# If $F_1$ and $F_2$ are distributions with compact support , how to well define the convolution of them , and how to show $F_1*F_2=F_2*F_1$

Suppose $$F_1$$ and $$F_2$$ are given distributions with $$F_2$$ having compact support , then we define the convolution $$F_1*F_2$$ as the distributions $$(F_1*F_2)(\varphi)=F_1(F_2^{a}*\varphi)$$ where $$F_2^a(\varphi)=F_2(\varphi^a)$$ and $$\varphi^a(x)=\varphi(-x)$$ ...... Then we can show that $$F_1*F_2$$ is also a distribution and if $$F_1$$ is compact , we have $$F_1*F_2(\varphi)=F_2*F_1(\varphi)$$

The discussion above was in Stein's functional analysis Page$$_{105}$$ and the author state that to show $$F_1*F_2$$ is a distribution is quite straightforward so is left to the reader but I can not show it .

My attempt :
Since $$F_2$$ and $$\varphi$$ has compact support , then so is $$(F_2^a*\varphi)(x)=F_2(\varphi(x+y))$$ and it belongs to $$C_0^{\infty}$$. We can see $$F_1*F_2$$ is well defined for every $$\varphi \in C_0^{\infty}$$ .
To show $$F_1*F_2$$ is a distribution , we need to show whenever $$\varphi_n \to \varphi$$ in $$D$$ , we have $$\lim_{n\to \infty}(F_1*F_2)(\varphi_n)=(F_1*F_2)(\varphi)$$ note that $$F_1*F_2(\varphi)=F_1(F_2(\varphi(x+y)))$$ , it suffice to show $$F_2(\varphi_n(x+y))\to F_2(\varphi(x+y))$$
in $$D$$ . For each $$x_0$$ , we have $$\varphi_n(x_0+y)\to \varphi(x_0 +y)$$ , so $$\lim_{n\to \infty}LHS=RHS$$ for each $$x$$ , however , I can not show this limitation is also uniform convergence , nor does its derivatives.

My question:
$$(1)$$ I need some hint to show $$F_2(\varphi_n(x+y))$$ converges to $$F_2(\varphi(x+y))$$ uniformly .
$$(2)$$ If $$F$$ is a distribution with respect to $$f$$ , I mean $$F(\varphi)=\int f\varphi \, d\mu$$ , then by fubini theorem I can show $$F_1*F_2=F_2*F_1$$ , but how to do this in the general case ?

• (1) If $\varphi_n \to \varphi$ in $C_0^\infty$ then $\varphi_n(x+y) \to \varphi(x+y)$ in $C_0^\infty$. – md2perpe Apr 17 '19 at 21:16
• For each $x$ , we have $\varphi_n(x+y)\to \varphi(x+y)$ , but can we prove $F(\varphi_n(x+y))\to F(\varphi(x+y))$ ? – J.Guo Apr 17 '19 at 22:48
• What we here write as $\varphi_n(x+y)$ and $\varphi(x+y)$ are just $\varphi_n$ and $\varphi$ translated. You seem to already accept that $\varphi_n(x+y) \to \varphi(x+y)$ in $C_0^\infty$, so since $F$ is a distribution you do have $F(\varphi_n(x+y)) \to F(\varphi(x+y))$ by definition of a distribution. What you should show is that $F(\varphi(x+y))$ is a $C_c^\infty$ function. – md2perpe Apr 18 '19 at 4:29
• @ md2perpe In Stein's book , we say $\varphi \to \varphi_n$ whenever $\varphi$ is a complexed valued function on$R^d$ with just one variable $x\in R^d$ . So we can not say $\varphi_n(x+y) \to \varphi(x+y)$ . If we define $g_n(x)=F(\varphi_n (x+y))$ and $g(x)=F(\varphi(x+y))$ and we need to prove $g_n(x)\to g(x)$ in $D$ . – J.Guo Apr 18 '19 at 11:11
• My last comment was an answer to your question in the comment before: "but can we prove $F(\varphi_n(x+y))\to F(\varphi(x+y))$"? – md2perpe Apr 18 '19 at 12:14

By the help of @md2perpe , I find that we can prove a lemma for this problem first .

Lemma:
Let $$F$$ denote a distribution on $$C_c^{\infty}$$ , then we can find some positive integer $$N$$ such that $$|F(\varphi)| \leq C \|\varphi\|_N = C \sup_{x \in K ,{|\alpha|
Indeed , assume otherwise . Then for each $$n$$ we can find $$\varphi_n$$ with $$\|\varphi_n\|_n=1$$ , while $$|F(\varphi_n)|\ge n$$ . Then let $$\phi_n=\frac{\varphi_n}{n^{\frac12} }$$ , we can see $$\phi_n \to 0$$ in $$C_c^{\infty}$$ while $$F(\phi_n)\to \infty$$ , contradicting that continuity of $$F$$ .

With the lemma above , we can see $$F(\varphi_n(x+y))\to F(\varphi(x+y))$$ in $$C_c^{\infty}$$ since $$\varphi_n \to \varphi$$ , so we complete the proof of the first problem .

For the second problem , Let $$\phi \in C_c^{\infty}$$ with $$\int \, \phi(x) \, dx=1$$ . Then we note that $$\phi_n(x)=n^d\phi(nx)$$ is an approximation to the identity . Let $$f_n(x)=\phi * F(x)$$ , we can see that $$\lim_{n\to \infty} \int f_n(x)\varphi(x) \,dx=F(\varphi(x))$$ For every $$\varphi \in C_c^{\infty}$$ and we can write $$f_n \to F$$. The proof of this part was in Stein's functional analysis Page$$_{103}$$ Corollary $$1.2$$ .

Now suppose $$F$$ and $$G$$ are two compact distributions with $$f_n \to F$$ and $$g_m \to G$$ , Then we have $$F(G(\varphi(x+y)))=F(G(\varphi(x+y)))-\int f_n(x)G(\varphi(x+y)) \,dx+\int f_n(x)G(\varphi(x+y)) \,dx$$ and $$\int f_n(x)G(\varphi(x+y)) \,dx=\int f_n(x)(G(\varphi(x+y))-\int g_m(y)\varphi(x+y)+\int g_m(y)\varphi(x+y) \, dy) \,dx$$ Note that for suffice large $$N$$ , $$\int f_n(x) \, dx$$ is bounded . Indeed $$\{f_n \}$$ are supported in some compact set which is contained in an open set $$O$$ , so we can find a function $$\varphi_0 \in C_c^{\infty}$$ , when $$x\in O$$ we have $$\varphi_0(x)=1$$ . Then we have $$\lim_{n\to \infty} \int f_n(x)\,dx=\lim_{n\to \infty} \int f_n(x)\varphi_0(x) \,dx=F(\varphi_0(x))=A$$ Next ,Since $$|G(\varphi(x+y))-\int g_m(y)\varphi(x+y)|=|G(\varphi(x+y)-\phi_m*\varphi(x+y))|$$
By the lemma above and a limit argument we can get the desired conclusion since $$\int\int f_n(x)g_m(y) \varphi(x+y) \,dx \,dy=\int\int f_n(x)g_m(y) \varphi(x+y) \,dy \,dx$$

• You had missed $| \bullet |$ around $F(\varphi)$ so I added it. I also replaced || with \| that looks a little bit better. – md2perpe May 4 '19 at 21:37
• The inequality is sometimes useful. Many authors define distributions to be linear functionals that satisfy the equality instead of defining them to be continuous w.r.t. limit of test functions. – md2perpe May 4 '19 at 21:39