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I would like help proving this elementary result:

Let $f\in L^{1}_{loc}(a,b)$. Let $x_0 \in (a,b)$ Let $F(x)=\int^{x}_{x_0} f$. Then $F'=f$ in the sense of distributions.

i.e How do I show $\langle F',\phi\rangle=\langle f,\phi\rangle$?

All I know is that $\langle F',\phi\rangle = -\langle F, \phi'\rangle$. I do not see how the right hand side is equal to $\langle f,\phi\rangle$

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1 Answer 1

up vote 2 down vote accepted

Let $\phi$ be a test function. Let $K$ be a compact interval in $(a,b)$ containing the support of $\phi$. Then $f\in L^1(K)$, as it's locally integrable and $K$ is compact.

We can approach $f$ in $L^1(K)$ by continuous functions $f_k$, e.g. $\lVert f-f_k\rVert\leq k^{-1}$. We have, integrating by parts: $$\int_a^b\int_{x_0}^xf_k(t)dt\phi'(x)dx=-\int_a^bf_k(x)\phi(x)dx.$$ Then take the limit $k\to +\infty$.

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sanity check: Integral of a continuous function is always differentiable? Yes, it is. thank you. I actually have a very similar proof in my book, I didn't understand why it was taking these limits. It is because you used CONTINUOUS functions. thanks so much –  Lost1 Oct 16 '12 at 20:17

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