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Let $f(k)$ be a function on the integers with the properties

  • $f(k)\geq 0$ $\forall k$
  • $f(k)\geq p$ for at least $t$ values of $k$

Let $g(k)=\sum_{k'} f(k')f(k-k')$ be the self-convolution of $f$.

What is the minimal number of non-zero entries of $g(k)$? Can something be said about how large they have to be at least (e.g. $\geq p^2$)?

2 Answers 2

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From the answer of Stella Biderman I managed to find the answer I was looking for.

There are at least $2t-1$ entries of $g(k)$ s.t. $g(k)\geq p^2$.

Proof:

Let $S=\{k:f(k)\geq p\}$.
Since $g(k)=\sum_{k'}f(k')f(k-k')$ and $f(k)\geq 0$, we know that $g(k)

Thus $g(k)

From the answer to this question, we know that $|S'|\geq 2|S|-1$.

By definition, $t\geq|S|$ which proves the statement.

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It's true that $g(x)\geq p^2$ happens at at least $t$ points. Suppose $f(k)=0$ except at the $t$ points that fall in a set $S$, and that $\forall s\in S, g(s)\geq p$.

$$g(k)=\sum_{s\in S}f(s)f(k-s)$$ holds by restricting to the terms where the $f(k)$ piece of the product is non-zero. For $g(k)$ to be non-zero, it would have to be that $\exists s$ such that $k-s\in S$. How many ways this could happen depends on the set $S$. If we have a function $h(x)$ that is a lower bound on the number of $s$ for which $x-s\in S$, then we can conclude that $g(x)\geq h(x)p^2$.

We see that there are at least $t$ values for which $h(x)>0$ by writing $S=\{s_1,s_2,\ldots, s_t\}$ where $i< j\Rightarrow s_i

We can see that this is sharp at $t=1$ by taking $S=\{0\}$ and $f(0)=p$ and $f(x)=0$ for $x>0$. Then $g(0)=p^2$ and $g(x)=0$ for $x>0$.

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    Thank you very much. Is this sharp for $t\geq 2$? I cannot think of a function with 2 non-zero values that does not self-convolute to at least 3 non-zero values. In fact I think that the number of non-zero values has to grow as soon as $t\geq 2$. – 2017-01-23