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I think no. And I am looking for examples. I would like a sequence $y_n$ in $Y$ such that $||y_n-x||\rightarrow d(x,Y)$ while $y_n$ do not converge.

Can anyone give a proof or an counterexample to this question?

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  • $\begingroup$ It's difficult to give a counterexample to a question. Make a statement and I'll try to find a counterexample. $\endgroup$ – Fly by Night Feb 6 '13 at 16:14
  • $\begingroup$ The statement is "given a point $x$ and a closed subspace $Y$ of a normed space, the distance from $x$ to $Y$ is achieved by some $y\in Y$" $\endgroup$ – Spook Feb 6 '13 at 16:16
  • $\begingroup$ I think that this is true for all closed $Y$ in a $T_4$ space. $\endgroup$ – Fly by Night Feb 6 '13 at 17:33
  • $\begingroup$ @DavidMitra Can you think of a counterexample when $X=l_\infty$ and $Y=c_0$? I tried to build one but I was not successful. One cannot use the construction in your link, as $c_0$ cannot be the kernel of any linear functional on $l_\infty$. $\endgroup$ – Theo Feb 6 '13 at 20:20
  • $\begingroup$ @DavidMitraThank you for the link. $\endgroup$ – Spook Feb 6 '13 at 21:23
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This is a slight adaptation of a fairly standard example.

Let $\phi: C[0,1]\to \mathbb{R}$ be given by $\phi(f)=\int_0^{\frac{1}{2}} f(t)dt - \int_{\frac{1}{2}}^1 f(t)dt$. Let $Y_\alpha = \phi^{-1}\{\alpha\}$. Since $\phi$ is continuous, $Y_\alpha$ is closed for any $\alpha$.

Now let $\hat{f}(t) = 4t$ and notice that $\phi(\hat{f}) = -1$ (in fact, any $\hat{f}$ such that $\phi(\hat{f}) = -1$ will do). Then $$\inf_{f \in Y_0} \|\hat{f}-f\| = \inf \{ \|g\|\, | \, g+\hat{f} \in Y_0 \} = \inf \{ \|g\|\, | \, \phi(g) =1 \} = \inf_{g \in Y_1} \|g\|$$ It is clear that $g_n$ is an infimizing sequence for the latter problem iff $g_n+\hat{f}$ is an infimizing sequence for the initial problem.

It is well known that $Y_1$ has no element of minimum norm, consequently there is no $f \in Y_0$ that mnimizes $\|f-\hat{f}\|$.

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  • $\begingroup$ We seem to have come to different conclusions. Have a look at my "proof" and see if you can find an error. $\endgroup$ – Fly by Night Feb 6 '13 at 17:42
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There cannot be an example in $c_0 \subset \ell^\infty$. In fact $c_0$ is a closed ideal in the $C^*$-algebra $\ell^\infty$. For such pairs distance to ideal always attained be an element in the ideal.

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  • $\begingroup$ It helps reability to format answers using Mathjax $\endgroup$ – Moo Oct 5 '16 at 3:59

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