This is actually Bass's Real Analysis Problem 18.14. It says let $A$ be the set of real-valued continuous functions on $[0,1]$ such that for $f \in A$,
$\int_0^\frac{1}{2} f(x)dx -\int_\frac{1}{2}^1 f(x)dx= 1$.
I am asked to prove that $A$ is closed and convex subset of $C([0,1])$, and there does not exist $f \in A$ such that $\lVert f \rVert = inf \lVert g \rVert$, where the infimum is taken among any $g \in A$. The norm is the $sup$ norm.
I can prove that $A$ is closed and convex, but I have completely not idea about proving that the infimum norm is not attained. Do we need to compute the infimum? Does it have anything to do with convexity? Any hint is appreciated, thanks!
The point is that to minimize $\|f\|$ you'd like $f(x)$ to be equal to $+1$ when $x < 1/2$ and $-1$ when $x > 1/2$, but of course that would make your function discontinuous. Still, you can get arbitrarily close...