a convex closed set of a Banach space contains no closest point

Question

I need to prove that:

For the Banach space $E=C([0,1])$ with $||.||_{\infty}$ and the closed convex set $A=\{f \in E : f(1)=0 \}$, can we find an element $f \in E$ such that $A$ contains no closest point to $f$.

Answer 1

I think this is not true.

Consider the functional $\delta_1 \colon E \to \mathbb R $ defined via $\delta_1(f)=f(1)$ (the dirac functional). Then $ ||δ_1||=1$. Notice that $A=\ker \delta_1$. Thus for all $ f \in E$ $$d(f,A) = d(f, \ker \delta_1) =\frac{|\delta_1(f)|}{||\delta_1|| } = |f(1)|$$ (see here). Let $ f \in E$ be given. If we pick $g \in E$ such that $g(x)=f(x)-f(1)$ we have that $g \in A $. Then $$||f-g||= \sup | f(x)-f(x)+f(1)| =|f(1)| = d(f,A)$$ and so the distance is always attained.

Edit: If $X$ is a nonreflexive Banach space (like $C([0,1]))$, then by James's theorem, there exists $x^* \in X^*$ that is not norm attaining (that is, there is no $x \in B_x$ such that $ ||x^*||=|x^*(x)|$. It is not hard to see that the subspace $ V=\ker x^*$ doesn't have the best approximation property. To see this, note that for all $x \in X$ $$d(x,V)= \frac{|x^*(x)|}{||x^*||} .$$ Arguing by contradiction, suppose that there exists $x \in X$ such that $ d(x,V)=||x-v||$ for some $v \in V$. Then, $ |x^*(x-v)| = |x^*(x)|= ||x^*|| \ || x-v||$. In other words, $$ ||x^*|| = |x^* \left ( \frac{ x-v}{||x-v||}\right )|$$ which is impossible since $x^*$ is not norm attaining (note that $\frac{ x-v}{||x-v||} \in B_X$)