Define a continuously differentiable function to be a function $f: \mathbb{R}^n \to \mathbb{R}$ which has a continuous derivative. Define a weakly differentiable function to be a function $f: \mathbb{R}^n \to \mathbb{R}$ which is locally integrable and there exists $n$ locally integrable functions $g_1, \dots, g_n$ which satisfy the integration by parts formula, $$\int_{\mathbb{R}^p} f(x) \frac{\partial \varphi}{\partial x_j} (x) \, \mathrm{d}x = - \int_{\mathbb{R}^p} g_j(x) \varphi(x) \, \mathrm{d}x,$$ for all $j \in \{1, \dots, n\}$ and for any function $\varphi$ that is any infinitely differentiable function with compact support.

I've come across the claim that "if a function $f$ is continuously differentiable, then it is weakly differentiable." How can that be true? Continuously differentiable functions needn't be locally integrable it seems.

  • $\begingroup$ Any continuous function is bounded on compact sets, and hence locally integrable. $\endgroup$
    – Jeff
    Nov 30, 2017 at 14:13

1 Answer 1


Continuously differentiable implies continuous implies measurable and bounded on compact sets which is equivalent to locally integrable.

  • $\begingroup$ Is the last equivalence correct? f(x)=|x|^{-1/2} is locally integrable on (-1,1) but not bounded on every compact subset of (-1,1). $\endgroup$
    – jlewk
    Jan 26, 2020 at 3:11
  • $\begingroup$ That function is not continuous at $0$. $\endgroup$
    – user510186
    Feb 19, 2022 at 21:01

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