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Minimum principle is following: Let $M$ be a closed convex nonempty subset of Hilbert space. Then there exists $x\in M$ which have a minimum norm.

Assume that $M$ is not convex subset. What is a counterexample, when there have not the element with minimum norm in space $\ell^{2}$?

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I think you mean subset, not subspace... In fact subspaces are always convex. – Pacciu Dec 18 '11 at 12:57
I'm guessing you still want $M$ to be closed, but will you please clarify so we don't have to guess? – Jonas Meyer Dec 18 '11 at 17:15

3 Answers 3

In infinite dimensional $\ell_2$, take $M=\{ \,{n+1\over n} e_n : n=1, 2,3, \ldots\,\}$, where $e_n$ is the standard $n^{\rm th}$ unit vector defined by: $e_n(m)=\cases{1,& m=n \cr 0, &\text{ otherwise}}$.

$M$ is closed, since the distance between any two of its elements is greater than $\sqrt 2$ (and thus the only convergent sequences from $M$ are those that are eventually constant). $M$ clearly is non-empty and has no element of minimal norm.

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In the Hilbert space $\ell^{2}=\{x=(c_1,c_2,...)|\displaystyle\sum_{i=1}^\infty|c_i|^2<\infty\}$ with inner product $\langle x,y\rangle=\displaystyle\sum_{i=1}^\infty c_i\overline{d_i}$, we can consider $M=\ell^{2}-\{0\}$, which is not convex: since for $x\in M$, $\displaystyle\frac{x+(-x)}{2}=0\notin M$. On the other hand, there does not exist element of minimum norm, for $x_n=(1/n,0,0,...)\in M$ and $\|x_n\|=1/n$ which tends to zero as $n\rightarrow\infty$.

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Take $H=\mathbb{R}^2$ and let $M$ be the subset of $\mathbb{R}^2$ built by adding to the circumference $C_0:=\{ (x,y)\in \mathbb{R}^2:\ x^2+y^2=1\}$ the needles: $$C_n:=\left\{(x,y)\in \mathbb{R}^2:\ x^2+y^2\leq 1,\ y=\frac{1}{n}\ x \text{ and } x\in [1/n,1]\right\}$$ i.e. $M=\bigcup_{n=0}^\infty C_n$. Then $X$ is neither closed, nor convex, and it has no element with minimun norm.

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