# Convolution of piecewise continuous function is continuous

Let $f\left(x\right)$ and $g\left(x\right)$ two piecewise continuous function. I would like to prove (but I'm not sure it is true) that $$\left(f*g\right)\left(y\right)=\int_{0}^{y}f\left(x\right)g\left(y-x\right)dx$$ is continuous. I tried to search in some books but I find nothing.

My attempt: I fix $y>0$. Since $f,\,g$ are piecewise continuous then exists a partition $\left(a_{m},\,a_{m+1}\right),\,m=1,\dots,M$ of the interval $\left(0,y\right)$, with $a_{1}=0$ and $a_{M}=y$ such that $f\left(x\right)$ and $g\left(y-x\right)$ are continuous and since $$\int_{0}^{y}f\left(x\right)g\left(y-x\right)dx=\sum_{m=1}^{M-1}\int_{a_{m}}^{a_{m+1}}f\left(x\right)g\left(y-x\right)dx$$ then $\int_{0}^{y}f\left(x\right)g\left(y-x\right)dx$ is continuous in $y$ because is a finite sum of continuous function.

Is this proof correct?

• There is a slight innacuracy (if that) that can be corrected easily, why do you assume $f$ and $g$ are continuous in the same partition of the interval? Oct 29 '17 at 8:35
• @Smurf Because in those sub-intervals I'm sure that the integral is continuous. I don't know if it is a overabundant hypothesis. Oct 29 '17 at 8:37
• I mean before that, you pick $f$ and $g$ piecewise-continuous, that means that for each of them exists a partition of the interval (one for each, so they might very well be different) such that they are continuous on each subinterval, my question is, why do you assume both partitions are exactly the same? Oct 29 '17 at 8:46
• @Smurf I'm not assuming that the partition are the same. Since there are two (probably different) partitions where the function are continuous, then exists another partition where both functions are continuous. Oct 29 '17 at 8:50
• That is correct! Maybe I am thinking too much as a teacher and that's something that would lower some scores if not mentioned.. :D Oct 29 '17 at 8:54

If I remember correctly, the matching conditions at the discontinuities imposed on piecewise continuous functions (i.e. the limit of a piecewise continuous function at a discontinuity must exist from both sides) imply that a piecewise continuous function is bounded. Since the product of two piecewise continuous (hence bounded) functions is bounded we have $|fg| <M$ for all $x$, therefore $$\left| \int_0^{y-\delta} fg \ \text{d}x -\int_0^{y} fg \ \text{d}x \right|=\left| \int_{y-\delta} ^yfg \ \text{d}x \right|\le \int_{y-\delta} ^y|fg| \ \text{d}x< \int_{y-\delta} ^y M \ \text{d}x<M\delta$$ and the same can be done approaching $y$ from the right.
• Wait. Since $g$ depends on $y$ can be $M$ a function in $y$? And can be this a problem? Oct 29 '17 at 8:51
• $M$ is not a function of $y$, $M$ is a constant. Since $g$ is piecewise continuous it is bounded over its whole domain by some real number, no matter on what point you choose to evaluate $g$. The same goes for $f$. $M$ is just any number bigger than the product of the two values bounding $g$ and $f$. Oct 29 '17 at 9:03
• But in this case the domain is $(0,y)$. For example let us take the floor function $f(x)=\left\lfloor x\right\rfloor$, which is piecewise continuous. Then the constant $M$ increase when we increase $y$. Am I wrong? Oct 29 '17 at 9:09
• You are right, I did not explain very clearly. What I mean to say is that once $y$ is fixed, both $f$ and $g$ are piecewise continuous over $(0,y)$, which means that they are bounded over $(0,y)$ and in fact over any finite interval. So while $M$ does depend on which interval you are considering, once you fix that interval, $M$ should not change, and you can make $\delta$ small enough to be within that interval. Oct 29 '17 at 9:14
• So the proof holds since for all $y$ fixed (and so the interval is fixed) we have that the integral is continuous in $y$, right? Oct 29 '17 at 9:41