# Show that the function $f$ is a tempered distribution

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 $$\{f_n\}$$?

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.