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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.

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  • $\begingroup$ Any continuous function is bounded on compact sets, and hence locally integrable. $\endgroup$ – Jeff Nov 30 '17 at 14:13
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Continuously differentiable implies continuous implies measurable and bounded on compact sets which is equivalent to locally integrable.

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