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Suppose that $f$ is continuously differentiable on $\mathbb{R}$ except at $x_1,\cdots,x_m$ where $f$ has jump discontinuities, and that its pointwise derivative $df/dx$ (defined except at $x_1,\cdots,x_m$) is in $L_{loc}^1(\mathbb{R})$. Prove that the distributional derivative of $f$ is given by: $$f' = (df/dx)+\sum_{j=1}^m[f(x_j+)-f(x_j-)]\tau_{x_j}\delta$$

So let $f \in L_{loc}^1(\mathbb{R})$ and wlog, let $x_1 < \cdots < x_m$. I know the following:

$$<f',\phi> = -<f,\phi'> = -\int_\mathbb{R} f\phi' dx = -\int_{-\infty}^{x_1}f\phi' dx - \int_{x_1}^{x_2}f\phi' dx - \cdots -\int_{x_m}^{\infty}f\phi' dx $$

Now using integration by parts for each term, I (think) I get:

$$\int_{-\infty}^{\infty} \frac{df}{dx}\phi(x) dx + \sum_{j=1}^m [f(x_j+)-f(x_j-)]\phi(x_j)$$

It's close, but I don't know how to make it equal to the above expression. Any help would be greatly appreciated!

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  • $\begingroup$ I would substract $\sum_{j=1}^m[f(x_j+)-f(x_j-)]\tau_{x_j} H$ to $f$, where $H$ is the Heaviside function. the obtained function $g$ is continuous and piecewise $C^1$, hence $g(x) = g(0) + \int_0^x g'(t)dt$, and the distributional derivative of $\sum_{j=1}^m[f(x_j+)-f(x_j-)]\tau_{x_j} H$ is clearly $\sum_{j=1}^m[f(x_j+)-f(x_j-)]\tau_{x_j} \delta$ $\endgroup$ – reuns May 18 '16 at 20:03
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You are basically finished - note that you can write $$ \sum_{j = 1}^{m}[f(x_j^+) - f(x_j^{-})]\phi(x_j) = \langle \sum_{j = 1}^{m}[f(x_j^+) - f(x_j^{-})]\tau_{x_j}\delta, \phi\rangle $$ where $\langle\tau_{x_j}\delta, \phi\rangle = \phi(x_j)$ denotes the Dirac measure at $x_j$. Now combine with $$ \int_{-\infty}^{\infty}\frac{df}{dx}\phi(x)d x = \langle \frac{df}{dx}, \phi\rangle. $$

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