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

Question

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 $\Phi(g)=\int \bar{f} g$?

Answer 1

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 $L^1_{\text{loc}}(\mathbb{R})$.

Answer 2

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.

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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$}$