MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $T \notin L^1_{loc}(\mathbb{R})$?

share|cite|improve this question
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
up vote 3 down vote accepted

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 $F+\alpha$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.