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I want to calculate the first and second distributional derivate of the $2\pi$-periodic function $f(t) = \frac{\pi}{4} |t|$, it is $$ \langle f', \phi \rangle = - \langle f, \phi' \rangle = -\int_{-\infty}^{\infty} f \phi' \mathrm{d}x $$ and $$ \langle f'', \phi \rangle = \langle f, \phi'' \rangle = \int_{-\infty}^{\infty} f \phi'' \mathrm{d}x $$ so have to evaluate those integrals, but other than $\phi$ is smooth and has compact support i know almost nothing about $\phi$, so it is enough to integrate over finite bound, but how does i use the definition of $f$? So how could i calculate those integrals?

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Integrate by parts, and use the fact that $\phi(x)\to 0$ when $x\to \infty$ –  Norbert Jun 23 '12 at 12:37
    
for the first term i got $-\int_{-\infty}^{\infty} f \phi' \mathrm{d}x = -[ f \phi ]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} f' \phi \mathrm{d}x$, the first summand vanishes because of the compact support ($\phi(x) \to 0$ for $x\to \infty$). the derivate of $f$ is $f''(t) = -\pi/4$ for $t \in [-\pi,0]$ and $f''(t) = \pi/4$ for $t\in [0,\pi]$, and $f(t) = f(t + 2\pi)$. so now i have difficulty to bring my knowledge of $f'$ in an derivation of an expression for the derivational derivate? –  Stefan Jun 23 '12 at 12:54
    
Your conclusions is wrong. After you have integrated by patrs you get must get derivative of $f$, not a second derivative. –  Norbert Jun 23 '12 at 12:58
    
oh yes, i am sorry that was a typo, i mean the first derivative $f'$. –  Stefan Jun 23 '12 at 13:01
    
So have you heard of Heaviside function and its derivative? –  Norbert Jun 23 '12 at 13:05

2 Answers 2

Let $G(t)=\max(t,0)$. The $G'=H$ where $H$ is the Heaviside function ($H(t)=1$ if $t>0$ and $0$ otherwise). Indeed, for any test function $\phi$ we have $$\langle G, \phi'\rangle = \int_0^{\infty} t\phi'(t)\,dt = - \int_0^{\infty} \phi(t)\,dt = - \langle H, \phi\rangle$$ in agreement with the definition of the distributional derivative. Also, $H'=\delta_0$ by a similar argument: $$\langle H, \phi'\rangle = \int_1^{\infty} \phi'(t)\,dt = -\phi(0) = -\langle \delta_0, \phi\rangle$$

Any piecewise linear function can be written in terms of the translates of $G$. To do this for your function, you can begin with $f_0(t)=\frac{\pi}{4}G(t)-\frac{\pi}{2}G(t-\pi)+\frac{\pi}{4}G(t-2\pi)$ which agrees with $f$ on $[0,2\pi]$ and is zero elsewhere. Note that $f(t)=\sum_{n\in\mathbb Z}f_0(t+2\pi n )$, where the sum converges in a very strong sense: for any finite interval $I$, there exists $N$ such that $f(t)=\sum_{|n|\le N}f_0(t+2\pi n )$ for $t\in I$. This kind of convergence implies convergence of all distributional derivatives, since test functions have compact support.

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ok, i tried to solve, for the first and second derivate i got $$ f'(t) = \frac{\pi}{4}(2H(t)-1) $$ (where $H(t)$ denotes the Heaviside-Function) on $[-\pi, \pi]$ and $f'(t) = f'(t+2\pi)$. For the second derivate i got $$ f''(t) = \frac{\pi}{4}\delta_0(t) $$ (where $\delta_0$ ist the Dirac-Delta at $0$) on $[-\pi, \pi]$ and $f''(t) = f''(t+2\pi)$. So for the distibutional derivate $$ \langle f', \phi \rangle = - \langle f, \phi' \rangle = -\int f \phi' = -\int f'\phi = \langle \frac{\pi}{4}(2H-1), \phi \rangle $$ and $$ \langle f'', \phi \rangle = \langle \frac{\pi}{4}\delta_0, \phi \rangle $$ is this right and is my notation acceptable?

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The second derivative should have a negative multiple of Dirac mass at $\pi+2\pi n$. –  user31373 Jun 23 '12 at 20:38
    
so i should be $$f''(t) = \frac{\pi}{4}\delta_0(t) - \frac{\pi}{4}\delta_0(t + \pi)$$ –  Stefan Jun 23 '12 at 21:06
    
Yes, and then by periodicity. The angle at the endpoints of the initially defined period is easy to miss. –  user31373 Jun 23 '12 at 21:08

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