An extremal type problem on segments 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?
 A: 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.
A: 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.
