2
$\begingroup$

Let $f$ be a locally integrable function on $\mathbb{R}^n$ such that $f$ is of polynomial growth at infinity. Prove that $f$ is a tempered distribution.

Here the distribution is defined as $f(\phi)=\int_{\mathbb{R}^n}f\phi.$

I think that I have no reason to be confused of this problem, but I am suddenly stuck.

The linearity is immediate, but I'm figuring out to prove the continuity.

The continuity is equivalent to proving that $\int f\phi_k\to 0$ whenever $\phi_k\to 0$ in the Schwarz class $\mathcal{S}(\mathbb{R}^n)$, i.e. the sequence of functions $\phi_k$ itself and all of the sequences of partial derivatives converge uniformly to zero.

Of course I can insert the limit into the integral by uniform convergence if the integral is done in a bounded set, but $\mathbb{R}^n$ is unbounded, so I'm finding another way.

Another way I'm considering is to apply DCT, but how should I find an integrable majorant of $\{f_n\}$?

Thanks in advance!

$\endgroup$
5
$\begingroup$

Actually, convergence to zero in the Schwartz space refers to something stronger, namely, that for all $\alpha, \beta\in\mathbb N^n$, $$ \lim_{k\to +\infty}\sup_{x\in\mathbb R^n}\left\lvert x^\alpha \left(D^{\beta}\phi_k\right)(x)\right\rvert=0. $$ All we need is that for all $p\in\mathbb N$, $$ \lim_{k\to +\infty}\sup_{x\in\mathbb R^n}\left\lVert x\right\rVert^p \left\lvert \phi_k (x)\right\rvert=0. $$ Indeed, from the polynomial growth assumption, we know that there exists $q\in \mathbb N $ and $c\gt 0$ such that for all $x\in\mathbb R^n$, $\left\lvert f(x)\right\rvert\leqslant c\left(1+\left\lVert x \right\rVert^q\right)$. Therefore, $$ \left\lvert \int_{\mathbb R^n} f\phi_k \right\rvert\leqslant c\int_{\mathbb R^n} \left(1+\left\lVert x \right\rVert^q\right)\left\lVert x\right\rVert^{-p} \left\lVert x\right\rVert^p \left\lvert \phi_k (x)\right\rvert\mathrm dx. $$ It thus suffices to choose $p$ such that the integral $\int_{\mathbb R^n} \left(1+\left\lVert x \right\rVert^q\right)\left\lVert x\right\rVert^{-p}\mathrm dx$ is finite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.