Supremum over an inner product space $V$: $\sup\limits_{v\in V}(\langle v',v\rangle-\frac 12 \langle v,v\rangle)=\frac 12\langle v',v'\rangle$

Question

Let $(V,\langle\cdot,\cdot\rangle)$ be an inner product space and take $v'\in V$. Then I want to show the identity $$ \sup_{v\in V}(\langle v',v\rangle-\frac{1}{2}\langle v,v\rangle)=\frac 12\langle v',v'\rangle. $$ We can rewrite what is in the supremum as $$ \langle v',v\rangle-\frac 12\langle v,v\rangle=\langle v',v\rangle-\langle \frac 12 v,v\rangle=\langle v'-\frac 12v,v\rangle $$ How do I see that the supremum is now attained at $v=v'$?

Answer 1

If $v\in V$, then\begin{align}\frac12\langle v',v'\rangle-\left(\langle v',v\rangle-\frac12\langle v,v\rangle\right)&=\frac12\bigl(\langle v',v'\rangle+\langle v,v\rangle\bigr)-\langle v',v\rangle\\&=\frac12\langle v-v',v-v'\rangle\\&\geqslant0\end{align}and therefore$$\langle v',v\rangle-\frac12\langle v,v\rangle\leqslant\frac12\langle v',v'\rangle.$$