Is there an example of a closed convex non-empty subset $A$ of an inner product space $X$ that does not contain an element of minimal norm?
I have absolutely no clue how to come up with an example for which the theorem does not hold when $X$ is not a Hilbert space. This is a practice test question, so I do apologize for the lack of work.
Any hints would be much appreciated.
Consider the space of continuous functions from $[0, 1]$ to $\mathbb{R}$ with the usual $L^2$ inner product. Let $A$ be the set of functions $f$ in this space with $f(x) = 1$ when $0 \le x \le 1/2$. It is simple to verify that $A$ is non-empty and convex. (In fact, it is an affine subspace.) Furthermore, $A$ is closed; whenever a function $f$ is not identically $1$ on $[0, 1/2]$, there must be an open subset of $[0, 1/2]$ on which the values of $f$ are bounded away from $1$, which guarantees that $f$ is bounded away from $A$ with the $L^2$ norm.
However, while no element of $A$ has $L^2$-norm $1/2$, there are elements of $A$ with norms arbitrarily close to $1/2$; $A$ has no element of minimal norm.
(Indeed, $C[0,1]$ is not complete; it lives as a dense but proper subspace of the Hilbert space $L^2[0, 1]$.)