$\max F(u)=\min G(u)$?

Question

Let $a>0$ be a fixed real number and consider the semi-circle $y(x)=\sqrt{(a/2)^2-(x-a/2)^2}$ for $x\in [0,a]$.

Let $Y=\{u\in C([0,a]):\ u(0)=u(a)=0,\ u\geq 0,\ \ell(u)=2a\}$, where $\ell(u)$ denotes the length of the graph of the curve.

Consider the two functional $F: Y\to [0,\infty)$ and $G:Y\to [0,\infty)$ defined by $$F(u)=\int_0^au(x)dx,\ G(u)=\|u-y\|_{C([0,a])}$$

Is true that the maximum of $F$ is the minimum of $G$?

Thank you

@Mercy, for the first question we have that $u\geq 0$, hence $F(u)\geq 0$. For the second question, I dont know the answer. Maybe it is better take it $C^1$ by parts. — Tomás, Jul 10 '13 at 15:50
Ok, sorry I didn't see the condition $u \ge 0$. Have you tried to solve your problem on $W^{1,2}_0(0,a)$? maybe it will give a hint! — HorizonsMaths, Jul 10 '13 at 15:52

score 0 · Answer 1 · answered Jul 03 '13 at 11:50

If I see it correctly, both functionals do not attain their extremals.

For $F$, the extremal value should be attained by $y(x) + (2-\pi/2)\,a$, which is not feasible, but could be approximated with functions in $Y$.

The infimum of $G$ should be zero, just approximate $y$ by oscillating functions. Also, this extremum is not attained.

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