general definition of the norm of an element $x^{\star }$ of the dual space $X^{\star }$ of a Banach space. This is $\left\|x^{\star } \right\|=sup\frac{<x^{\star },x>}{\left\|x \right\|}$. Which is the same (easy to prove) with $sup\left\{<x^{\star },x>, over\, x:\,\left\| x\right\|=1 \right\}$. In $\mathbb{R^{n}}$ the dual space is the same with the original space and we get the result since $<x^{\star },x>$ is (in this case) an inner product!!