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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 ?

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    $\begingroup$ (1) If $\varphi_n \to \varphi$ in $C_0^\infty$ then $\varphi_n(x+y) \to \varphi(x+y)$ in $C_0^\infty$. $\endgroup$ – md2perpe Apr 17 '19 at 21:16
  • $\begingroup$ For each $x$ , we have $\varphi_n(x+y)\to \varphi(x+y)$ , but can we prove $F(\varphi_n(x+y))\to F(\varphi(x+y))$ ? $\endgroup$ – J.Guo Apr 17 '19 at 22:48
  • $\begingroup$ 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. $\endgroup$ – md2perpe Apr 18 '19 at 4:29
  • $\begingroup$ @ 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 $D$ . $\endgroup$ – J.Guo Apr 18 '19 at 11:11
  • $\begingroup$ My last comment was an answer to your question in the comment before: "but can we prove $F(\varphi_n(x+y))\to F(\varphi(x+y))$"? $\endgroup$ – md2perpe Apr 18 '19 at 12:14
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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 $$|F(\varphi)| \leq C \|\varphi\|_N = C \sup_{x \in K ,{|\alpha|<N}} |\partial^\alpha \varphi(x)|$$
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 $$\int\int f_n(x)g_m(y) \varphi(x+y) \,dx \,dy=\int\int f_n(x)g_m(y) \varphi(x+y) \,dy \,dx$$

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  • $\begingroup$ You had missed $| \bullet |$ around $F(\varphi)$ so I added it. I also replaced || with \| that looks a little bit better. $\endgroup$ – md2perpe May 4 '19 at 21:37
  • $\begingroup$ 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. $\endgroup$ – md2perpe May 4 '19 at 21:39

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