# Distributional derivative is locally integrable - then the distribution as well?

Given a distribution $T \in D'(\mathbb{R})$ such that the distributional derivative $\partial T \in L^1_{loc}(\mathbb{R})$. Can one deduce that $T \in L^1_{loc}(\mathbb{R})$ as well? Or can anyone give me an example where $T \notin L^1_{loc}(\mathbb{R})$?

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It's not just in $L^1_{loc}$; it's actually continuous. You should be able to prove this yourself. – Deane Yang Mar 26 '12 at 21:48

Let $T$ be your distribution. By hypothesis $$T'(\phi)=-T(\phi')=\int_\mathbb R f\phi'dx,$$ where $\phi$ is a test function and $f\in L^1_{\text{loc}}$.

Fix $x_0\in\mathbb R$ and define $$F(x):=\int_{x_0}^x f(x)dx.$$ Then $F$ is a continuous function, which is almost everywhere differentiable; moreover, if $F'(x)$ is defined at some point $x$, then it equals $f(x)$, so that $F'$ and $f$ define the same element in $L^1_{\text{loc}}$.

Then, the distribution $T$ is given by integration against $F$ up to some additive constant. Indeed, by integration by parts one obtains $$\int_\mathbb R F\phi' dx=-\int_\mathbb R F'\phi dx=-\int_\mathbb R f\phi'dx=T(\phi')$$ This shows that the distribution $T_F$ defined by $F$ equals $T$ on every test function which is the derivative of a test function (the point is that a priori a primitive of a test function is no longer a test function in general). To conclude you just need the following elementary lemma:

Lemma. Let $S$ be a distribution such that $S'=0$. Then, $S(\bullet)=\alpha\int_\mathbb R\bullet dx$, for some $\alpha\in\mathbb R$.

Proof. Let $\alpha:=S(\phi_0)$, where $\phi_0$ is a test function such that $\int_\mathbb R\phi_0 dx=1$. Let $\phi$ be any test function and write it as $\phi=(\phi-c\phi_0)+c\phi_0$, where $c=\int_\mathbb R\phi dx$. Then, $\int_\mathbb R(\phi-c\phi_0)dx=0$ so that $\phi-c\phi_0$ admits a primitive which is actually a test function, for instance $$\Phi_c:=\int_{-\infty}^x(\phi(t)-c\phi_0(t))dt.$$ Therefore, $S(\phi-c\phi_0)=S(\Phi_c')=-S'(\Phi_c)=0$. But then, $$S(\phi)=c S(\phi_0)=\alpha\int_\mathbb R\phi dx.\qquad\qquad\square$$ Thus, since $(T_F-T)'(\phi)=T(\phi')-T_F(\phi')=0$ for every test function $\phi$, you have that $T=T_F+\alpha\int_\mathbb R\bullet dx$, for some $\alpha\in\mathbb R$.

Therefore, $T=T_{F+\alpha}$ and $T$ is the distribution associated to the continuous almost everywhere differentiable function $F+\alpha$.

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