# A question regarding least norm element of a closed, convex subset of a Hilbert space

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?

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

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.
As pointed out in the comments, 'letting $$t = 1$$' is imprecise and should be replaced with 'let $$t \uparrow 1$$'.
• 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). Commented Mar 25, 2021 at 18:34
• 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 $t \nearrow 1$"? Commented Feb 23, 2023 at 9:42
• @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. Commented Feb 23, 2023 at 9:49