# Show that convolution of two $L^1(\mathbb{R})$ functions is continuous

Suppose $$f, g \in L^1(\mathbb{R})$$. I want to show that their convolution is continuous.

I can show continuity if one of the functions were in $$L^\infty(\mathbb{R})$$.

I have tried to approximate one of the functions, say $$f$$ with $$f_M = f\cdot \mathbb{I}\{ |f(x)| \leq M\}$$. But I have problem bounding the residual, which would be of form:

$$\int_\mathbb{R} |f(x-z) - f_M(x-z)|\cdot |g(z)| dz$$

Any help/hint would be greatly appreciated!

This is not true. The counterexample I know is:

$$f(x) = g(x) = \begin{cases} x^{-3/4}&0

then $$\lim_{x \to 0^+}f*f(x) = \infty$$ and $$\lim_{x \to 0^-}f*f(x)=0$$.

I guess you need at least one of $$f$$ or $$g$$ to be $$L^{\infty}.$$

• $x^{-3/4}\notin L^1.$ – zhw. Feb 22 at 0:56
• So I'll just cut it at $1$. – Matematleta Feb 22 at 1:04
• Thanks! Do you happen to have a simple way of showing that the $\lim_{x \to 0^+} f*f(x) = \infty$? – is it normal Feb 22 at 1:25
• I think we can do this: Note that if $0<x<1,\ f*f(x)=\int^x_0f(x-y)f(y)dy$. Then, since $y<x$, we compute $\frac{1}{x-y}\cdot \frac{1}{y}=\frac{1}{x}\left(\frac{1}{x-y}+\frac{1}{y}\right)\ge \frac{1}{x}\cdot \frac{1}{y}$ and so we get $f*f(x)\ge \frac{1}{x^{3/4}}\int^x_0y^{-3/4}dy=\frac{4}{x^{3/4}}\cdot x^{1/4}=\frac{4}{x^{1/2}}\to \infty$ as $x\to 0^+$. – Matematleta Feb 22 at 1:56
• That's perfect! I did an underestimate with rectangles and got $1/\sqrt{x}$ too. – is it normal Feb 22 at 2:07

$$f*g$$ is not a well defined function from $$\mathbb R$$ to itself, in general. For example, if $$f(x)=g(x)=\frac 1 {\sqrt {|x|}}$$for $$0<|x| \leq1$$ and $$0$$ elsewhere then $$(f*g)(0)=\infty$$.

• Thanks for the reply! I can see that convolution of two continuous functions is continuous but I'm not sure how to use the uniformity of limits. I don't think approximation by continuous function is uniform on $L^1(\mathbb{R})$? – is it normal Feb 22 at 0:18
• The point is that if you convolve two $L^1$ functions then you still get an a.e. defined $L^1$ function. This follows directly from Young's inequality with $p=q=r=1$. – Shalop Feb 22 at 23:27