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

Question

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?

Answer 1

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 $(1+|t|)^{-k}\in L^{p'}$.

Answer 2

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 $\mathcal S(\Bbb R^n)$.