Sobolev and distribution

Question

If $u \in L^{p}$ and continuous. And if $u' \in L^{p}$ (derivation for distribution) and continuous then $u$ is $C^{1}$ ?

EDIT : $L^{p}(\Omega)$ with $\Omega$ a open.

Thanks and regards.

Answer 1

The answer is yes. We have the following result:

Let $\Omega \subset \mathbb{R}^n$ be an open set. Assume that for $u \in C^0(\Omega) \subset L^1_{\text{loc}}(\Omega)$ there exists a function $v \in C^0(\Omega;\mathbb{R}^n)$ such that \begin{equation} \int_{\Omega} v \cdot \varphi = - \int_{\Omega}u\: \text{div} \,\varphi \quad \text{ for all } \varphi \in C_c^{\infty}(\Omega;\mathbb{R}^n)\, . \end{equation} Then $u \in C^1(\Omega)$ and $\nabla u = v$.

Smoothness is a local property; it is thus sufficient to show that for every $x \in \Omega$ there exists $K\Subset \Omega$ with $x \in K$ and $u \in C^1(K)$.

Fix $x \in \Omega$ and let $(\rho_{\varepsilon})_{\varepsilon}$ be a sequence of standard mollifiers. Choose $r > 0$ such that the ball $B_r(x)$ of radius $r$ around $x$ is compactly contained in $\Omega$. For all $0 < \varepsilon < r$ we can define the functions $u_{\varepsilon} := u * \rho_{\varepsilon} \in C^{\infty}(\Omega)$. Using basic properties of the convolution and the fact that $u\in C^0(\Omega)$ and $v \in C^0(\Omega;\mathbb{R}^n)$, we have \begin{align*} (a)&\quad u_{\varepsilon} \to u \quad \text{uniformly on every compact set } K \subset \Omega\, ,\\ (b)& \quad \nabla u_{\varepsilon}(y) = v * \rho_{\varepsilon} (y) \quad \text{ for all } y \in B_r(x)\, ,\\ (c)& \quad \nabla u_{\varepsilon} \to v \quad \text{uniformly on every compact set } K\subset \Omega\, . \end{align*}

Set $K := \overline{B_r(x)}$. Thanks to $(a)$ and $(c)$ the sequence $(u_{\varepsilon})_{\varepsilon}$ is a Cauchy sequence in $C^1(K)$. As $C^1(K)$ is complete, we find a function $f \in C^1(K)$ such that $u_{\varepsilon} \to f$ in $C^1(K)$. Again by $(a)$, we deduce that $u = f \in C^1(K)$. The fact that $\nabla u = v$ follows then from the fundamental lemma of the calculus of variations.