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Let $f$ be a continuous function from $\mathbb{R} \rightarrow \mathbb{R}$. Let $g \in C^1(\mathbb{R})$ with compact support. Prove that the convolution function $$(f*g)(x)=\int f(x+t)g(t) \,dt$$ is well-defined and lies in $C^1(\mathbb{R})$. Furthermore, prove that if $g \in C^k(\mathbb{R})$ then $f*g \in C^k(\mathbb{R})$.

To prove that $f*g$ is well-defined, do I show that if $x_1=x_2$ then $(f*g)(x_1)=(f*g)(x_2)$? Besides this, I am unsure how to approach this problem. I have never learned the concept of convolution before. Any idea how to do this problem?

Michael Hardy
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user112358
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  • To be well defined means that the integral is well defined. – John B Feb 09 '16 at 23:18
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    I think the term "well-defined" here has a different meaning. I think you're supposed to show that the integral is finite for every $x$. – Tunococ Feb 09 '16 at 23:18
  • Well-defined means here that $f *g$ is actually a function, i.e. the integral actually exists. – Roland Feb 09 '16 at 23:18

1 Answers1

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To prove that the convolution is well-defined you must show that it's finite for every $x \in \mathbb{R}$. I'll assume that the integration is over $\mathbb{R}$. So, because $g$ has compact support, there is a $K \subset \mathbb{R}$ compact such that $g(x)=0, \forall x \in K^c$. So we have that $$ (f*g)(x)= \int_{\mathbb{R}}f(x+t)g(t)\,dt=\int_{K}f(x+t)g(t)\,dt$$

Because both functions are continuous they are bounded in compacts, so there is $M, N >0$ such that $ f(x+t)<M$, $g(t)<N, \forall t \in K$. Furthermore, because $K$ is compact, there is some $R >0$ such that $K \subset [-R,R] $ So:

$$\int_K f(x+t)g(t)\,dt < MN\int_K\,dt <MN \int_{[-R,R]}dt<2RMN < +\infty$$

Now we have to show that the convolution of our functions defines a $C^1(\mathbb{R})$ function. Well, if we do $u=x+t$ :

$$(f*g)(x)=\int_{\mathbb{R}}f(u)g(u-x)\,du$$

Now, You can calculate the derivative of $(f*g)(x)$ using the definition!

$$\lim_{x\rightarrow x_0}\frac{(f*g)(x)-(f*g)(x_0)}{x-x_0}= \lim_{x\rightarrow x_0}\int_K\frac{f(u)(g(u-x)-g(u-x_0))}{x-x_0} \, du $$

We can pass the limit inside the integral in this case, by the dominated convergence theorem, so

$$(f*g)'(x_0)= -\int_Kf(u)g'(u-x_0)\,du = -\int_K f(x_0+t)g'(t) \, dt$$

This is a continuous function, you can see that using an $\epsilon-\delta$ argument with the help of the inequality $|\int f \, d\mu| \leq \int|f|\,d\mu$ and noting that you are in a compact set.

Remark 1: Note that is a minus sign there, it appeared because of your definition of convolution that forced my choice of change of variables above. If you change your definition for a $f(x-t)$, there will be no sign at all!

Remark 2: To the case where $g \in C^k(\mathbb{R})$ you can continue calculating the derivatives.

Michael Hardy
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Affonso
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