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 $\{f_n\}$?
Thanks in advance!
 A: 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.
