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Let $(a,b)$ and $f\in L_{\text{loc}}^1(a,b)$. For $x_0\in (a,b)$, consider $$ F(t)=\int_{x_0}^{t}f(s)ds. $$ Prove that, $DF=f$ (towards theoretical distribution).

I thought of the following: Let $\varphi$, we had $$ \left\langle DF, \varphi \right\rangle = \left\langle F, \dfrac{d}{dx}\varphi \right\rangle=\int_a^bF(x)\dfrac{d}{dx}\varphi (x)dx = \int_a^b \int_{x_0}^xf(s)ds\dfrac{d}{dx}\varphi (x)dx... $$ but it leads to nothing. I also thought about, $$ \left| \left\langle DF, \varphi \right\rangle - \left\langle f, \varphi \right\rangle \right| = \left| \left\langle F, \dfrac{d}{dx}\varphi \right\rangle - \left\langle f, \varphi \right\rangle \right|=\int_a^bF\dfrac{d}{dx}\varphi-\int_a^bf\varphi... $$

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First assume that $f$ is continuous. Then $F$ is $C^1$ and $F' = f$. Also \begin{align} -\int_a^b F(x) \varphi'(x) dx &=\int_a^b F'(x) \varphi(x) dx = \int_a^b f(x) \varphi (x)dx. \end{align}

Thus $DF = f$ in the distribution sense. Now for any $f\in L^1_{loc}(a, b)$ and any test function $\varphi$, let $I \subset (a, b)$ be a compact interval which contains the support of $\varphi$ and $x_0$. Let $\{f_n\} \in L^1(a, b)$ be a sequence of continuous functions to converges to $f$ in $L^1(I)$. Let $F_n(x) = \int_{x_0}^x f_n (t) dx$. One can check that $F_n$ converges to $F$ in $L^1(I)$. Thus

\begin{align} -\int_a^b F(x) \varphi'(x) dx &= -\lim_{n\to \infty} \int_a^b F_n(x) \varphi'(x) dx \\ &= \lim_{n\to \infty} \int_a^b F'_n(x) \varphi(x) dx \\ &=\lim_{n\to \infty} \int_a^b f_n(x) \varphi(x) dx \\ &=\int_a^b f(x) \varphi(x) dx. \end{align}

Thus $DF = f$ also in the distribution sense.

Arctic Char
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