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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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    @Rajesh: Both of my examples can be restricted to the case $f: \mathbb{R} \to \mathbb{R}$.2011-07-27

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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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    @Rajesh: You can generalize Jyrki's example by starting with a whole set of functions and closing it under convolutions; if they're all centred on the same side of the origin, the resulting set won't be closed under addition. You can generalize my example to various other constraints on the total integral/sum $S$, for instance $0\le S\le c\le1$ or $|S|\le c\le1$. There might be further ones; I haven't thought about how one might characterize these sets systematically.2011-07-27
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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 :-) ${}{}$2011-07-27