# Convolution Integral only defined Almost Everywhere?

My professor has mentioned in class that if we have two $2\pi$-periodic functions $f$ and $g$ that are both in $L_1(\mathbb{T})$, then

$$(f*g)(t) := \frac{1}{2\pi}\int_{-\pi}^{\pi}f(s-t)g(s)ds$$ is only guaranteed to be defined for almost every $t$.

I thought I followed it at the time, but now that I'm sitting at home looking at the remark, I just don't see it. If we took $f_{1}$ and $g_{1}$ such that $f=f_{1}$ almost everywhere and $g = g_{1}$ almost everywhere, wouldnt we have

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}f_{1}(s-t)g_{1}(s)ds = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(s-t)g(s)ds$$ ?

... or was I missing his point completely? I thought it had something to do with the fact that functions in the $L_{1}(\mathbb{T})$ are actually equivalence classes.

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Fix $f$ and fix $g$. Your professor meant that the convolution integral converges for almost all $t$, not necessarily for all $t$. If you switch $f$ and $g$ to be other functions that are equal to them almost everywhere then the convolution integral of the new pair of functions is also defined for almost all $t$, but the measure 0 set where the second integral is not defined need not be the same as the measure 0 set where the first one is not defined. –  KCd Oct 3 '11 at 23:43
OK. So now I more precisely understand what his remark is. But why is is it not defined for all $t$? We are assuming that both $f$ and $g$ are integrable, so why would a shift in the argument of $f$ change this? –  Kyle Schlitt Oct 3 '11 at 23:54
Lightbulb - According to Fubini's Theorem (at least the version in my notes), we have that if $f(s,t)\in L_{1}(\mathbb{T}^{2})$, then $f(\cdot,t) \in L_{1}(\mathbb{T})$ for almost every $t$ and and $f(s, \cdot) \in L_{1}(\mathbb{T})$ for almost every $s$. If we treat $f(s-t)$ as a function of two variables, then this result would explain the issue I think? –  Kyle Schlitt Oct 3 '11 at 23:59
Another remark: $f(s-t)$ and $g(s)$ are both in $L^1$ as functions of $s$, but that does not guarantee that their product is in $L^1$. –  GEdgar Oct 4 '11 at 12:11

The only way to show that there exists some $t$ such that $(f*g)(t)$ exists is often to consider the integral of the function $|f*g|$. As you know, Fubini tells you that the result is bounded by the product of the integrals of $|f|$ and $|g|$, hence that it is finite. This proves that $f*g$ is integrable, hence finite almost everywhere, but cannot give you the finiteness of $f*g$ at any given point.
By the way, the natural objects here are not functions but equivalence classes of functions coinciding almost everywhere, and one often defines the convolution of these objects rather than the convolution of functions. Then the result is, by nature, such a class and the assertion, for example, that the convolution of the functions $f$ and $g$ is a given function $h$ often simply means that $f*g$ is in the class of $h$, in other words, that $f*g=h$ almost everywhere and it could be rephrased as the fact that $\tilde f*\tilde g$ is in the class of $h$ for every $\tilde f$ in the class of $f$ and every $\tilde g$ in the class of $g$. Sometimes the space of integrable functions is written $\mathcal L_1$ and the space of classes $L_1$.
Here is an example. Function $$f(x)=\sum_{n=1}^\infty\frac{\cos n x}{\sqrt n}$$ is continuous on $\mathbb{T}$ with the exception of $x=0$. Since $\lim_{x\to0}f(x)|x|^{1/2}=\sqrt{{\pi }/{2}}\ \$ we have $|f(x)|\le C|x|^{-1/2}$ and $f\in L_{1}(\mathbb{T})$. The convolution $f$ with itself is $$f*f(x)=\sum_{n=1}^\infty\frac{\cos n x}n=-\log |x|+\frac{x^2}{24}+O(x^3),\quad x\to0,$$ so it is not defined at the origin.