# prove $\mathcal F(f) = c_1\delta + c_2 \delta' + c_3\delta'' + T_g$ with $f(x)=|x^2 -1|$

let $f(x)=|x^2 -1|$ be a tempered distribution (i proved it) , and calculated its 3rd derivation (as a distribution) and then this stopped me :

prove that we have :

$$\mathcal F(f) = c_1\delta + c_2 \delta' + c_3\delta'' + T_g$$

with $T_g$ is a Tempered distribution and $c_1,c_2,c_3 \in \mathbb R$ and $<\delta,\phi>= \phi(0)$ (the dirac distribution) , calculate $c_1, c_2 , c_3 .$

i tried to do the following :

they asked me before this to calculate $(f)'''$ (as a distribution) , and i thought there will be a connection between $(f)'''$ and the fact that $\mathcal F(f^{(m)})(x)= (2i\pi x)^m. \mathcal F(f(t))$ soo i calculated $(f)'''$ and i got :

$$f''' = 4 \delta'_{-1} + 4\delta'_1 - 4\delta_{-1} + 4\delta_1$$

soo we have :

$$\mathcal F(f''')=(2i\pi x)^3 . F(f)$$ $$\Rightarrow F(4 \delta'_{-1} + 4\delta'_1 - 4\delta_{-1} + 4\delta_1) = (2i\pi x)^3 \mathcal F(f)$$

and i stopped here

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Could you include your attempts? – Davide Giraudo May 26 '13 at 12:49
i also tried to apply $\overline F$ to the equation and played with it a bit but didnt failed – Lofaif May 26 '13 at 14:16
@Lofaif In your last equation you wrote $f''' = (2i\pi x)^3\mathcal{F}(f)$ and not $\mathcal{F}(f''')$? Maybe this stopped you? – Vobo May 26 '13 at 20:01
yeah that was a mistake , however .. when i corrected it it just got more complicated . i will have to do an exam on distributions tomorrow and im almost sure that this question will come up , however i couldnt find the damn answer -_- – Lofaif May 26 '13 at 20:14
@Lofaif You know that e.g. $\mathcal{F}\delta'_{-1} = (2i\pi x)e^{-2i\pi x}$? – Vobo May 26 '13 at 20:26

Hint: you can calculate the derivatives directly, without resorting to Fourrier transform. Just apply the following formula:

If $g$ is distribution represented by piece-wise $\mathcal C^1$ function on $\mathbb R$ with jumps at points $x_n<x_{n+1}$, then its derivative is equal to $f'= \sum_n \{f(x_n)\}\delta_{x_n} + \sum_n \left(f\large|_{(x_n,x_{n+1})}\right)'$, where $\{f(x_n)\}$ is a jump of $f$ in the point $x_n$.

It is known as formule des sauts (formula of jumps) in French but I don't know its name in English.

For example, if you have a distribution $g = \begin{cases}-x,&x\le 0,\\1,& x>0\end{cases}$, then by this formula $g'= \delta_0 -\mathbf{1}_{x\le 0}$.

In your case, the function is continuous, then its first derivative can be expressed directly

$f'= 2x\,\mathbf{1}_{x<-1} +2x\,\mathbf{1}_{x>1} -2x\,\mathbf{1}_{|x|\le 1}$. You can continue with Leibnitz's rule and this jump formula until you arrite to $f''' = 4(\delta_1'+\delta_1-\delta_{-1}+\delta'_{-1})$.

Now we need to apply the Fourier transform to these functions: it's well-known that $\mathcal F(\delta_a) = e^{2\pi i a\xi}$ and $\mathcal F(\delta'_a) = -2\pi i \xi e^{2\pi i a\xi}$ (signs and constants may vary depending on the conventions in the definition of the transform). Hence we get

$\mathcal F(f''') = 4(e^{2\pi i \xi}-e^{-2\pi i \xi}-2\pi i \xi e^{2\pi i \xi} -2\pi i \xi e^{-2\pi i \xi}) = 8i\sin (2\pi\xi) - 16\pi i\xi \cos(2\pi\xi) = -8\pi^3i\xi^3\mathcal F(f).$

Now, whenever we solve in the sense of distributions the equation $xT=S$, we obtain $T= c\delta_0 + \frac 1x S$ with $c$- an arbitrary constant (it's a good exercise to see why we have this family of solutions). Similarly, $x^3T=S$ would produce $\delta'$ and $\delta''$, because $\{\delta^{(k)}_0\}_{k=0..n-1}$ are linearly independent and generate all solutions of $x^nT=0$.

Hence, we can write $\mathcal F(f) =\frac{ \sin (2\pi\xi) - 2\pi \xi \cos(2\pi\xi) }{ - \pi^3 \xi^3 }+c_0\delta +c_1\delta'+c_2\delta''$. Apparently, the first term is a continuous, even, and integrable function on $\mathbb R$.

In order to deal with constants $c_i$, we take well-chosen test functions $\varphi$. It's known, that a function and its Fourier transform are even/odd simultaneously. So, if $\varphi$ is odd, then $\mathcal F(\varphi)$ is odd and this function and its second derivative are both zero in $\xi=0$. We now can write $0=<f,\varphi>=c_1<\delta',\mathcal F(\varphi)>$. This assures that $c_1=0$.

The final step involves the following statement: if $f$ is even, then $\mathcal F(\mathcal F(f))=f$. Hence, $\mathcal F\left( \frac{ \sin (2\pi\xi) - 2\pi \xi \cos(2\pi\xi) }{ - \pi^3 \xi^3 }+c_0\delta +c_2\delta''\right) = g(x)+c_0-4\pi^2 c_2 x^2$ where $g\in \mathcal C\cap\mathcal L^2\cap\mathcal L^\infty$ and $g(x)\to 0$ as $|x|\to\infty$ because it's an image of a $\mathcal C^{\infty}$ function. As we know the behaviour of $f$ on infinity, we conclude that $c_0=-1$ and $c_2 = -\frac{1}{4\pi^2}$.

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Looks good. I removed my parallel answer as I messed up convolution and division by x. The key idea are the general solutions of $x^3 S = T$, where the $\delta$ and its derivatives come into the game. – Vobo May 26 '13 at 20:54