Let $0<a<b<1$ and $f\in L^2[0,a]$ be a real-valued function with $\int_0^af^2=1.$ Define its logarithmic energy by $$\mathcal{E}_a(f)=\int_0^a\int_0^af(x)f(y)\log\frac{1}{|x-y|}dxdy$$ Q. Does there exist $g\in L^2[0,b]$ with $\int_0^bg^2=1,$ such that $$\mathcal{E}_a(f)<\mathcal{E}_b(g),$$ where $\mathcal{E}_b(g)=\int_0^b\int_0^bg(x)g(y)\log\frac{1}{|x-y|}dxdy$?
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1$\begingroup$ There are smooth functions on $[0,b]$ with unit $L^2$ norm and arbitrarily large energy, so what is the question? $\endgroup$– Alexandre EremenkoApr 25, 2016 at 13:37
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1$\begingroup$ @AlexandreEremenko I disagree. $\sup\int_0^b\int_0^bf(x)f(y)\log\frac{1}{|x-y|}$ taken over all $f\in L^2[0,b]$ with $\|f\|_2=1$ is the operator norm of a Hilbert-Scmhidt operator. So it is indeed finite. $\endgroup$– BigMApr 25, 2016 at 14:04
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$\begingroup$ But the norm of this operator will be very big if $b$ is small. Your $b$ is not related to $a$, is it? $\endgroup$– Alexandre EremenkoApr 25, 2016 at 21:31
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$\begingroup$ @AlexandreEremenko: No, it will not be large, it's bounded by the HS norm, which equals (the square root of) the $L^2$ of the kernel. $\endgroup$– Christian RemlingApr 25, 2016 at 21:47
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1$\begingroup$ @BigM: Fine. Now estimate the same from below and you will answer your question. $\endgroup$– Alexandre EremenkoApr 26, 2016 at 1:31
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