Product of $W^{1,p}_0$ functions Let $p>n$, and let $f,g\in W^{1,p}_0(\mathbb{R}^n)$ be two sobolev functions. Prove that $fg\in W^{1,p}_0(\mathbb{R}^n)$.
I was able to prove the Leibniz formula for weak derivativatives, but still I do not understand why the function should be in $L^p$. Probabily it is Morrey inequality or something like that. But I cannot figure out.

Just a slight correction to @Diesirae92's answer, for posteriority's sake:
we can't really write
$$||f(g_n - g)||_{W^{1,p}(\mathbb{R}^n)} \leq ||f||_{\infty} ||(g_n - g)||_{W^{1,p}(\mathbb{R}^n)}$$
as the $W^{1,p}$ also involves the derivative, which may not be bounded.
Instead, what we should write is
\begin{equation}
||f(g_n - g)||_{W^{1,p}(\mathbb{R}^n)} = ||f(g_n - g)||_{L^p} + \sum_i || \frac{\partial}{\partial x_i} f(g_n - g)||_{L^p}
\end{equation}
We can take care of the first term as done above, by taking out the infinity norm of $f$, but the second one needs more work:
$$|| \frac{\partial}{\partial x_i} f(g_n - g)||_{L^p}\leq || \frac{\partial f}{\partial x_i} (g_n - g)||_{L^p} + ||f \frac{\partial}{\partial x_i}(g_n - g)||_{L^p} \leq ||g_n - g||_{L^\infty} ||\frac{\partial f}{\partial x_i}||_{L^p} + ||f||_{L^\infty} ||\frac{\partial}{\partial x_i}(g_n - g)||_{L^p}$$
and this goes to $0$, as $||g_n - g||_{L^\infty} \rightarrow 0$ (by continuity of the embedding $W^{1,p} \hookrightarrow L^\infty$) and $||\frac{\partial f}{\partial x_i}||_{L^p} \leq \infty$, and $|f||_{L^\infty} \leq \infty$ and $||\frac{\partial}{\partial x_i}(g_n - g)||_{L^p} \rightarrow 0$.
Doing the same for the other term, $||(f_n - f)g_n||_{W^{1,p}}$, we get the result
