# On distributions over $\mathbb R$ whose derivatives vanishes

Let $I \subset \mathbb R$ be open, $u \in \mathcal D'(I)$ be a distribution whose distributional derivatives vanishes (i.e. is zero for all test functions, which we may assume to be complex valued ).

We show $\forall c \in \mathbb C: \forall \phi \in \mathcal D(I) : u(\phi) = \int c\cdot\phi dx$. (EDIT: Correctly, $c$ should be quantified with $\exists$. My question has been why the following proof doesn't allow for arbitrary complex $c$, which explains the preceding statement.)

Proof:

Let $\Psi \in D(I)$. $\Psi$ is the derivative of a test function iff $\int \phi dx = 0$. In that case $u(\Psi) = 0$.

Let $h \in D(I)$ be arbitrary with $\int h dx = 1$. Now for every test function $\phi \in D(I)$ we see:

$\phi - \int \phi dx \cdot h \in D(I)$ and $\int ( \phi - \int \phi dx h ) dx = 0$.

therefore

$u( \phi - \int \phi dx h ) = 0$, i.e. $u(\phi) = u(h) \int \phi dx$

$\square$

If this is not wrong, how can I interpret the fact that $h$ has been arbitrary?

Note: This is part of a larger proof, which shows the same for non-one-dimensional domains.

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## 2 Answers

The function $h$ is not arbitrary; it is arbitrary with respect to the condition that $\int h dx = 1$. (And this latter condition is certainly necessary for the proof to go through.)

The proof given (which seems correct to me) shows that $u(\phi)$ only depends on the value of $\int \phi$. In particular, one sees (after the proof is done) that $u(h)$ only depends on the value of $\int h dx$, which was fixed to be $1$. Thus $u(h)$ is independ of the choice of $h$ (as long as $\int h dx = 1$), and is equal to the constant $c$ in the statement of the theorem.

[Added: As Theo points out in his answer, you have an incorrect universal quantifier on the constant $c$ in the statement of the theorem; it should read for some $c$. I didn't notice this when I read the question!]

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The statement you're actually proving is: If the distributional derivatives of $u$ vanish then there exists $c$ such that $u(\phi) = c \int \phi$. (You're stating a completely different property: $\forall c$...) If you suppose for a moment that $u(\phi) = c \int \phi$, how can you recover $c$? Well, by evaluating $u$ at any test function $h$ with $\int h = 1$ (this is a restriction but not a severe one). Apart from the slip of confusing $\forall$ with $\exists$ your argument is correct.

In fact, the trick of using $h$ with $\int h = 1$ is used quite often in the theory of distributions and the fact that it doesn't matter which $h$ you take is one of the big strengths of the theory of distributions. For instance, if $h$ is such that $\int h = 1$ then $u_{n}(\phi) = \int nh(nx) \phi(x)$ approximates the Dirac $\delta_{0}$-distribution in the sense that $u_{n}(\phi) \to \phi(0) = \delta_{0}(\phi)$.

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