6

Let $(a_n)_{n≥0}$ and $(b_n)_{n≥0}$ be two sequences of a nomed algebra such that $\sum{\| a_n\|}$ and $\sum{\| b_n\|}$ converge, and$ $$$\forall n, \ a_n, b_n \neq 0$$

Show that $(\forall k \in \mathbb N^*, \ \sum_{n=0}^{\infty}{a_n}^k = \sum_{n=0}^{\infty}{b_n}^k) \implies \exists\ \sigma \in{\mathfrak S}_{\mathbb N} \ ,\ (a_n) = (b_{\sigma(n)})$

The vector-space should be a finite-dimensional $\mathbb R$-vector-space.

I do not see how to deal with the hypothesis. If you have ideas/hints...

It would be relevant to deal with complex sequences even if it is less general. But I don't think one can generalize using real and complex cases.

Ѕᴀᴀᴅ
  • 35,369
  • First you need to solve $a_1^k+a_2^k+a_3^k=b_1^k+b_2^k+b_3^k$ then they are equal after a permutation. – Rene Schipperus Mar 03 '18 at 20:02
  • @ReneSchipperus: But $a_1^k+a_2^k+a_3^k=b_1^k+b_2^k+b_3^k$ is not given, only the equality of the infinite series $\sum_{n=0}^{\infty}a_n^k = \sum_{n=0}^{\infty}b_n^k$. – Martin R Mar 03 '18 at 20:08
  • Yeah , but just assume that all values beyond $a_2$ are zero, so this is an easier problem. – Rene Schipperus Mar 03 '18 at 20:10
  • Oh, I see that you have this extra condition. Why ? How do you expect to solve this question if you cannot solve the simpler problem ? – Rene Schipperus Mar 03 '18 at 20:14

2 Answers2

5

Assume without loss that $|b_i|\leq |b_0|$ for all $i>0$. You have $$\sum \left(\frac{a_i}{b_0}\right)^{2k}=\sum \left(\frac{b_i}{b_0}\right)^{2k}\to n$$

Where $n$ is the number of $b_i^2$'s equal to $b_0^2$. Now let $k\to \infty$. The right hand side limits to $n$. The left hand side can go to $n$ if and only if for there are $n$, $i$'s such that $a_i^2=b_0^2$.

Now, to get the signs $$\sum \left(\frac{a_i}{b_0}\right)^{2k+1}=\sum \left(\frac{b_i}{b_0}\right)^{2k+1}$$ and both sides can have the same limits as $k\to \infty$ iff there are the same number of $1$,s as $-1$$'s.

Rene Schipperus
  • 40,478
2

Here is an approach which works for sequences in $\Bbb C$, obtained by a small variation of Sangchul Lee's answer at $a_n =0$ for all $n \in \mathbb N$ if $\sum a_n^k =0$ for all integers $k \ge 1$.

Assume that for some $w_0 \in \Bbb C$ $$ \tag{*} \# \{ j \mid a_j = w_0 \} \ne \# \{ j \mid b_j = w_0 \} \, . $$ and set $z_0 = 1/w_0$.

Choose any $R > |z_0|$. We have that $ |a_n|R \leq \frac{1}{2}$ and $ |b_n|R \leq \frac{1}{2}$ for all sufficiently large $n \ge N$, and therefore $$ \left| \frac{a_n}{1 - a_n z} \right| \leq 2|a_n| \, , \quad \left| \frac{b_n}{1 - b_n z} \right| \leq 2|b_n| \, . $$ for $|z| < R$ and $n \ge N$. Then $$ f(z) = \sum_{n=1}^{N-1} \left( \frac{a_n}{1 - a_n z} - \frac{b_n}{1 - b_n z}\right) + \sum_{n=N}^{\infty} \left( \frac{a_n}{1 - a_n z} - \frac{b_n}{1 - b_n z}\right) $$ is well-defined in $B_R(0)$ as the sum of a rational function and a uniformly converging series of analytic functions, and a meromorphic function in that disk (with at most finitely many poles).

Now $(*)$ implies that $f$ has a pole at $z_0$. On the other hand, $$ \frac{f^{(k)}(0)}{k!} = \sum_{n=1}^{\infty} (a_n^{k+1} - b_n^{k+1}) = 0 $$ for each $k$, so that $f$ is identically zero, which is a contradiction.

Martin R
  • 128,226