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Let $V$ be a real normed vector space. Denote by $V_1$ the subset of $V$ of vectors with norm $\le 1$. Consider a positive-semidefinite symmetric bilinear form $\langle \cdot, \cdot\rangle : V\times V\to \Bbb{R}$, not necessarily the one giving the norm.

Is it true, in general, that $$ \sup_{v,w\in V_1} \langle v, w \rangle = \sup_{v\in V_1} \langle v, v \rangle $$ or do we just have an inequality?

Edit: clarified a bit.

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    $\begingroup$ A somewhat obvious obstruction is that if $\langle\cdot,\cdot\rangle$ is negative semi-definite and not constant zero, then the RHS is $0$ and the LHS is positive. $\endgroup$
    – user562983
    Commented Oct 21, 2021 at 9:57

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Since $\{(v,v) | v\in V_1 \}\subset \{(v,w) | v\in V_1 w \in V_1 \} $ you get the one direction. The other direction is Cauchy Schwarz.

Edit: $\langle v,w\rangle\leq \sqrt{\left\|v\right\|\left\|w\right\|}\leq \sqrt{\sup_{v\in V_1}\left\|v\right\|^2}=\sup_{v\in V_1}\sqrt{\left\|v\right\|^2}=\sup_{v\in V_1}\left\|v\right\|$ where I used that the square root is monotonic to get the supremum out of it.

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  • $\begingroup$ There are two issues: 1) the question doesn't state that $\langle \cdot,*\rangle$ is a scalar product, i.e., that it is positive definite (or even that it might be positive semi-definite, which is a minimal condition for Cauchy-Schwarz to hold); 2) even if the blinear form were a scalar product, the question does not say that the the norm is the one induced by that scalar product (in point of fact, it is strongly implied that that isn't the case). $\endgroup$
    – user562983
    Commented Oct 21, 2021 at 12:31
  • $\begingroup$ I might also add that it isn't all that clear why $\sqrt{\lVert v\rVert\lVert w\rVert}\le\sqrt{\sup_{v\in V_1}\lVert v\rVert^2}$ should hold for all $v,w\in V_1$: I could agree that $\sqrt{\lVert v\rVert\lVert w\rVert}\le\sqrt{\sup_{v,w\in V_1}\lVert v\rVert\lVert w\rVert}$ for all $v,w\in V_1$, but your assertion seems to beg the question. $\endgroup$
    – user562983
    Commented Oct 21, 2021 at 12:37
  • $\begingroup$ I've edited the question to add some clarification. If you want, you can edit your answer. $\endgroup$
    – geodude
    Commented Oct 21, 2021 at 13:20

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