# Differentiability of convolution

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## Main Question or Discussion Point

If f and g are continuous functions on the right half-line, [0,∞], then f✶g, the convolution of f and g, is defined by
f✶g(x) = ∫[0,x] f(t)g(x-t)dt.
I would like to know if f✶g is a differentiable function of x.
If, for example, g(t) = 1 for t ≥ 0 then f✶g(x) = ∫[0,x]f(t)dt has a derivative equal to f(x). But what about in general?

If either $f$ or $g$ is differentiable, then so is the convolution.

If either $f$ or $g$ is differentiable, then so is the convolution.
Great, thanks. I now see that. But what happens if both f and g are nowhere differentiable?

Samy_A
Homework Helper
The following paper from 1951 constructs a continuous function x, for which the convolution x*x is only differentiable in 0.

Jarník, V. "Sur le produit de composition de deux fonctions continues." Studia Mathematica 12.1 (1951): 58-64

https://eudml.org/doc/216531

The following paper from 1951 constructs a continuous function x, for which the convolution x*x is only differentiable in 0.

Jarník, V. "Sur le produit de composition de deux fonctions continues." Studia Mathematica 12.1 (1951): 58-64

https://eudml.org/doc/216531
Merci beaucoup. It seems that I am in good company since Mikusinski asked the question as well The question came to me as I was reading his "operational calculus". I worked on it for a day and gave up. Now I'll pour over the article. Thanks again.