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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: http://math.stackexchange.com/questions/80604/why-in-uniformly-convex-banach-space-every-non-empty-closed-convex-subset-cont2011-12-29
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    See also http://mathoverflow.net/questions/81979/projection-exists-uniformly-convex2011-12-29
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    Thank you very much David Mitra.2011-12-29
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    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 http://mathprelims.wordpress.com/2008/07/17/rieszs-lemma/2011-12-29
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    @Thomas I am afraid, sorry. How does lemma of Riesz you have linked have connection with the projection I am talking about.2011-12-29
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    @David Mitra: What I understand from the links you have mentioned is the following. The necessary condition for the result I quoted to be true is that the Banach space has to be reflexive. The sufficient condition is that it is uniformly convex Banach space. Am I right?2011-12-29
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    Yes. This may also help: http://math.stackexchange.com/questions/80773/a-vector-without-minimum-norm-in-a-banach-space2011-12-29
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    So, now I'm wondering if there is an example of a reflexive space without the minimum norm property...2011-12-29
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    @David: doesn't existence of a point with minimal norm in a reflexive space follow from weak lower semi-continuity of the norm, the fact that norm-closed convex sets are weakly closed and weak sequential compactness of the closed unit ball? Uniqueness can fail of course, already in finite dimensions: equip $\mathbb{R}^2$ with the maximum norm $\|(x,y)\| = \max{\{|x|,|y|\}}$ and consider the closed ball of radius $1$ around $(2,0)$. The points $(1,y) \in B$ with $-1 \leq y \leq 1$ are all at the same distance $1$ of $(0,0)$ and $d((0,0),B) = 1$.2011-12-29
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    @Ashok The vector the existence of which is assured in that lemma has some properties which resemble somewhat those of the $y_0$ you are reffering to - such a $y_0$ need not exist in general. It depends on what you are up to whether you can use it or not.2011-12-29

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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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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.2011-12-30
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    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.2011-12-30
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    Thanks Jonas. Nice example.2011-12-31