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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?

Michael Hardy
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Will Burke
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  • Have you tried invoking the definition of distributional derivative? –  Jun 26 '12 at 14:24
  • See also: http://math.stackexchange.com/q/147748 and http://math.stackexchange.com/q/28018/ – t.b. Jun 26 '12 at 14:51
  • Also related: http://math.stackexchange.com/questions/118525/a-sufficient-condition-for-a-function-to-be-of-class-c2-in-the-weak-sense/120894#120894 – mrf Jun 26 '12 at 15:53

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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.

Siminore
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