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?
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.