# Proving Inner Product Space Induced Norm Property

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?

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 $$\|y\| = 1$$.