I have these related questions here that I could really use some help on. I believe there is a related question here although I don't think it is exactly the same...

1) Let $f\in L^1(\mathbb R)$ and $\varphi$ a smooth compactly supported function. Prove that $f\ast \varphi$ is a smooth function whose derivative is $f\ast \varphi'$.

Any help would be much appreciated. I have been trying to work on this for the past couple days but all my ideas didn't lead me very far. Namely, I was trying to use the definition of a convolution. I am not sure though how the smooth and compactly supported $L^1$ function fits into this...

  • 1
    For the first part, use the definition of the convolution (and maybe the identity $f \ast g = g \ast f$), and then differentiate under the integral sign (you need to show that this is allowed). – PhoemueX Dec 7 at 8:47
  • The direct approach is to find a bound for $f \ast \varphi' - f \ast (\varphi(.+y)-\varphi)/y$. The less direct one is to look at $\int f \ast \varphi'$ – reuns Dec 7 at 16:44
up vote 1 down vote accepted

I have an answer for the first question: recall: $$f \ast \phi= \phi \ast f=\int_\mathbb{R}{\phi(x-t)f(t)dt}$$ Note that in our case the convolution is defined everywhere since $\phi$ is compactly supported and thus bounded, and $f \in L^1$. Now look at the definition of the derivative: $$\frac{f \ast \phi (x+h) - f \ast \phi (x)}{h} = \int_\mathbb{R}{\frac{\phi(x-t+h) - \phi(x-t)}{h}f(t)dt}$$ By the mean value theorem for each fixed $t$ there exists $\theta \in [x-t+h,x-t]$ with $\phi'(\theta)=\frac{\phi(x-t+h) - \phi(x-t)}{h}$. But since it is compactly supported we know that all derevatives of $\phi$ are bounded by some constant $M$, thus we have that the absolute value of the integrand on the right is bounded (without dependence on h) by: $$\vert f(t)\phi(\theta)\vert\le M\vert f(t)\vert$$ which is an integrable function. Now apply the Dominated convergence theorem on some sequence $h_n\rightarrow 0$ and get: $$\underset{n\rightarrow \infty}{lim}\frac{f \ast \phi (x+h_n) - f \ast \phi (x)}{h_n}=\int_\mathbb{R}{\underset{n\rightarrow \infty}{lim}\frac{\phi(x-t+h_n) - \phi(x-t)}{h_n}f(t)dt}$$ $$ = \int_\mathbb{R}{\phi'(x-t) \ast f(t)dt} = \phi' \ast f (x)$$.

Great, we found the derevative of $\phi \ast f$. Now since easily $\phi'$ is again a compactly supported $C_\infty(\mathbb{R})$ function, we can apply the above argument again and again to get that $f \ast \phi$ is in $C_\infty(\mathbb{R})$.

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  • Thank you very much. I really appreciate the help on this. – MathIsHard Dec 9 at 6:35
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    Thanks very much for accepting my first answer on this site! – pitariver Dec 9 at 7:10

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