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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!

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This is not true. The counterexample I know is:

$f(x) = g(x) = \begin{cases} x^{-3/4}&0<x< 1\cr 0& \text{otherwise} \end{cases}$

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 $L^{\infty}.$

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    $\begingroup$ $x^{-3/4}\notin L^1.$ $\endgroup$ – zhw. Feb 22 at 0:56
  • $\begingroup$ So I'll just cut it at $1$. $\endgroup$ – Matematleta Feb 22 at 1:04
  • $\begingroup$ Thanks! Do you happen to have a simple way of showing that the $\lim_{x \to 0^+} f*f(x) = \infty$? $\endgroup$ – is it normal Feb 22 at 1:25
  • $\begingroup$ 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 $x\to 0^+$. $\endgroup$ – Matematleta Feb 22 at 1:56
  • $\begingroup$ That's perfect! I did an underestimate with rectangles and got $1/\sqrt{x}$ too. $\endgroup$ – is it normal Feb 22 at 2:07
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$f*g$ 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 $(f*g)(0)=\infty$.

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  • $\begingroup$ 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 $L^1(\mathbb{R})$? $\endgroup$ – is it normal Feb 22 at 0:18
  • $\begingroup$ 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 $p=q=r=1$. $\endgroup$ – Shalop Feb 22 at 23:27

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