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

Question

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 $y\in Y$, $\lVert f-y \rVert > dist(f,Y)$?

Answer 1

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.

Answer 2

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.