a question on distributions. Suppose $L\in \mathcal{S}$ is a tempered distribution, and for each $\phi \in \mathcal{S} :\phi \geq 0 \implies L(\phi) \geq 0$.
Prove that there exists a borel measure $\mu$ with polynomial growth such that $L=L_{\mu}$, where $L_{\mu}(\phi) = \int \phi \:d\mu$.
I think I need to use here Riesz Representation Theorem, but not sure how.
Any hints?
Thanks.
 A: 1. Proof of the existence of Borel measure. Consider restriction $\tilde{L}$ of $L$ on the space of smooth compactly supported functions $C_c^\infty(\mathbb{R}^p)$. This is a positive linear functional, hence using ideas of this answer one can show that $\tilde{L}$ is continuous. Since $C_c^\infty(\mathbb{R}^p)$ is dense in the space of continuous compactly supported functions $C_c(\mathbb{R}^p)$ with respect to the $\sup$-norm, then there exist unique continous linear extension $\hat{L}$ of $\tilde{L}$ to the space $C_c(\mathbb{R}^p)$. 
Now we want to prove that $\hat{L}$ is positive. Take arbitrary non-negative function $f\in C_c(\mathbb{R}^p)$, and consider its convolutions with molifiers $\omega_{1/n}$, i.e. consider functions $f_n=f*\omega_{1/n}$. It is known that they are smooth, and also they are compactly supported becase $f$ is compactly supported. Since $f$ is non-negatie, then does $f_n$. Since $f$ is compactly supported and continuous, then $f_n$ converges uniformly to $f$, i.e. in $\sup$-norm. Since $\hat{L}$ is the continuous extension of $\tilde{L}$, then
$$
\hat{L}(f)=\hat{L}\left(\lim\limits_{n\to\infty} f_n\right)=\lim\limits_{n\to\infty} \hat{L}(f_n)=\lim\limits_{n\to\infty} \tilde{L}(f_n)\geq 0
$$
Since $f\in C_c(\mathbb{R}^p)$ is arbitrary, $\hat{L}$ is positive. Then by Reisz representation theorem there exist Borel regular measure $\mu$ such that
$$
\hat{L}(f)=\int\limits_{\mathbb{R}^p} f(x)d\mu(x)\quad\text{ for }\quad f\in C_c(\mathbb{R}^p)
$$
In particular
$$
\tilde{L}(f)=\hat{L}(f)=\int\limits_{\mathbb{R}^p} f(x)d\mu(x)\quad\text{ for }\quad f\in C_c^\infty(\mathbb{R}^p)
$$
Since $\tilde{L}$ is continuous with respect to $\sup$-norm of $C_c^\infty(\mathbb{R}^p)$, then it is continuous with respect to the locally convex topology in $C_c^\infty(\mathbb{R}^p)$ coming from $S(\mathbb{R}^p)\supset C_c^\infty(\mathbb{R}^p)$.
Note that locally convex space $C_c^\infty(\mathbb{R}^p)$ is dense in $S(\mathbb{R}^p)$, so you can uniquely extend $\tilde{L}$ to the continuous linear functional $\overline{L}$ on the whole space $S(\mathbb{R}^p)$.  Now we want to prove that $\overline{L}$ acts on the functions as integral. Consider arbitrary $f\in S(\mathbb{R}^p)$. Since $C_c(\mathbb{R}^p)$ is dense in $S(\mathbb{R}^p)$ there exist sequence $\{f_n:n\in\mathbb{N}\}$ of compactly supported functions that uniformly converges to $f$, i.e. converges in $\sup$-norm. Since $\hat{L}$ is the continuous extension of $\tilde{L}$, then using dominated convergence theorem we get
$$
\overline{L}(f)=\overline{L}(\lim\limits_{n\to\infty} f_n)=
\lim\limits_{n\to\infty}\overline{L}(f_n)=
\lim\limits_{n\to\infty}\tilde{L}(f_n)=
$$
$$
\lim\limits_{n\to\infty}\int\limits_{\mathbb{R}^p} f_n(x)d\mu(x)=
\int\limits_{\mathbb{R}^p} \lim\limits_{n\to\infty}f_n(x)d\mu(x)=
\int\limits_{\mathbb{R}^p} f(x)d\mu(x)
$$
Thus,
$$
\overline{L}(f)=\int\limits_{\mathbb{R}^p} f(x)d\mu(x)\quad\text{ for }\quad f\in S(\mathbb{R}^p)
$$
By definition $\tilde{L}=L|_{C_c(\mathbb{R}^p)}$ and by construction $\tilde{L}=\overline{L}|_{C_c(\mathbb{R}^p)}$, then from the uniqueness we conclude that $L=\overline{L}$, i.e.
$$
L(f)=\int\limits_{\mathbb{R}^p} f(x)d\mu(x)\quad\text{ for }\quad f\in S(\mathbb{R}^p)
$$
2. Proof of polynomial gowth property. Now we proceed to the proof that $\mu$ is of polynomial growth. Since $L$ is a continuous functional on $S$, then it is continuous with respect to the semi-norm of Schwartz space
$$
p_{\alpha,\beta}(f)=\sup\limits_{|\alpha|\leq k}\sup\limits_{x\in\mathbb{R}^p}|x^\alpha(\partial^\beta f)(x)|\quad\text{ for }\quad f\in S(\mathbb{R}^p)
$$
for some $k\in\mathbb{Z}_+$. Hence for some $C>0$ we have
$$
|L(f)|\leq C \max\limits_{i=1,\ldots,m} p_{\alpha_i,\beta_i}(f)\quad\text{ for }\quad f\in S(\mathbb{R}^p)
$$
Consider sequence of smooth functions $\{\eta_n:n\in\mathbb{N}\}\subset S(\mathbb{R}^n)$ such that


*

*for all $n\in\mathbb{N}$ function $\eta_n$ is compactly supported

*for all $n\in\mathbb{N}$ and $x\in \mathbb{R}^p$ we have $0\leq\eta_n(x)\leq 1$. 

*for all $n\in\mathbb{N}$ and $x\in \mathbb{R}^p$ such that $|x|<n$ we have $\eta_n(x)=1$.

*for all $n\in\mathbb{N}$ we have $\max\limits_{i=1,\ldots,m} p_{\alpha_i,\beta_i}((1+|x|^2)^{-\alpha}\eta_n)\leq 1$


Speaking informally $\eta_n$ is a smooth analogue of the characteritic functions of the ball $B(0,n)$. Paragraph 4 means that derivatives of $\omega_n$ varies
very slowly. The proof of the existence is very messy, but if you need it I'll write it later.
Then define $g_n=(1+|x|^2)^{-\alpha}\eta_n$. From paragraph 1 we see that $\{g_n:n\in\mathbb{N}\}\subset C_c(\mathbb{R}^p)\subset S(\mathbb{R^p})$. From paragraph 4  for all $n\in\mathbb{N}$ we obtain
$$
\left|\int\limits_{\mathbb{R}^p} \eta_n(x)\frac{d\mu(x)}{(1+|x|^2)^\alpha}\right|=|L(g_n)|\leq C p_{\alpha,\beta}(g_n)=C p_{\alpha,\beta}((1+|x|^2)^{-\alpha}\eta_n)\leq C
$$
Paragraph 2 and inequality above allows us to apply dominated convergence theorem. Then we get
$$
\left|\int\limits_{\mathbb{R}^p}\frac{d\mu(x)}{(1+|x|^2)^\alpha}\right|=
\left|\int\limits_{\mathbb{R}^p}\lim\limits_{n\to\infty} \eta_n(x)\frac{d\mu(x)}{(1+|x|^2)^\alpha}\right|
$$
$$
\lim\limits_{n\to\infty}\left|\int\limits_{\mathbb{R}^p} \eta_n(x)\frac{d\mu(x)}
{(1+|x|^2)^\alpha}\right|\leq C
$$
Thus measure $\mu$ is of polynomial growth.
