Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

If a real function $f\colon[a,b]\to\mathbb{R}$ is differentiable and its derivative $f'$ is zero, then $f$ is constant. Does this result still hold when $f$ has a weak derivative?

Explicitly, suppose $f\colon[a,b]\to\mathbb{R}$ is an integrable function such that its distributional derivative $Df$ is zero. Does this mean that $f$ is constant?

share|cite|improve this question
Have you tried invoking the definition of distributional derivative? – Hurkyl Jun 26 '12 at 14:24
Also related:… – mrf Jun 26 '12 at 15:53

The following corollary of the celebrated Du Boys-Reymond Lemma holds true.

Corollary. If $u \in L^1_{\mathrm{loc}}(a,b)$ is such that $$\int_a^b u(x) \varphi'(x)\, dx=0$$ for every $\varphi \in C_0^\infty(a,b)$, then $u$ is almost everywhere constant.

I wrote the statement of B. Dacorogna, Introduction au calcul des variations.

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.