Given an inner product space $(H, \langle .,.\rangle)$ and a vector $x \in H$, prove that $$\lVert x \rVert = \sup_{\lVert y \rVert = 1} | \langle x,y \rangle|.$$
My attempt:
Begin with the Cauchy-Schwarz inequality $$|\langle x,y \rangle| \leq \lVert x \rVert \ \lVert y \rVert$$ Take $\lVert y \rVert = 1$, then $$ \lVert x \rVert \geq |\langle x,y \rangle |$$ Now this last part is where I struggle. I don't see how taking the supremum of the RHS guarantees equality seeing as the supremum is a least upper bound and thus does not have to be attained for any $y \in H$. In my mind taking the max of the RHS instead would be more natural. Thoughts?
Best Answer
For $x=0$ there is nothing to prove. For $x \ne 0$ take $y = \frac{x}{\|x\|}$ which is possible, that proves that equality is attainable. Indeed, $$ \left|\left\langle x, \frac{x}{\|x\|} \right\rangle\right| = \frac{1}{\|x\|}|\langle x, x \rangle| = \frac{\|x\|^2}{\|x\|} = \|x\| $$ Of course you still have $\|y\| = 1$.