# A family of functions which is closed under convolution operation but not under addition and multiplication. operations

The set of $\mathcal{L}^p$ functions is closed under addition, multiplication and convolution operations. I'd like to know an example of a set of functions which is closed under convolution and not under addition and multiplication.

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The easiest example is the standard orthonormal basis in $\ell^2(\mathbb{N})$, I presume. – t.b. Jul 27 '11 at 8:12
@Theo Buehler : I do not know of such a thing, could you please give a link or explain. – Rajesh Dachiraju Jul 27 '11 at 8:28
@Rajesh: See en.wikipedia.org/wiki/Sequence_space. – joriki Jul 27 '11 at 8:37
Let $e_n$ be the sequence that is zero except at place $n$. The convolution $c=(c_n) = a \ast b$ of two sequences $a=(a_n)$ and $b=(b_n)$ is given by $c_n = \sum_{k} a_k \cdot b_{n-k}$. Using this formula you have $e_{m+n} = e_m \ast e_n$. But obviously the set $\{e_m\} \subset \ell^p$ is neither closed under addition nor multiplication. This convolution arises when looking at the Fourier transform $\mathcal{F}(f\cdot g) = \mathcal{F}(f) \ast \mathcal{F}(g)$ of periodic functions, for example. See e.g. here. – t.b. Jul 27 '11 at 8:39
@Theo Buehler : thank you for the example. I wonder there are any examples of set of functions of the form $f: \mathbb{R} \to \mathbb{R}$. – Rajesh Dachiraju Jul 27 '11 at 8:40

Two examples come to mind -- the set of all functions normalized to $1$, and the set of Gaussians. (The latter is closed under multiplication -- don't know whether your "and" was intended to exclude that.)

[Edit in response to the comment:]

The product of two Gaussians is $\mathrm e^{a_1x^2+b_1x+c_1}\mathrm e^{a_2x^2+b_2x+c_2}=\mathrm e^{(a_1+a_2)x^2+(b_1+b_2)x+(c_1+c_2)}$, again a Gaussian.

By a function normalized to $1$, I mean a function whose sum/integral over its entire domain is $1$; for instance a probability distribution function.

[Edit in response to further comment:]

The space of functions normalized to $1$ is closed under convolution:

$$\begin{eqnarray} \int_{-\infty}^\infty \left(\int_{-\infty}^\infty f(x)g(y-x)\mathrm dx\right)\mathrm dy &=& \int_{-\infty}^\infty f(x)\left(\int_{-\infty}^\infty g(y-x)\mathrm dy\right)\mathrm dx \\ &=& \int_{-\infty}^\infty f(x)\left(\int_{-\infty}^\infty g(y)\mathrm dy\right)\mathrm dx \\ &=& \int_{-\infty}^\infty f(x)\mathrm dx \\ &=& 1\;, \end{eqnarray}$$ and analogously for sums or multi-dimensional integrals.

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my 'and' is a logical one although i didn't have such an example in mind. But I do not understand how the set of Gaussians is closed under multiplications. Also please let me know what is meant by set of all functions normalized to 1. – Rajesh Dachiraju Jul 27 '11 at 8:22
I didn't doubt that your 'and' was a logical one, but linguistically, since the repetition of "closed" and "under" was implied, it wasn't quite clear whether the repetition of "not" was also implied :-) – joriki Jul 27 '11 at 8:34
The set of all functions normalized to 1 is not be closed under convolution unless you re-normalize it. – Rajesh Dachiraju Jul 27 '11 at 8:36
thank you for the edit. – Rajesh Dachiraju Jul 27 '11 at 8:44
just out of curiosity, do you think the examples given here are the only type of examples, or there could be more...and very different from these. – Rajesh Dachiraju Jul 27 '11 at 9:09

Convolution is associative, so pick you favorite function $f$ and start iterating. Define $f=f^{*[1]}$ and then recursive for all positive integers $n$ define $f^{*[n]}=f*(f^{*[n-1]})$. By associativity the set $$S=\{f^{*[n]}\mid n\in\mathbf{Z}_+\}$$ is closed under convolution. If $f$ is not identically zero, then it cannot be closed under scalar multiplication by virtue of being a countable set. If we concentrate the mass of $f$ around 1 (for example use a Gaussian like joriki), then the mass of $f^{*[n]}$ will be concentrated (though not as sharply) around $n$, so set $S$ cannot be closed under multiplication or addition either.

Edit: This generalizes Theo Buehler's suggestion. Set $f=e_1$, and you get his example.

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Nice :-) ${}{}$ – joriki Jul 27 '11 at 8:47