Continuously differentiable and weak partial derivatives

Question

$A=\{f:\mathbb R^n \to \mathbb R | f \text{ is continuously differentiable}\}$

$B=\{f:\mathbb R^n \to \mathbb R | f \text{ has weak partial derivatives}\}$

I think that $B \subset A$ is true. Am I right?

Answer 1

Suppose $f$ is continuously differentiable. Then the partial derivatives exists and are continuous, hence $L^1_{\text{loc}}$ and the weak derivative are well defined.

A weak derivative $v$ of $f$ with respect to $x_i$ satisfies, for every test function $\phi$ :

$$\int_{\mathbb{R}^n} f \partial_{x_i} \phi = - \int_{\mathbb{R}^n} v \phi.$$

But since $\partial_{x_i} f$ is well defined (in its usual sense), integrating by parts shows that taking $v = \partial_{x_i} f$ a.e. is enough.

To sum up : a continuously differentiable function necessarily has weak partial derivatives, so $A \subset B$. And of course it is a proper subset (consider the absolute value). Meaning that functions with weak derivative does not necessarily have continuous derivatives (it may not have -usual- derivatives at all). Note to finish that weak derivatives coincide with derivatives as soon as $f$ is differentiable (you do not need continuity of the derivatives).

Answer 2

Counterexample - $\sqrt{|x|}$ as a function $\mathbb R^1 \to \mathbb R$ is once weakly differentiable and is not $C^1$