# Semigroup property for convolution

Summary: Show that for this family of functions the following semigroup property with respect to convolution holds.

Hi.

My task is to prove that for the family of functions defined as:

$$f_{a}(x) = \frac{1}{a \pi} \cdot \frac{1}{1 + \frac{x^{2}}{a^{2}} }$$
The following semigroup property with respect to convolution holds:
$$f_{a}(x) * f_{b}(x) = f_{a + b}(x)$$
I suspect it has to do with their fourier transforms, my intuition tells me that:
$$F_{a}( j \omega) \cdot F_{b}( j \omega) = F_{a \cdot b}( j \omega)$$
Am I on the right track, so far?

A and B are nonzero positive.

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fresh_42
Mentor
How is the convolution defined? What do you have to show for the semigroup property?

The convolution is defined on $$l^{1}(\mathbb{Z})$$ and $$l^{1}(\mathbb{R} )$$ spaces.
Respectively, for continuous and discrete (absolutely integrable and absolutely summable functions):
$$x[n] * y[n] = \displaystyle \sum_{k \in \mathbb{Z} } x[k] \cdot y[n - k]$$
And
$$x(t) * y(t) = \displaystyle \int_{-\infty}^{\infty} x(\tau) \cdot y(t - \tau) \,\,\,\, \text{d} \tau$$
You have to show that for the family of the functions below:
$$f_{a}(x) = \dfrac{1}{a \pi} \cdot \dfrac{1}{ 1 + \dfrac{x^{2}}{a^{2}} }$$
Forms a semigroup, under convolution:
$$f_{a}(x) * f_{b}(x) = f_{a + b}(x)$$
I am on the path, to use the foiurier transform. Please see my intuition and reasoning for this, in my first post.

fresh_42
Mentor
It could be that the Fourier transform helps, but I don't see it yet. You can use the formula for ##\mathcal{F}(f*g)## but you have to undo the transformation again, and ##\mathcal{F}(f)\cdot \mathcal{F}(g)## doesn't look much better. Why don't you calculate the integrals?

Hi

The fourier transform of:
$$\mathcal{F} \bigg\{ \frac{1}{a \pi} \cdot \frac{1}{1 + { \frac{x}{a} }^{2} } \bigg \} =\frac{1}{ a^2 \pi} \cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-| \frac{\omega}{a} |}$$
My idea is that, since, the fourier transform of the convolution results in a mulitplication of their individual fourier transforms:
$$\mathcal{F} (f_{a} * f_{b} ) = \mathcal{F}(f_{a}) \cdot \mathcal{F}(f_{b} )$$
Our original claim was that:
$$f_{a} * f_{b} = f_{a + b}$$
So, if we multiply their fourier transforms, we get:
$$\frac{1}{ a^2 \pi} \cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-| \frac{\omega}{a}| } \cdot \frac{1}{b^2 \pi} \cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-| \frac{\omega}{b} |} = \frac{1}{(ab\pi)^{2} } \cdot \frac{\pi}{2} \cdot e^{ - ( \frac{\omega}{a} + \frac{\omega}{b} )}$$
This equation above, is simply a given, due to convolution and frequency multiplication duality.

My question is, how can I positively show that:
$$\mathcal{F} \bigg \{ f_{a + b} \bigg \} = \frac{1}{(ab\pi)^{2} } \cdot \frac{\pi}{2} \cdot e^{ - ( \frac{\omega}{a} + \frac{\omega}{b} )}$$
If I can poisitively show that, then I can complete my proof.-

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fresh_42
Mentor
Where is the integral? If we rewrite
$$f_a(x)=\dfrac{1}{a\pi} \cdot \dfrac{1}{1+\frac{x^2}{a^2}}= \dfrac{1}{\pi}\cdot \dfrac{a}{a^2+x^2}$$
then this function says tangent to me. E.g. ##\dfrac{d}{dx} \operatorname{arctan}(x)=\dfrac{1}{1+x^2}## Now if we fold ##f_a*f_b## then we could get rid of the integral. How did you remove it?

• AVBs2Systems
strangerep
$$\mathcal{F} \bigg\{ \frac{1}{a \pi} \cdot \frac{1}{1 + { \frac{x}{a} }^{2} } \bigg \} =\frac{1}{ a^2 \pi} \cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-| \frac{\omega}{a} |}$$
Where did you get this from? It looks wrong to me.

Where did you get this from? It looks wrong to me.
Scaling property and linearity property. Note a and b are greater than zero.

The fourier transform of
$$\mathcal{F} \bigg \{ \frac{1}{1 + {(\frac{x}{a})}^{2} } \bigg \} = \frac{1}{|a|}\cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-|\frac{\omega}{a} | }$$
Hence:
$$\frac{1}{a \pi} \cdot \mathcal{F} \bigg \{ \frac{1}{1 + {(\frac{x}{a})}^{2} } \bigg \} = \dfrac{1}{a^2 \pi} \cdot X( \frac{\omega}{a} )$$

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strangerep
That's not a sensible answer. You have an "##a##" on the RHS (of 1st equation), but not on the LHS.
Show your work in more detail.

(Btw, shouldn't this be in one of the homework forums?)

fresh_42
Mentor
In case you will end up in an integration formula, and I still don't see how to avoid this, the substitutions ##x=a\tan \varphi## and ##t-x=b\tan \vartheta## could be helpful.

strangerep
[...] I still don't see how to avoid [integration formula]
Maybe I'm missing something but,... are you sure you're not overthinking this?

Imho, if the OP could get his/her Fourier transform expression right (i.e., use the scaling rule correctly), then I think the original problem becomes trivial.

• AVBs2Systems
MathematicalPhysicist
Gold Member
I am sure if you search for "proof of semigroup property of convolution for Cauchy distribution" you'll get something hopefully meaningful.

• AVBs2Systems
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I submitted both the approaches, one with the incorrect fourier transform and the other with the convolution integral. The convolution integral, reduced using partial fraction decomposition or the residue theorem (neither of which I was able to produce as a solution) reduces it to the form required.

Thank you everyone for your assistance.
$$\textbf{Sources for the solution}$$
## \text{For future reference, if anyone needs it:} ##

1. https://math.stackexchange.com/ques...ion-of-cauchy-density-function-int-infty-inft2. https://www.tandfonline.com/doi/abs/10.1080/00029890.1985.11971537

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• fresh_42