Prove that $K,$ a closed convex subset of $\left(C[0,1],\|\cdot \|_{\infty}\right)$ contains no element of minimum norm.

Question

Let $E=\left(C[0,1],\|\cdot \|_{\infty}\right)$ and $K$ be a closed convex subset of $E$ which consists the set of all $f\in E$ such that $$\int^{1/2}_{0}f(s)ds-\int^{1}_{1/2}f(s)ds=1.$$ Prove that $K$ contains no element of minimum norm.

My trial

Suppose for contradiction, that it contains an element of minimum norm. This element can either unique or not. Assume that it is unique. Then, there exists $f_0\in K$ such that $$\|f_0\|=\inf\limits_{f\in K}\|f\|.$$ By characterization of $\inf$, there exists $(f_n)_n\subseteq K$ such that $\|f_n\|\to \|f_0\|\in K.$ However, $(f_n)_n\subseteq K$ implies $(f_n)_n\subseteq \left(C[0,1],\|\cdot \|_{\infty}\right)$ and

$$\int^{1/2}_{0}f_n(s)ds-\int^{1}_{1/2}f_n(s)ds=1.$$

From here, I don't see how to arrive at a contradiction. Can anyone further help?

Answer 1

By definition of $K$ we see that $1 \leq ||f||$ for all $f \in K$. Let $f_n(x)= 1$ for $x \leq \frac 1 2$, $-2nx+1+n$ for $\frac 1 2 \leq x \leq \frac 1 2 +\frac 1 n$ and $-1$ for $ x\geq \frac 1 2 +\frac 1 n$> Let $c_n=\int_0^{1/2}f_n(x)\, dx-\int_{1/2} ^{1}f_n(x)\, dx$ and $g_n(x) =\frac {f_n(x)} {c_n}$. You can easily verify that $c _n \to 1$ and $g_n \in K$ for all $n$. Also $\|f_n\| =1$ for all $n$. Putting these together we see that $\inf \{\|f\|:f \in K\}=1$.

Now suppose there is an element of norm $1$ in $K$. Then $1=\int_0^{1/2}f(x)-\int_{1/2} ^{1}f(x) \leq \int_0^{1/2}1+\int_{1/2} ^{1}1=1$. Thus equality must hold throughout. This implies that $f(x)=1$ for $x <\frac 1 2$ and $f(x)=-1$ for $x >\frac 1 2$ . But this contradicts the fact that $f$ is continuous.

Answer 2

Show that $\|f\|>1$ for all $f\in K$, but for every $c>1$, you can find $f\in K$ with $\|f\|=c$. (The idea behind this is that there is an "obvious" minimizer if you drop the demand for continuity)