Is this element in $\ell_2$?

Let $(y_1,y_2,\cdots, y_n,\cdots)$ be an element of $\ell_2$. Is it true that for any complex number $\lambda$ , the sequence defined recursively by $$x_1=\lambda y_1, x_n=\lambda\left(y_n+\frac{x_{n-1}}{n-1}\right), n=2,3,\cdots$$ is also in $\ell_2$?

• it seems that you are in process of finding some spectra Commented Aug 21, 2013 at 20:26
• Hint: for $\lambda \neq 0$, how do you get $y$ from $x$? Commented Aug 21, 2013 at 20:30

We have $$\sum_{i=2}^N|x_i|^2=\sum_{n=1}^{N-1}|x_{n+1}|^2\leqslant 2\sum_{n=1}^{N-1}|\lambda|^2|y_n|^2+2\sum_{n=1}^{N-1}|\lambda|^2\frac{|x_{n}|^2}{n^2},$$ so it's enough to show that $(x_n,n\geqslant 1)$ is bounded.

Let $R:=\sup_n|y_n|$. Let $n_0>2$ be such that $\frac{|\lambda|}{n_0-1}<1$. Choose $M$ large enough such that $\max\{|x_j|,1\leqslant j\leqslant n_0-1\}\leqslant M$ and $|\lambda|\cdot R+\frac{|\lambda|}{n_0-1}M\leqslant M$. Let $n\geqslant n_0$ and assume that $|x_m|\le M$ for $m<n$. Then $|x_n|\leqslant |\lambda|\cdot R+\frac{|\lambda|}{n_0-1}M\leqslant M$, proving that $|x_n|\le M$ for all $n$ by induction.

• I don't follow your proof about boundedness. Your M seems to depend on n, so it does not work. How do you fix it?
– TCL
Commented Aug 22, 2013 at 2:34
• I think I can prove that $|x_n|\le R|\lambda| e^{|\lambda|}$.
– TCL
Commented Aug 22, 2013 at 3:21
• Since $\lim_{M\to \infty}\frac{|\lambda|R+\frac{|\lambda|}{n_0-1}}M\to \frac{|\lambda|}{n_0-1}\lt 1$, we can pick $M$ large enough so that that $\frac{|\lambda|R+\frac{|\lambda|}{n_0-1}}M\to \frac{|\lambda|}{n_0-1}\lt 1$. Commented Aug 22, 2013 at 8:37
• Your limit above is obviously 0. But that does not help.
– TCL
Commented Aug 22, 2013 at 12:56
• I missed a $M$, I meant, denoting $l:=|\lambda|$, $\lim_{M\to \infty}\frac{lR+\frac l{n_0-1}M}M$. Commented Aug 22, 2013 at 12:58