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.
 A: 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.
A: 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.
