4
$\begingroup$

I am interested in the following extremal-type problem.

Let us define $\Psi$ by $$\Psi(x)=\max_{f\in L^2[0,x] \,\,\text{with}\,\,\|f\|_2=1}\Bigg|\int_0^x\int_0^xf(t)f(s)\ln|t-s|dsdt\Bigg|$$ on $(0,\infty)$. Is $\Psi$ injective? In particular can it be represented explicitly?(perhaps $\Psi(x)$ there is some connections with logarithmic capacity of a segment of length $x$)

$\textbf{Edit 1:}$ It is obvious that if $x<y$ then $\Psi(x)\leq\Psi(y)$, what is not apparent is that whether the inequality is strict or not.

$\textbf{Edit 2:}$ Is the maximum attained?Or should it be replaced with supremum?

$\endgroup$
2
  • $\begingroup$ Why is it obvious that $\Psi$ is non decreasing? $\endgroup$ Oct 22, 2015 at 7:46
  • $\begingroup$ @JeanDuchon If $x<y$ then extremal function on $[0,x]$ extends to a function on $[0,y]$.But Im having a hard time to show that this extended function is not external on $L^2[0,y]$ $\endgroup$
    – ISO
    Oct 22, 2015 at 11:57

2 Answers 2

2
$\begingroup$

If you denote by $T_x$ the integral operator with kernel $\ln |s-t|$ on $L^2(0,x)$, then you are asking about $\max_{\|f\|=1}|\langle \overline{f}, T_x f\rangle |$. Now $T_x$ is compact (the kernel is in $L^2$) and self-adjoint, and since the kernel is real-valued, the eigenfunctions can be required to be real as well. So you are really asking about the largest (in absolute value) eigenvalue, and it follows that the $\max$ exists.

$\endgroup$
1
$\begingroup$

Let $x<1$.

Then $\Psi(x)=\max_{f\in L^2[0,x],||f||_2\le1}\Phi_x(f)=\Phi_x(f_x)$ where $\Phi_x(f):=\int_0^x f(t)L_x f(t)\ dt$ with $L_xf(t):=\int_0^x f(s)\ln\frac1{|t-s|}\ ds$, and the maximizer (which I assume is unique, which is more or less obvious in this case) $f_x\ge0$ (otherwise $\Phi_x(|f_x|)\ge\Phi_x(f_x)$ because the kernel $\ln\frac1{|t-s|}$ is positive).

Now, consider $g\in L^2[0,y]$, $x<y\le1$, $g=f_x$ on $[0,x]$ and $g=0$ on $(x,y]$, clearly $||g||_{L^2[0,y]}=||f_x||_{L^2[0,x]}$ and $\Psi(y)\ge \Phi_y(g)=\Phi_x(f_x)=\Psi(y)$.

But $g$ cannot be the maximizer $f_y$ of $\Phi_y$ because then it would be an eigenfunction of $L_y$ as a (compact) operator on $L^2[0,y]$, but $L_yg(t)$ -- unlike $g(t)$ -- cannot vanish for $x<t<y$ (it is positive there, as $g$ is nonnegative and not identically zero on $0<s<x$, and $\ln\frac1{|t-s|}>0\ \forall s\in[0,x]$).

The more general case $x>0$ seems harder, as the kernel is no longer positive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.