Show that f $f\in C(\mathbb{R}^n)$ and $\partial^{dist}_{j} f\in C(\mathbb{R}^n)$ then $\partial_j f$ exists at any point and is equal to $\partial^{dist}_{j} f$.
Where $\partial^{dist}_{j}$ is the derivaet in the distributional sense and $\partial_{j}$ is the derivate in the classic sense.
My approach: iIf $h$ is a mollifier, then $f*h_{\epsilon}$ is $C^{\infty}$. then $$\partial_{j}(f*h_{\epsilon})=\partial^{dist}_{j}(f*h_{\epsilon})=(\partial^{dist}_{j} f)*h_{\epsilon}$$
Now if take $\epsilon\to 0$
$$\partial_{j}(f*h_{\epsilon})=(\partial^{dist}_{j} f)*h_{\epsilon}\to \partial^{dist}_{j} f$$
But how can prove that this implies $\partial_{j} f$ exists at any point and is equal to $\partial^{dist}_{j} f$
The mentioned convergence holds uniformly on compact sets (at least if the support of the mollifier is compact). Thus, we can interchange integration and the limes $\varepsilon \downarrow 0$ to find \begin{align*} \int_a^b (\partial_j^{\mathrm{dist}} f )(x) \, dx_j &= \lim_{ \varepsilon \downarrow 0} \int_a^b \partial_j(f \ast h_\varepsilon)(x) \, d x_j \\ & =\lim_{ \varepsilon \downarrow 0} f \ast h_\varepsilon (x_1,\ldots,x_{j-1},b,x_{j+1},\ldots, x_n) - f \ast h_\varepsilon (x_1,\ldots,x_{j-1},a,x_{j+1},\ldots, x_n) \\ &= f (x_1,\ldots,x_{j-1},b,x_{j+1},\ldots, x_n) - f (x_1,\ldots,x_{j-1},a,x_{j+1},\ldots, x_n). \end{align*} Now the one-dimensional fundamental theorem of calculus states that the derivate in the $j$-th direction exists and is exactly $(\partial_j^{\mathrm{dist}} f )(x)$. Since the partial derivates are continuous, we conclude that $f$ is continuously differentiable.