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Suppose you have two arbitrary sequences of real numbers, say $\{ \gamma_n \}$ and $\{ c_{n,m} \}$. Suppose also that $c_{n,m}$ is not identically 0 for all pairs $(m,n)$, and also that for fixed $n$, $\sum_{m=0}^n c_{n,m} \neq 0$.

If

$$\sum_{m=0}^\infty \sum_{n=0}^\infty \gamma_n c_{n,m} = 0$$

Does it necessarily follow that $\gamma_n = 0$ for all $n$?

My intuition tells me yes but I do not know how to prove it, and it is also entirely possible that it is not actually true. Any help appreciated. Thanks.

Rellek
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1 Answers1

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The answer is negative. Consider the case where $$ c_{n,m}=1 \qquad (n,m\geq0), $$ $$ \gamma_0=-1,\qquad \gamma_n=2^{-n} \qquad (n\geq 1). $$ Then, for fixed $m$, we have $$\sum_{n=0}^\infty\gamma_nc_{n,m}=-1+\sum_{n=1}^\infty2^{-n}=0. $$ Then summing over $m$ is just a sum of infinitely many zeros, yielding zero.

Aweygan
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  • Good example. Indeed this choice for $\gamma_n$ will always work for a constant sequence. What about nonconstant sequences? – Rellek Jul 03 '16 at 17:53
  • That will require more work, given your condition that $\sum_{m=0}^n c_{n,m} \neq 0$ for fixed $n$. But I still think one can find a counterexample. – Aweygan Jul 03 '16 at 18:11
  • Yes, I came across this condition while trying to prove a certain set of polynomials were a basis. The $c_{n,m}$ are in fact the $n$th Legendre Polynomial's coefficient of the term $x^m$ in my case, it is very counter intuitive. – Rellek Jul 03 '16 at 19:00
  • Interesting! Now I see why you don't want constant sequences. – Aweygan Jul 03 '16 at 19:11