Counterexamples to a theorem in Rudin's book on elements of smallest norm in a closed convex sets in a Hilbert space The starting point of this question is the following theorem.

Theorem (4.10 in Rudin's) Every nonempty, closed, convex set $E$ in a
Hilbert space $H$ contains a unique element of smallest norm.

Consider $(C([0,1], \mathbb{K}), \Vert \cdot \Vert_\infty)$. Let
$$A = \left\{f \in C([0,1]) \text{ such that } \int_{0}^\frac{1}{2} f(t)\, dt - \int_{\frac{1}{2}}^{1} f(t)\, dt= 1\right\}$$
Consider $L^1([0,1])$. Let
$$B = \left\{f \in L^1([0,1]) \text{ such that } \int_{0}^1 f(t)\, dt = 1\right\}$$
$A$ and $B$ are nonempty.

*

*How can I prove that $A$ is convex and closed but has no element of smallest norm?


*How can I prove that $B$ is convex and closed but has infinitely many elements of smallest norm?
 A: Hint: In the case of $A$, any function in $A$ trying to minimize its norm must try very hard to look like
$$\begin{cases}\frac12&0\le t<\frac12,\\-\frac12&\frac12\le t\le1.\end{cases}$$
However, that is not continuous.
In other words, show that the infimum of norms of elements of $A$ is $\frac12$, but that infimum is not achieved.
For $B$, use the triangle inequality to show that $\|f\|\ge1$ for all $f\in B$. But if $f\ge0$, that becomes an equality. There will be lots of nonnegative functions with $\int_0^1 f(t)\,dt=1$!
In both cases, proving convexity and closedness should be straightforward.
A: Convexity means that if $f$ and $g$ are element of your sets then $tf+(1-t)g$ is also in your set for $t\in[0;1]$. Just apply this definition and you'll seee that $A$ and $B$ are convex.
For closedness you can write each set as $\phi^{-1}(\{0\})$ with a well choosen, continuous $\phi$.
Now to show that $A$ has no element of smalest norm : take some $f$ in $A$ and find $g\in A$ such that $\|f\|>\|g\|$. 
For $B$ one have $f\in B \Rightarrow \|f\|_1\geq 1$, can you find an infinity of distincts functions $f$ such that $\int_0^1 fd \lambda = 1$ and $\|f\|_1=1$ ?
