### Author Topic: Bonus Web Problem--4  (Read 9513 times)

#### Victor Ivrii

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##### Bonus Web Problem--4
« on: October 27, 2012, 06:21:01 AM »
Don't post until November 7, 21:30

Prove the following properties of convolution:

• (a) $f*g=g*f$
• (b) $(f*g)'=f'*g = f*g'$, where $'$ denotes the derivative in one variable
• (c) $f*(g*h)=(f*g)*h$
• (d) Let $x_+^\lambda := x^\lambda$ as $x>0$ and $0$ as $x<0$. Prove that for $f$ which fast decays as $x\to -\infty$ and $n=1,2,\ldots$
\begin{equation*}
\frac{x_+^{n-1}}{(n-1)!} * f =\underbrace{\int_{-\infty}^x \int_{-\infty}^{x_1} \ldots \int_{-\infty}^{x_{n-1}}}_{\text{$n$ integrals}} f(x_n)\, dx_n \cdots dx_1
\end{equation*}

#### Calvin Arnott

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##### Re: Bonus Web Problem--4
« Reply #1 on: November 07, 2012, 09:31:19 PM »
[\thicksim \left(f \ast g\right) = \left(g \ast f\right) \thicksim]
[\text{Let:  }t=\left(x-y\right) \implies y=\left(x-t\right),    dy=\left(-dt\right).   \text{Then:}]
[\left(f \ast g\right)\left(x\right) = \int_{-\infty}^{\infty}f\left(x-y\right)g\left(y\right)dy = \int_{\infty}^{-\infty}f\left(t\right)g\left(x-t\right)\left(-dt\right) = -\int_{\infty}^{-\infty}g\left(x-t\right)f\left(t\right)dt = \int_{-\infty}^{\infty}g\left(x-t\right)f\left(t\right)dt = \left(g \ast f\right)\left(x\right)    \blacksquare]

[\thicksim \left(f' \ast g\right) = \left(f \ast g\right)' = \left(f \ast g'\right) \thicksim]
[\text{We have:    }\left(f \ast g\right) = \left(g \ast f\right),   \text{so it suffices to show:    }\left(f \ast g\right)' = \left(f' \ast g\right)    \text{as:    }\left(f \ast g\right)' = \left(g \ast f\right)' = \left(g' \ast f\right)   \text{since both    }\{f,g\}   \text{arbitrary.}]
[\text{Let:  }\{f,g\}\left(x\right)   \text{be nice enough functions that we can differentiate them under the sign. Then:}]
[\left(f \ast g\right)'\left(x\right) = \partial_x\int_{-\infty}^{\infty}f\left(x-y\right)g\left(y\right)dy = \int_{-\infty}^{\infty}\partial_x \left(f\left(x-y\right)g\left(y\right)\right)dy = \int_{-\infty}^{\infty}f'\left(x-y\right)g\left(y\right)dy = \left(f' \ast g\right)\left(x\right)    \blacksquare]

[\thicksim \left(f\ast\left(g \ast h\right)\right) = \left(\left(f \ast g\right)\ast h\right) \thicksim]
[\text{Let:  }\{f,g,h\}\left(x\right)   \text{be nice enough functions that the order of integration is arbitrary for any combination of them.}]
[\left(f\ast\left(g \ast h\right)\right)\left(x\right) = \left(f\left(x\right)\ast\int_{-\infty}^{\infty}g\left(x-s\right)h\left(s\right)ds\right) = \int_{-\infty}^{\infty}f\left(x-t\right)\int_{-\infty}^{\infty}g\left(t-s\right)h\left(s\right)dsdt = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x-t\right)g\left(t-s\right)h\left(s\right)dsdt]
[\text{Let:  }u=\left(t-s\right) \implies t=\left(u+s\right),    dt=du.    \text{Then:  }]
[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x-\left(u+s\right)\right)g\left(u\right)h\left(s\right)dsdu = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(\left(x-s\right)-u\right)g\left(u\right)h\left(s\right)duds = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f\left(\left(x-s\right)-u\right)g\left(u\right)du\right)h\left(s\right)ds]
[= \left(\int_{-\infty}^{\infty}f\left(x-u\right)g\left(u\right)du \ast h\left(x\right)\right) = \left(\left(f \ast g\right)\ast h\right)\left(x\right)    \blacksquare]

#### Victor Ivrii

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##### Re: Bonus Web Problem--4
« Reply #2 on: November 07, 2012, 09:51:29 PM »
Calvin: I would like to see the last property

#### Calvin Arnott

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##### Re: Bonus Web Problem--4
« Reply #3 on: November 07, 2012, 11:33:30 PM »
[\thicksim \text{For}  f   \text{which decays fast as}   x \rightarrow \infty: \forall n \in \mathbb{N}: (\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_1 \thicksim]
[\text{We proceed by induction on  } n.  \text{ Let:  }f  \text{be a nice function on } \mathbb{R}   \text{which decays fast as}   x \rightarrow -\infty.  \text{Base case where   }n = 1:]
[(\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = (\frac{x^{0}_+}{(0)!} \ast f(x)) = \int_{-\infty}^{\infty}(x-x_1)^{0}_+ f(x_1)dx_1 = \int_{-\infty}^{x}(x-x_1)^{0}_+ f(x_1)dx_1 + \int_{x}^{\infty}(x-x_1)^{0}_+ f(x_1)dx_1]
[= \int_{-\infty}^{x}1 f(x_1)dx_1 + \int_{x}^{\infty}(0) f(x_1)dx_1 = \int_{-\infty}^{x} f(x_1)dx_1,    \text{as for:}    x_1 > x,   (x-x_1)_+ = 0]
[\text{So for   } n = 1: (\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = (\frac{x^{0}_+}{(0)!} \ast f(x)) = \int_{-\infty}^{x} f(x_1)dx_1 = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_1]
[\text{This integrals converges because } f(x) \text{ decays fast and vanishes as } x \rightarrow \infty]
[\text{Now suppose for some}    n\in\mathbb{N}: (\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_1.    \text{Then for}    n + 1 \in \mathbb{N} \text{ we show:}]
[(\frac{x^{(n+1)-1}_+}{( (n+1)-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-2}} \int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_2 dx_1]
[\text{Recall the previous property that: } (f' \ast g) = (f \ast g)' = (f \ast g') ]
[\text{Let: } h(x) = \frac{x^{n}_+}{(n)!},    h'(x) = \frac{x^{n-1}_+}{(n-1)!}]
[\text{And let: } g(x) = \int_{-\infty}^{x}f(y)dy,    g'(x) = f(x)]
[\text{Then we have: } (h \ast g')(x) = (h' \ast g)(x) ]
[ \text{Because } f(x) \text{ is nice and decays fast, we have that } g(x) \text{is nice and decays fast as } x \rightarrow -\infty]
[\text{... and we can use our induction hypothesis on } g(x) \text{, namely:}]
[(h' \ast g)(x) = (\frac{x^{n-1}_+}{( n-1)!} \ast g(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}(\int_{-\infty}^{x_{n}}f(y)dy)dx_{n}\cdots dx_1]
[= \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}} \int_{-\infty}^{x_{n}}}_{n+1\text{ integrals}}f(x_{n+1})dx_{n+1}\cdots dx_2 dx_1 = (h \ast g')(x) = (\frac{x^{(n+1)-1}_+}{( (n+1)-1)!} \ast f(x))]
[\text{And for } (n+1) \text{ we have } (\frac{x^{(n+1)-1}_+}{( (n+1)-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}} \int_{-\infty}^{x_{n}}}_{n+1\text{ integrals}}f(x_{n+1})dx_{n+1}\cdots dx_2 dx_1]
[\text{By induction then: } \forall n \in \mathbb{N}: (\frac{x^{n-1}_+}{( n-1)!} \ast f(x)) = \underbrace{\int_{-\infty}^{x}\cdots\int_{-\infty}^{x_{n-1}}}_{n\text{ integrals}}f(x_n)dx_n\cdots dx_1 \blacksquare]
« Last Edit: November 10, 2012, 01:26:04 PM by Calvin Arnott »

#### Victor Ivrii

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##### Re: Bonus Web Problem--4
« Reply #4 on: November 08, 2012, 02:35:16 AM »
Nice! $\renewcommand{\Re}{\operatorname{Re}}$

BTW one can consider for $\lambda \in \mathbb{C}\setminus \{-1,-2,-3,\ldots\}$  function $f_\lambda(x):= \frac{1 }{\Gamma (\lambda+1)}x_+^\lambda$ where $\Gamma$ is Euler's gamma-function (both $\Gamma(\lambda+1)$ and $f_\lambda(x)$ are first defined for $\Re \lambda >-1$ and then analytically continued to $\mathbb{C}$ with the poles ($f_\lambda$ is here a distribution rather than an ordinary function) and one can consider "fractional integration" $I_\lambda u:=f_{\lambda-1} * u$.

However $I_\lambda$ as analytic continuation makes sense even for $\lambda=-n-1$: $f_{-1} = \delta$, $f_{-2}=\delta'$ etc $f_{-n-1}=\delta^{(n)}$.

#### Calvin Arnott

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##### Re: Bonus Web Problem--4
« Reply #5 on: November 10, 2012, 02:26:11 PM »
Does fractional integration make sense in the language of differential forms?

#### Victor Ivrii

No. Furthermore it is basically not integral but primitive of order $\lambda$ (or derivative of order $-\lambda$) and thus does not make a lot of sense in any dimension but $1$.