# a variational problem involving $L^p$ norm

Is there any way to prove that there are only finitely many maximizers to the following variational problem:

$$\max\left\{||f||_{L^p[0,1]}-\int_0^1(f'(t))^2dt\right\}$$ over all functions $f$ which are absolutely continuous and $f(0)=0$? We can choose a suitable $p$ as needed. Any suggestion is appreciated...

(note that the above quantity is bounded above, and the existence of a maximizer can be also proved)

The functional you are maximizing is weird: it lacks homogeneity. So the first thing to look at is what happens when $f$ is multiplied by a positive constant $\lambda$. Write $$A=\|f\|_{L^p[0,1]},\quad B = \int_0^1 (f'(t))^2\,dt$$ For $\lambda f$, the functional is $\lambda A-\lambda^2 B$. This is maximized when $\lambda=A/(2B)$, the maximum being $A^2/(2B)$.
Conclusion: maximizers of your functional are appropriately scaled maximizers of the functional $$\Psi(f)=\frac{\|f\|_{L^p[0,1]}^2}{\int_0^1 (f'(t))^2\,dt}$$
You say that $p$ can be chosen... looking at $\Psi$, I want $p=2$. Then the matter is essentially reduced to a form of Wirtinger's inequality. Indeed, extend $f$ to the interval $[0,2]$ by $f(x)=f(2-x)$ for $1< x\le 2$. The extended function is absolutely continuous and vanishes at both endpoints of $[0,2]$. Wirtinger's inequality (second form) yields $$\int_0^2 f^2 \le \frac{4}{\pi^2} \int_0^2 (f')^2$$ with equality only for multiples of $\sin (\pi x/2)$ By symmetry of our $f$, we have $$\int_0^1 f^2 \le \frac{4}{\pi^2} \int_0^1 (f')^2$$
Returning to the original functional, conclude: with $p=2$, your functional has maximum $2/\pi^2$, attained only by $$f(x)=\pm \frac{2}{\pi^2} \sin \frac{\pi x}{2}$$
• Thanks a lot for the suggestions! Yes, the case $p=2$ seems to be solvable in the way suggested, which is relatively easy. Any hint for more general $p$? Dec 24, 2013 at 19:15
• @user108871 I don't really know. Since $\Psi$ is homogeneous, you can impose the constraint $\|f\|_p=1$ and look for constrained minimum of $\int (f')^2$ using Lagrange multiplier. To find $f$, you'll need to solve some ugly ODE. Perhaps by writing that ugly ODE you'll see a reason why there are finitely many solutions? Dec 24, 2013 at 19:32