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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?

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Have you tried invoking the definition of distributional derivative? –  Hurkyl Jun 26 '12 at 14:24
    
See also: math.stackexchange.com/q/147748 and math.stackexchange.com/q/28018 –  t.b. Jun 26 '12 at 14:51
    
Also related: math.stackexchange.com/questions/118525/… –  mrf Jun 26 '12 at 15:53
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1 Answer

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.

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