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I know thats the following statement is true.

$f,g$ are continuous function $[a,b]$.Suppose $\int\limits_a^bf(t)h(t)+g(t)h'(t) \, dt=0$ for every $h$ belonging to $C_0^{\infty}[a,b]$, then $g$ is differentiable and $\dot{g}(t)=f(t),\text{ }t\in[a,b] \text{ a.e.}$

Is the theorem also true for functions $f,g \in L^1[a,b]$ and can you give me a reference for it?

Michael Hardy
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John
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1 Answers1

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Let $g, f \in L^1[a,b]$, and suppose that $\nu$ is a weak derivative for $g$. That means $$\int_a^b g h' dx = - \int_a^b \nu h dx$$ for all $h \in C^\infty_0[a,b]$.

If $\int_a^b fh + gh' dx = 0$ for all $h \in C^\infty_0$, then $$\int_a^b h fdx = \int_a^b h \nu dx.$$

Therefore when viewed as measures $f$ and $\nu$ agree after integration against $C^\infty_0$ functions. Which means that $f = \nu$ almost everywhere. This comes from measure theory.

Joel
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  • What I really proved here is the uniqueness of a weak derivative (a.e.). You can find this in Evan's Partial Differential Equations book section 5.2.1. – Joel Apr 21 '14 at 21:44
  • The DuBois-Reymond Lemma tells us that weak differentiability of a continuous $g$ implies differentiability in the normal sense. – Joel Apr 22 '14 at 13:07
  • ... which makes the $L_1$ counterpart of the lemma kind of useless since it does not give us more differentiability than the original one. It is necessary to put more conditions on the derivative to make it interesting. – A.Γ. Jul 12 '15 at 14:18