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Let $H$ be a Hilbert space and $C$ be a non empty closed convex subset of $H$ and let $x\notin C$. We know that there exists a unique $y_0$ in $C$ such that $\|x-y_0\|=\inf_{y\in C}\|x-y\|$. Call $y_0$, the projection of $x$ onto $C$.

The proof of this result heavily depends on the parallelogram law which holds only in Hilbert spaces. Is the result true for just normed spaces also? Have people already studied about this?

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It's not generally true in Banach spaces. But:… – David Mitra Dec 29 '11 at 14:07
See also… – David Mitra Dec 29 '11 at 14:14
Thank you very much David Mitra. – Ashok Dec 29 '11 at 14:31
There is something which may sometimes serve as substitute, when I learned this it was given the name 'lemma of almost orthogonal element', but I guess the name 'lemma of Riesz' is more common. You can find it, for example, at – user20266 Dec 29 '11 at 14:42
@Thomas I am afraid, sorry. How does lemma of Riesz you have linked have connection with the projection I am talking about. – Ashok Dec 29 '11 at 14:51
up vote 3 down vote accepted

For Banach spaces, see the links (the MO link in particular) in my comments for positive results: A uniformly convex Banach space $X$ has your property and a Banach space with your property is reflexive. In the MO post, a link to a space with your property that is not uniformly convex is given.

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I accept David's answer though Jonas post is also equally informative. – Ashok Dec 30 '11 at 7:25

Because you ask whether it is true for just normed spaces, with no mention of completeness, I will mention that it is false for every incomplete inner product space. Suppose $X$ is an incomplete inner product space, and let $\overline X$ be its (Hilbert space) completion. Let $y$ be an element of $\overline X\setminus X$, and let $C=\{x\in X:\|x-y\|\leq \frac{1}{2}\|y\|\}$. Then $C\subset X$ is closed, convex, and nonempty, but has no element of smallest norm.

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Using the parallelogram law of $\bar{X}$, we can show that any sequence $x_n\in C$ such that $\|x_n\|\to d=\inf \{\|x\|:x\in C\}$ is Cauchy. I don't know how to show $C$ has no element of smallest norm. – Ashok Dec 30 '11 at 7:03
Ashok: Let $\overline{C}=\{x\in\overline{X}:\|x-y\|\leq \frac{1}{2}\|y\|\}$. Then $\overline{C}$ has a unique element of smallest norm, and that element is $\frac{1}{2}y$, which is not in $C$. So $C$ has no element of norm $\frac{1}{2}\|y\|$, but by density of $X$ in $\overline X$, $C$ has elements of norm arbitrarily close to $\frac{1}{2}\|y\|$, and therefore it has no element of smallest norm. – Jonas Meyer Dec 30 '11 at 17:42
Thanks Jonas. Nice example. – Ashok Dec 31 '11 at 11:26

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