Let $C$ denote the set of continuos functions on $[0,1]$ with the supremum norm.
$M\subset C$ such that
$$\displaystyle\int_{0}^{1/2}f(t)\, dt-\int_{1/2}^{1}f(t)\, dt=1,\; \forall f\in M$$
Show that:
i) $M$ is a closed convex subset of $C$ which contains no element of minimal norm.
ii) $\inf\{\|f\|_\infty:f\in M\}=1/2$
Any suggestion for ii) and "$\textbf{$\ldots$ which contains no element of minimal norm.}$".
Thank you all
On
angryavian's answer should also point you into the right direction for your second question on why no norm-minimal element exists: For any $f \in C$ there holds $$\int_0^{1/2} f(x) \,\text{d}x \leq \frac 1 2 \| f \|_\infty \quad\text{and}\quad - \int_{1/2}^1 f(x) \,\text{d}x \leq \frac 1 2 \| f \|_\infty,$$ so a norm-minimal $f$, i.e. $\| f \|_\infty = 1/2$ according to (ii), can satisfy $f \in M$ just if both inequalities above are in fact equalities. But how does $f$ have to look like on $[0,1/2]$ and $[1/2, 1]$ then?
In my comment above, I noted $$\int_0^{1/2} f(t) \mathop{dt} - \int_{1/2}^1 f(t) \mathop{dt} \le \|f\|_\infty.$$
I have a hunch that the definition of $M$ should be $$\int_0^{1/2} f(t) \mathop{dt} - \int_{1/2}^1 f(t) \mathop{dt}=1/2$$ in order for part ii) to work. [Or more generally, the number in the definition of $M$ and the number in part ii) should be the same.]
Indeed, if $\|f\|_\infty<1/2$, then $f \notin M$ by our work above, so $\inf\{\|f\|_\infty: f \in M\} \ge 1/2$.
To show the reverse inequality $\inf\{\|f\|_\infty: f \in M\} \le 1/2$, it suffices to show that given any $\epsilon>0$, there exists $f \in M$ such that $\|f\|_\infty \le 1/2+\epsilon$. Loosely speaking, such an $f$ can be constructed by taking the step function $$\tilde{f}(x)=\begin{cases}1/2+\epsilon & x \in [0,1/2)\\ -1/2-\epsilon & x \in (1/2,1]\end{cases}$$ and "smoothing out" the discontinuity of $\tilde{f}$ at $1/2$ by making a steep decrease in such a way so that $f$ is continuous and satisfies the condition to be in $M$.