Let $T \in \mathcal{D}'(0,1)$ be such that $\frac{dT}{dx} \in L^2(0,1)$. Show that $T \in L^2(0,1)$.
My attempt:
Let $\frac{dT}{dx} =f$ and $g=\int_{0}^{x} f(x) dx$. Then $g'=f=\frac{dT}{dx}$. Hence, $T-g=c$.
How do I proceed from here?
If the derivative of distribution is square integrable, then so is the distribution
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functional-analysis
sobolev-spaces
distribution-theory
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0What's missing is to show that $g \in L^2(0,1)$. – 2017-02-14
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0@DanielFischer, $g$ is absolutely continuous hence continuous on $(0,1)$, hence square integrable? – 2017-02-14
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1Pretty much, but you need to show that $g$ doesn't grow too fast near $1$. That follows immediately from $f \in L^2(0,1) \subset L^1(0,1)$ for example. – 2017-02-14
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0And the lower limit of the integral ought to be $0$, $f$ isn't defined outside $(0,1)$. – 2017-02-14
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0I fixed that. Thanks. – 2017-02-14