# Is the derivative of a locally integrable function always a locally integrable function?

Let $$f$$ be a locally integrable function on $$\mathbb{R}$$, that is, $$f$$ is Lebesgue-integrable on every compact subset. Consider the application : $$\Phi : C_C(\mathbb{R}) \rightarrow \mathbb{R}$$ defined by $$\Phi(g)=-\int fg'$$. Is $$\Phi$$ necessarily a locally integrable distribution, that is, does there exist a locally integrable function $$\bar{f}$$ for which $$\Phi$$ is defined by $$\Phi(g)=\int \bar{f} g$$?

What you are seeking is the local Sobolev space $$W^{1,1}_{\text{loc}}(\mathbb{R})$$. This consists of locally integrable functions that admits a locally integrable weak derivative. And no this space is not equal to $$L^1_{\text{loc}}(\mathbb{R})$$.

• A concrete example will be better. Commented May 13, 2021 at 11:51

I was stuck here before but not any more. I will present you a concrete example rather than quoting something fancy.

First we note that $$g$$ vanishes at infinity and so the product $$fg$$ vanishes at infinity too. By integration by parts(or the Divergence Theorem in higher dimensions), we can conclude that, if $$f^\prime$$ exists, then $$\int \bar{f}g=-\int f g^\prime=\int f^\prime g$$, where the last equality follows from integration by parts. Since this holds for all $$g\in C_c(\mathbb R)$$, we have that $$f^\prime =\bar{f}$$ a.e.(This step requires some efforts. First we note that the set $$\{(f^\prime -\bar{f}\,)\neq0 \}$$ is a measurable set. Since $$\mathbb R$$ is $$\sigma-$$finite, we can assume that set is of finite measures. And note that every Lebesgue measurable set with finite measures can be approximated from inside by compact sets. We take $$g_n$$ as the indication functions of these sets respectively. And then applying the Lebesgue dominate convergence theorem, we finally obtain the conclusion that $$f^\prime =\bar{f}$$ a.e.). So $$f^\prime$$ is locally integrable if and only if $$\bar{f}$$ is locally integrable.

Now consider the example: $$f(x)=\frac{1}{x^s}$$ with any fixed $$s\in(0,1)$$.

From elementary calculus we know $$f$$ is locally integrable.

But $$f^\prime(x)=s \frac{1}{x^{(s+1)}}.$$

Note that $$s+1>1$$, and so $$f^\prime$$ is not locally integrable near the origin. Hence, $$\bar{f}$$ is not locally integrable too.

$$\tag*{\blacksquare}$$

• I don't believe that $f'$ is not locally integrable. Locally integrable means that function is integrable on any compact set. This together with $f'$ is continuous can show that it is locally integrable.
– xxxg
Commented Jun 26 at 11:15