distributional derivative with dirac

Question

Let $f:\mathbb{R}\to\mathbb{R}$ with, $f(x)=1-x/\pi$ if $x\in(0,\pi]$ and $f(x)=cos(x)$ if $x\in[-\pi,0]$ and 0 in $\mathbb{R}\backslash[-\pi,\pi]$. I want calculate the distributional derivate, So my idea was $$\langle f',\varphi\rangle=-\langle f,\varphi'\rangle=-\left(\varphi(-\pi)+\int_{-\pi}^{0}\sin(x)\varphi(x)\,dx+\int_{0}^{\pi}\dfrac{1}{\pi}\varphi(x)\,dx\right)$$ $$=-\left(\int_{\mathbb{R}}\delta(x+\pi)\varphi(x)\,dx+\int_{-\pi}^{0}\sin(x)\varphi(x)\,dx+\int_{0}^{\pi}\dfrac{1}{\pi}\varphi(x)\,dx\right)$$

but I have a confusion to say which is the derivate, i dont know is right say that $f'=-\delta(x+\pi)-\sin(x)$ if $x\in[-\pi,0]$ and $1/\pi$ if $x\in[0,\pi]$. Or maybe the derivative is equal to $$ \int_{\mathbb{R}}\left[-\delta_{-\pi}(x)\mathbb{1}_{\mathbb{R}\backslash(-\pi,\pi]}(x)-\sin(x)\mathbb{1}_{(-\pi,0]}(x)-\frac{1}{\pi}\mathbb{1}_{(0,\pi]}(x)\right]\varphi(x)dx $$

I don't know if the expression $\delta_{-\pi}(x)\mathbb{1}_{\mathbb{R}\backslash(-\pi,\pi]}(x)$ is right, because this dirac is 1 only on the point $-\pi$.

Thanks!!

Answer 1

Your expression at the end of your computation (your first equation) is correct. Now you do not need any indicator function near the Dirac, $\delta_{-\pi}$ is already supported at $x=\pi$. Hence you can write $$ f'(x) = -\delta_{-\pi}(x) - \sin(x)\,\mathbf{1}_{(-\pi,0]}(x) - \frac{1}{\pi}\,\mathbf{1}_{(0,\pi]}(x) $$ in the sense of distributions. Or even more rigorously, since this is an equality of distributions and not a pointwise identity, you can write $$ f' = -\delta_{-\pi} - \sin\,\mathbf{1}_{(-\pi,0]} - \frac{1}{\pi}\,\mathbf{1}_{(0,\pi]}. $$

Answer 2

With $$ f(x) = \begin{cases} \cos x, & -\pi\leq x\leq 0 \\ 1-x/\pi, & 0<x\leq\pi \\ 0, & \text{otherwise} \end{cases} $$ we get $$ \langle f', \varphi \rangle = -\langle f, \varphi' \rangle = - \left( \int_{-\pi}^{0} \cos x \, \varphi'(x) \, dx + \int_{0}^{\pi} (1-x/\pi) \, \varphi'(x) \, dx \right) \\ - \left( \left[\cos x \, \varphi(x) \right]_{-\pi}^{0} - \int_{-\pi}^{0} (-\sin x) \, \varphi(x) \, dx + \left[(1-x/\pi) \, \varphi(x)\right]_{0}^{\pi} - \int_{0}^{\pi} (-1/\pi) \, \varphi(x) \, dx \right) \\ = \varphi(-\pi) - \int_{-\pi}^{0} \sin x \, \varphi(x) \, dx + \int_{0}^{\pi} (-1/\pi) \varphi(x) \, dx \\ = \langle \delta_{-\pi} + g, \varphi \rangle, $$ where $$ g(x) = \begin{cases} -\sin x, & -\pi \leq x \leq 0 \\ -1/\pi, & 0 < x \leq \pi, \\ 0, & \text{otherwise} \end{cases} $$

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Using the rules

• $(fu)' = f'u + fu',$ where $f$ is a $C^\infty$ function and $u$ a distribution,
• $\mathbf{1}_{(a,b)}' = \delta_a - \delta_b,$
• $f\delta_a = f(a)\delta_a$ we can do the calculation directly on a distribution.

First we write $f$ as $$ f(x) = (\cos x) \mathbf{1}_{[-\pi,0]}(x) + (1-x/\pi) \mathbf{1}_{(0,\pi]}(x) $$ and then take the derivative using the rules above: $$ f' = (\cos x)' \mathbf{1}_{[-\pi,0]} + (\cos x) \mathbf{1}_{[-\pi,0]}' + (1-x/\pi)' \mathbf{1}_{(0,\pi]} + (1-x/\pi) \mathbf{1}'_{(0,\pi]} \\ = (-\sin x) \mathbf{1}_{[-\pi,0]}(x) + (\cos x) (\delta_{-\pi} - \delta_0) + (-1/\pi) \mathbf{1}_{(0,\pi]} + (1-x/\pi) (\delta_0 - \delta_\pi) \\ = (-\sin x) \mathbf{1}_{[-\pi,0]}(x) + (-\delta_{-\pi} - \delta_0) + (-1/\pi) \mathbf{1}_{(0,\pi]} + (\delta_0 - 0\delta_\pi) \\ = -\delta_{-\pi} + (-\sin x) \mathbf{1}_{[-\pi,0]}(x) + (-1/\pi) \mathbf{1}_{(0,\pi]}. \\ $$