# Showing that $T$ is a tempered distribution by showing $T(\phi) \in L^p$ for some p?

EDIT: Okay, I am rephrasing the question for more clarity:

Given a function $T(x)$, what are sufficient conditions to show that $T(x)$ is a tempered distribution? For instance, is it sufficient to show that for any $\phi \in \mathscr{S}(\mathbb{R}^n)$ that the action defined by

$$T(\phi) := \int T(x)\phi(x)dx$$

From reading I got the impression that if we can show that this action is in $L^p$ for some $p$, then this would imply that $T$ must be a tempered distribution. Is this wrong?

• How can $T(\phi)$ be linear if there is an absolute value inside the integral? I mean $T(a\phi)=|a|T(\phi)$. Other than that in your statement we just know that it is true for one $\phi$?
– chak
Dec 4, 2017 at 16:02
• @chak Sorry, that was a mistake. I meant that $T(\phi) = \int T(x)\phi(x)dx$, and if we can show that the integral with absolute values is bounded, then $T$ is in $S'(R^n)$ Dec 4, 2017 at 16:05
• What is $T(x)?$ Dec 4, 2017 at 16:09
• I don't understand your question. Did you mean $T \in L^p$ ? Otherwise take $T(x) = e^{|x|}$ it is a distribution but not tempered. @GuyFsone In this context it is clear it means $T$ is represented by a function. Dec 4, 2017 at 16:09
• I suppose the question is how to prove that "if $T$ is a measurable function such that $\int \lvert T(x)\phi(x)\rvert\,dx < +\infty$ for all $\phi \in \mathcal{S}(\mathbb{R}^n)$, then $\phi \mapsto \int T(x)\phi(x)\,dx$ is a tempered distribution"? Dec 4, 2017 at 16:16

## 2 Answers

The question is badly garbled in various ways. First, you're using the letter $T$ for two different things. Second, you haven't seen people say that if $T(\phi)\in L^p$ then $T$ is a tempered distribution; that makes very little sense, since $T(\phi)$ is a scalar if $T$ is a tempered distribution and $\phi\in\mathcal S$. Finally, I have no idea what "equivalence" you might be referring to.

Here's what you meant to ask about:

Suppose $K\in L^p(\mathbb R)$. Then $K\phi\in L^1$ for every $\phi\in\mathcal S$, and if we define $T\phi=\int K\phi$ then $T\in\mathcal S'$.

Proof: Let $$\rho_k(\phi)=\sup_t(1+|t|)^k|\phi(t)|.$$ It's easy to see that there exists $k$ such that $$||\phi||_{p'}\le c\rho_k(\phi).$$Hence $\mathcal S\subset L^{p'}$, so that $K\phi\in L^1$. And the inequality $$|T\phi|\le||K||_p||\phi||_{p'}\le c||K||_p\rho_k(\phi)$$shows that $T$ is a tempered distribution, since $\rho_k$ is one of the seminorms defining the topology on $\mathcal S$.

Hmm, looking at some of the comments, maybe you really meant to ask this:

If $K\phi\in L^1$ for every $\phi\in\mathcal S$, and if we define $T\phi=\int K\phi$ then $T\in\mathcal S'$.

That follows from the Closed Graph Theorem. (CGT is usually stated for Banach spaces, but luckily it's true for Frechet spaces as well: "Banach’s theorem states that when $E$ and $F$ are Frechet spaces and $u$ is linear, this map is continuous if, and only if its graph is closed ([3], p. 41, Thm. 7)". Define $T:\mathcal S\to L^1$ by $Tf=Kf$. If $T$ is continuous we're done. By CGT we need only show that the graph of $T$ is closed. So suppose $f_n\to f$ in $\mathcal S$ and $Tf\to g$ in $L^1$. Then $Kf_n\to g$ in $L^1$ and $Kf_n\to Kf$ pointwise, so $Kf=g$, which is to say $g=Tf$.)

Or maybe you actually meant this:

If $K\phi\in L^p$ for every $\phi\in\mathcal S$ then $K\phi\in L^1$ for every $\phi\in \mathcal S$, and hence if we define $T\phi=\int K\phi$ then $T\in\mathcal S'$.

That follows since if $\phi\in \mathcal S$ then $(1+|t|)^k\phi(t)\in\mathcal S$, and there exists $k$ so $(1+|t|)^{-k}\in L^{p'}$.

• Yes that is exactly what I wanted to know about! Sorry, the notation is commonly used in my classes. I guess the proper way to write it would be $$T_f(\phi) := \int f(x)\phi(x)dx$$ Thank you, that was exactly the result I have seen referred to a lot but never sence explicitly stated. Dec 4, 2017 at 16:32
• @david-c-ullrich Follow up: So my situation is the second one, that $K\phi \in L^1$. I am not very familiar with the closed graph theorem, I guess that the product integral being finite has to do with the graph being closed? What exactly is 'the graph', is it the image of $T\phi$ (for some phi)? Dec 7, 2017 at 15:02
• @Laplacinator I expanded on "This follows from CGT" in an edit - the link that I added explains what CGT says. Dec 7, 2017 at 15:39
• @david-c-ullrich Thank you so much. I may have to cite this result, do you know of any book/paper that mentions this result? Dec 8, 2017 at 23:07
• @Laplacinator I don't really know a reference, sorry. In a research paper you could probably get away with just saying that hence $T$ was a distribution, perhaps with a "By CGT" as in my original version. If this is like for school or something you could just cite this MSE post, or reproduce it with attribution. (Probably you want to mention that CGT does hold for Frechet spaces, citing Banach as in that paper I linked to.) Dec 8, 2017 at 23:19

Note the following

Lemma since $\mathcal S(\Bbb R^n)$ is dense in $L^p$ we have, by the Riesz- Representation theorem that: $$\|T\|_q = \sup\left\{ \left|\int T(x)\phi(x)dx\right|: \phi \in L^p, \|\phi\|_p= 1\right\}\\ = \sup\left\{ \left|\int T(x)\phi(x)dx\right|: \phi \in\mathcal S(\Bbb R^n), \|\phi\|_p= 1\right\}$$

where, $\frac{1}{q}+\frac{1}{p}=1.$

Therefore if for any $\phi\in \mathcal S(\Bbb R^n)$ we have, $$\left|\int T(x)\phi(x)dx\right|<\infty$$ then $T\in L^q$ for $1<q<\infty$ Hence the maps

$$T: \mathcal S(\Bbb R^n) \to \Bbb R$$ with

$$T(\phi)= \int T(x)\phi(x)dx$$ is continuous with respect to the topology of $\mathcal S(\Bbb R^n)$. Namely $T\in \mathcal S'(\Bbb R^n)$.

Indeed,

$$|T(\phi)|\le \|T|\|_q\|\phi\|_p$$

but $$\|\phi\|_p^p = \int |\phi|^pdx \le \sup\left[(1+|x|^2)^{n+1}|\phi(x)|^p\right] \int \frac{dx}{(1+|x|^2)^{n+1}} \\= C_n \sup\left[(1+|x|^2)^{n+1}|\phi(x)|^p\right]$$ with $$C_n=\int \frac{dx}{(1+|x|^2)^{n+1}}<\infty$$

Finally we have, $$|T(\phi)|\le \|T|\|_q\|\phi\|_p\le C_n^{1/p}\|T|\|_q\sup\left[(1+|x|^2)^{\frac{n+1}{p}}|\phi(x)|\right]$$

this prove the continuity of T on $\mathcal S(\Bbb R^n)$.