Nash inequality : does $f\in L^1$ and $\nabla f \in L^2$ implies $f\in L^2$? Let $f$ be any function that belongs to $L^1(\textbf{R}^d)\cap H^1(\textbf{R}^d)$ ($d$ a positive integer). Nash inequality applies in this case and gives us
$$\| f\|_{L^2}\leq C \| f\|_{L^1}^r \| \nabla f\|_{L^2}^{1-r}$$
for some real constants $C$ and $r$ such that $C>0$ and $0<r<1$ independant of $f$.
I know that this inequality holds when $f$ is assumed to be in $L^2$. But does $f\in L^1$ and $\nabla f \in L^2$ implies $f\in L^2$ ?
Many thanks !
 A: Suppose that $f : \newcommand{\R}{\mathbb R} \R^d \to \R$ is locally integrable and that $\{f_\epsilon\}_{\epsilon > 0}$ be a family of regulatizations of $f$; that is,
$$
 f_\epsilon(x) = \epsilon^{-d} \int_{\R^d} \phi(y/\epsilon) f(x-y) \, dy
$$
for a suitable regularizing kernel $\phi$. 
It is well-known that if $f \in L^p(\R^d)$ for some $1 \le p < \infty$ then $f_\epsilon \to f$ in $L^p(\R^d)$.  Moreover, if $f$ has a locally integrable weak gradient $\nabla f$, then the regularization of the gradient is the gradient of the regularization: $$\nabla (f_\epsilon) = (\nabla f)_\epsilon.$$
If $f \in L^1(\R^d)$ and $\epsilon > 0$ then $f_\epsilon \in L^1(\R^d) \cap L^\infty(\R^d)$ so that $f_\epsilon \in L^2(\R^d)$. If, in addition, $\nabla f \in L^2(\R^d)$ then $\nabla f_\epsilon \in L^2(\R^d)$. We may apply the Nash inequality to find that
$$
\|f_\epsilon - f_\delta\|_{L^2(\R^d)} \le C \|f_\epsilon - f_\delta\|_{L^1(\R^d)} \|\nabla f_\epsilon - \nabla f_\delta\|_{L^2(\R^d)}
$$
for any $\epsilon,\delta > 0$. It follows that $\{f_\epsilon\}$ is Cauchy in $L^2(\R^d)$ so that $\{f_\epsilon\}$ has a limit in $L^2(\R^d)$. This limit is unique and thus must be $f$. It follows that $f \in L^2(\R^d)$.
