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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.

  • 0
    $\mu$ is of ploynomial growth when: \int \frac{d\mu}{1+x^k} <\infty for some constant positive integer $k\in \mathbb{Z}$.2012-06-23

1 Answers 1

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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

  1. for all $n\in\mathbb{N}$ function $\eta_n$ is compactly supported
  2. for all $n\in\mathbb{N}$ and $x\in \mathbb{R}^p$ we have $0\leq\eta_n(x)\leq 1$.
  3. for all $n\in\mathbb{N}$ and $x\in \mathbb{R}^p$ such that $|x| we have $\eta_n(x)=1$.
  4. 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.