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There are many condition to make the best approximation exist.

For example,

If a subspace $A$ of a Hilbert space $H$ is complete and convex, then for any fixed point $x_0 \in H$, there exists an unique best approximation $y_0 \in H$, such that $$d(x_0,y_0)=\inf_{y\in A}d(x_0,y)$$

Now I'm wondering the following extension,

If $A$ is only complete, does the best approximation of $A$ exists (not necessarily unique)?

For a special case, say,

If $A$ is complete and finitely dimensioned, then the best approximation exists.

this is trivial by the fact that $f(y)=d(x_0,y)$ is continuous, that $A$ is closed and that compactness is equivalent to bounded closed in a finitely dimensioned normed space.

Now, how about $A$ is complete and infinitely dimensioned? Maybe some of you can help me.

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  • $\begingroup$ A complete Hilbert space? $\endgroup$
    – BigbearZzz
    Jul 22, 2016 at 6:32
  • $\begingroup$ sorry, I've corrected that. $\endgroup$ Jul 22, 2016 at 6:36

2 Answers 2

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If $e_n$ are an orthonormal basis of $H$, consider the sequence $a_n = (1+1/n) e_n$. The set $A = \{a_n : \; n = 1,2,3,\ldots\}$ is a closed (and thus complete) set, because the distance between any two members of $A$ is greater than $1$. There is no best approximation of $0$: $\|a_n - 0\| = 1 + 1/n \to 1$ as $n \to \infty$, but the infimum is not attained.

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The weakest conditions I know to have existence and unicity are $H$ is an inner product space (not necesserly complete) and $A$ is convex and complete.

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