Assume $f\in L^2[0,1]$ and let $g(x)=\int_0^1f(y)\ln|x-y|dy$. Is it true that $g\in C^\infty(0,1)$? Is it true that $g$ is analytic in $(0,1)$? Can you refer me to a right reference to look up such integrals and their properties.
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2 Answers
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No, if $f$ is the characteristic function of $[0,\frac12]$, $g$ will not be smooth at $\frac12$ by a simple explicit calculation.
We have $g=f*h$, where $h(x)=\ln|x|$ is in $L^1$ and has singular support ${0}$. Thus $g$ is $C^\infty$ whereever $f$ is. See e.g. Hörmander, The Analysis of Partial Differential Operators II, chap 16 (which perhaps should be considered overkill for this).
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The answer is no to the both questions. Let us define $$f(y):=\begin{cases} - \frac 1 2 ,\,\,& y \in [0,1/2];\\ \,\,\,\,\,\frac 1 2 ,\,\, &y \in (1/2,1]. \end{cases}$$ Then a straightforward calculation with Maple (see the output here) produces $$ g(x):=$$ $$ \begin{cases}-1/2\ln(x)x+1/2\ln(2)-1/2\ln(1-2x)-x\ln(2)+x\ln(1-2x)+\\1/2\ln(1-x)-1/2x\ln(1-x), &x \in [0,1/2];\\ -1/2x\ln(x)-\ln(2)x+1/2\ln(2)+x\ln(2x-1)-\\1/2\ln(2x-1)-1/2x\ln(1-x)+1/2\ln(1-x), &x \in (1/2,1]. \end{cases}$$ In particular, $\lim_{x \uparrow \frac 1 2} g'(x)= -\infty.$
