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Does there exist a Banach space $X$ such that in $X$ exists a closed subspace $Y$ and a $f \in X \backslash Y$ such that for any $y\in Y$, $\lVert f-y \rVert > dist(f,Y)$?

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  • $\begingroup$ @JpMcCarthy I want "strictly larger than", but the definition of $dist(f,Y)$ is the infimum of $\lVert f-y \rVert$, which allows "equality". $\endgroup$ – SHBaoS Feb 27 '17 at 13:11
  • $\begingroup$ @JpMcCarthy So you mean if $Y$ is closed, then the equality always exists? $\endgroup$ – SHBaoS Feb 27 '17 at 13:22
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Consider $X = C([0,1])$ with the closed subspace $Y = \{f \in X : f(0) = 0\}$ and $f \equiv 1$. Now, the trick is to equip $X$ with the norm $$\| g \| = \sup_{t \in [0,1]} |g(t)| + \int_0^1 |g(t)| \, \mathrm{d}t.$$

Then, it is quite easy to check that this satisfies your assumptions.

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EDIT: this only works in finite dimension.

I don't think so.

Consider the function: $D_{f} : X \rightarrow \mathbb{R}$ , $y \mapsto \|f-y\|$. This function is continuous.

Note that if the image of $Y$ by $D_{f}$ is closed, your hypothesis is not possible. Now, we would like $Y$ to be compact to conclude (because then the image of $Y$ is compact, closed in particular).

Note that $dist(f,Y)$ is equal to $dist(f,A)$ where $A$ is the intersection of $Y$ and the closed ball centered at $f$ With radius $dist(f,Y)+1$. Now $A$ is compact, and we conclude.

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  • $\begingroup$ Sorry, the closed ball is compact assumption only works in finite dimension, so this proof is wrong in infinite dimension. $\endgroup$ – Matias Puig Feb 27 '17 at 13:21
  • $\begingroup$ Yeah, that's what I got stuck in your proof. $\endgroup$ – SHBaoS Feb 27 '17 at 13:23

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