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Suppose $(V, (\cdot, \cdot))$ is a Hilbert space, $U$ is a nonempty closed convex subset of $V$, and $g \in U$ is the unique element of $U$ with smallest norm. Prove that $\Re(g, h)\ge \|g\|^2 \ \forall h\in U$.

I was thinking about considering $f=\frac{1}{2}(g+h)\in U$. So $\|f\|^2 \ge \|g\|^2$. Unable to get the desired inequality. Any help?

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  • $\begingroup$ Using your $f$, the proof can be completed as in the answer with my addendum...with $||h|| \gt ||g||$ then $2.1/2 \Re(g, h) = \Re(g, h) \ge (3/2)\|g\|^2 - (1/2)\|h\|^2 \ge ||g||^2$. $\endgroup$ Commented Mar 25, 2021 at 18:47
  • $\begingroup$ Where is this question from? Is it perhaps an exercise in a book? $\endgroup$ Commented Feb 23, 2023 at 9:55

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As you noticed, for $t \in [0,1]$, using the convexity of $U$, $$ \|tg + (1 - t)h\|^2 \geq \|g\|^2 $$ Which, when expanding the inner product and dividing by $(1 - t)$ gives $$ 2t\Re(g, h) \geq (1 +t)\|g\|^2 - (1 - t)\|h\|^2 $$ Letting $t = 1$ gives the result.

Edit

As pointed out in the comments, 'letting $t = 1$' is imprecise and should be replaced with 'let $t \uparrow 1$'.

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  • $\begingroup$ Nice construction. In fact, with $t \le 1$ and $||h|| \gt ||g||$ then $2t\Re(g, h) \ge (1 +t)\|g\|^2 - (1 - t)\|g\|^2 = 2t||g||^2$. So the inequality holds for all $t$ (as it would have to). $\endgroup$ Commented Mar 25, 2021 at 18:34
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    $\begingroup$ When you divide by $1 - t$, then the following results only hold for $t \in [0, 1)$. So should be substitute "Letting $t = 1$" by "For $t \nearrow 1$"? $\endgroup$ Commented Feb 23, 2023 at 9:42
  • $\begingroup$ @TomCollinge. I disagree. Expanding the inner product gives $t^2 \| g \|^2 + 2 t (1 - t) \Re(\langle g, h \rangle) + (1 - t)^2 \| h \|^2 \ge \| g \|^2$, and dividing by $(1 - t)$ yields $$2 t (1 - t) \Re(\langle g, h \rangle) \ge (1 + t) \| g \|^2 - (1 - t) \| h \|^2 \qquad \forall t \in [0, 1).$$ But because $\| g \| \le \| h \|$, the right hand side is $\le 2 t | g |_2^2$ and not "$\ge$" because $- (1 - t) < 0$, so the inequality is reversed. $\endgroup$ Commented Feb 23, 2023 at 9:49

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