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Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous but non-differentiable function. If $g: \mathbb{R} \to \mathbb{R}$ is a function that is differentiable almost everywhere on $\mathbb{R}$, or if it is differentiable on the interior of its support, then it is known that $f * g$ is differentiable with $(f*g)' = f*g'$.

Let $\mu = \frac{1}{2} \mathbb{I}_A$ be the density (w.r.t Lebesgue measure) of the uniform distribution on the interval $A = [-1, 1]$. So $\mu$ is differentiable almost everywhere on $\mathbb{R}$, and it is also differentiable on the interior of its support. Thus we should have that $$ (f*\mu)' = f*\mu' = 0, $$ where the last equality is because $\mu'$ exists almost everywhere and equals $0$ whenever it exists. Alternatively, we know that the above integral is only over the interior of the set $A$ (since $\mu$ is only supported on $A$ and we can discard the measure zero endpoints), and on this interior $\mu'$ always exists and equals $0$.

But surely this is wrong? Where does this "proof" break down? I spelled out the details of my thinking as much as possible. I appreciate your help!

learning_the_basics
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  • Best to write out the convolution and start from there to see what’s happening. – A rural reader Oct 13 '24 at 15:53
  • What is your source for the theorem about $g$ being differential on the interior of its support implies $(fg)'=fg'$? This appears to be incorrect, and the source of the problem. – Mike Earnest Oct 13 '24 at 16:40

2 Answers2

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It is correct to say that $(f*g)'=f*g'$, as long as you think of $g$ as a distribution, and consider $g'$ as its distributional derivative, or weak derivative. The distributional derivative of $g$ is $\tfrac12(\delta_{-1}-\delta_1)$, where $\delta_a$ is the Dirac delta distribution centered at $a$. Intuitively, $\frac12 {\mathbb I}_{[-1,-1]}$ is constant almost everywhere, but has a positive jump at $-1$ and a negative jump at $1$, and the derivative of a jump is infinite. This intuitive reasoning can be made precise if you go by the definition of a distribution, using compactly supported test functions. Therefore, $$ f*g'=\tfrac12f*(\delta_{-1}-\delta_1)=\frac{f*\delta_{-1}-f*\delta_1}{2}=\frac{f(x+1)-f(x-1)}{2}. $$ Then, if you directly compute $(f*g)'$, you will find the result is equal to $(f(x+1)-f(x-1))/2$, as shown below. $$ (f*\mu)(x)=\int_{y\in \mathbb R} \mu(y)f(x-y)\,dy=\frac12\int_{-1}^1f(x-y)\,dy=\frac12\int_{-1}^1f(x+y)\,dy, $$ so $$\begin{align} (f*\mu)'(x) &=\frac12\lim_{h\to 0} \frac{\int_{-1}^1f(x+h+y)\,dy-\int_{-1}^1 f(x+y)\,dy}{h} \\ &=\frac12\lim_{h\to 0} \frac{\int_{1}^{1+h} f(x+y)\,dy}{h}-\frac{\int_{-1}^{-1+h}f(x+y)\,dy}{h} \\ &\stackrel{\text{a.s}}=\frac{f(x+1)-f(x-1)}2. \end{align}$$ In the last step, we used the Lebesgue differentiation theorem, which is valid since $f$ is continuous.

Mike Earnest
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  • thank you! follow up: it seems we could derive the same result via the fundamental theorem of calculus? now let us generalize the question and say that $f: \mathbb{R}^n \to \mathbb{R}$, and $\mu$ is the uniform distribution on the unit sphere. is there a generalization of the fundamental theorem of calculus that will give us this result?

    i ask because i would prefer to avoid machinery like weak derivatives (which I have not learned), and stick to more "basic" ideas. i looked into Stokes theorem, but it is stated with the language of differential geometry (which is inaccessible to me).

     – learning_the_basics Oct 13 '24 at 20:42
  • perhaps there is a version of Stokes theorem that can be easily stated for $\mathbb{R}^n$ without any differential geometry language? – learning_the_basics Oct 13 '24 at 20:43
  • thanks for your time @MikeEarnest. i have made a separate question here https://math.stackexchange.com/questions/4984270/gradient-of-smoothed-version-of-function-f , and I have also included a reference to a paper which uses Stokes theorem. as mentioned in the question, i am looking for a proof which follows from standard measure theory techniques (such as Lebesgue Differentiation Theorem), or for a version of Stokes theorem that can be stated in $\mathbb{R}^d$ without any differential geometry. i appreciate your help! – learning_the_basics Oct 14 '24 at 01:58
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The "known" fact you are quoting is not true in general. Here is an article published in the journal Real Analysis Exchange where the authors construct differentiable and periodic functions which have a convolution that is not differentiable at any point of a perfect set. And here is a MathOverflow page where they discuss how to construct infinitely differentiable functions whose convolution is nowhere continuous.

If the function $g$ is integrable and differentiable everywhere with an integrable derivative, or a bounded derivative, then it is well-known that the convolution with integrable $f$ will be differentiable with $(f\ast g)' = f\ast g'$. But here, the function $g$ which is the indicator function of a closed interval does not have a derivative defined everywhere, only almost everywhere, so it does not fit into the assumptions of these theorems.

Alex Ortiz
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  • thank you for the response @AlexOrtiz. followup question: you mention that if $g$ is differentiable everywhere with integrable/bounded derivative, then for integrable $f$ we will have $fg$ is differentiable with $(fg)' = f * g'$. what if $f$ is not integrable, but is continuous? or if it is locally integrable? does your same claim hold? a reference would also be very helpful. – learning_the_basics Oct 14 '24 at 18:36
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    @learning_the_basics: If $f$ and $g$ are not both integrable, I am not sure how you would define $f\ast g$ other than as a distribution as in Mike Earnest's answer. – Alex Ortiz Oct 23 '24 at 21:40