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My knowledge of analysis is admittedly awful, and the following situation has come up in a paper I'm reading: Suppose that $f \in H^2([0,1])$ has the property that $f''(x) = ax + b\ $ a.e. on $(a,b) \subset [0,1]$, where $f''$ denotes the weak second derivative of $f$. The authors claim without further justification that this implies that $f \vert_{[a,b]}$ is smooth, but I'm not sure how to convince myself of this.

My best guess is that $f''$ being smooth a.e. implies that $f$ is smooth a.e. (is this correct?). But since $f \in H^2([0,1])$, Sobolev embeddings imply that it is absolutely continuous, and hence must in fact be smooth everywhere.

SubtleMechanism
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1 Answers1

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Define $g:[a,b] \to \mathbb R$ by $$ g(x) = \int_a^x \int_a^y (az+b)dz \ dy, $$ so that $g$ is $C^\infty$. Then weak and classical derivatives of $g$ agree, and $g'' = f''$ a.e. for the second weak derivative. This implies $g' = f' + c$ a.e., $c\in \mathbb R$, and $g = f + cx + d$ a.e., $d\in \mathbb R$. Then $f$ is smooth on a the interval $(a,b)$, in fact it is equal to a fourth-order polynomial in your case.

The latter depends on a technical result that a function with weak derivative zero is equal to a constant function, see, e.g., Weak derivative zero implies constant function

daw
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