If $f\in L^1(\mathbb{T})$ and $g\in L^\infty(\mathbb{T})$ where $\mathbb{T}$ is the circle, such that $\hat{f}\in L^{p}(\mathbb{Z})$ for some $1\leq p<\infty$, do we have that $f*g$ is differentiable almost everywhere?
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2$\begingroup$ Nope. The classical Weierstrass example of nowhere differentiable function is $h(z)=\sum_{k\ge 1} 4^{-k}z^{100^k}$ and that is $f*g$ with $f=g=\sum_{k\ge 1} 2^{-k}z^{100^k}$, both Holder continuous and with coefficients in all $\ell^p$, $p>0$. $\endgroup$– fedjaCommented Oct 31, 2013 at 4:00
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