Show that the series converges ($l^2$) I know that $\sum_{k=1}^\infty|y_n|^2=S<\infty$. I also have that $\lambda >1$.
I need to show that
$$
\sum_{k=1}^\infty \left|  \frac{y_1}{\lambda^k}
                        + \frac{y_2}{\lambda^{k-1}}
                        + \cdots \frac{y_k}{\lambda} \right|^2
$$
converges. (When $k=1$, there is one term in absolute value, when k=2, there are two terms etc...) Can someone please help?
I tried using https://en.wikipedia.org/wiki/Minkowski_inequality:
If I can show that
$$
\sum_{k=1}^N \left| \frac{y_1}{\lambda^k}
                   +\frac{y_2}{\lambda^{k-1}}
                   +\cdots \frac{y_k}{\lambda}\right|^2
$$
is bounded for any N, I am done.
By Minowski's inequality:
$$
\begin{split}
\sum_{k=1}^N \left| \frac{y_1}{\lambda^k}
                   +\frac{y_2}{\lambda^{k-1}}
                   +\cdots \frac{y_k}{\lambda}\right|^2
&= \left[ \left( \sum_{k=1}^N \left| \frac{y_1}{\lambda^k}
                                    +\frac{y_2}{\lambda^{k-1}}
                                    +\cdots \frac{y_k}{\lambda} \right|^2
           \right)^{1/2} \right]^2 \\
&\le \left[ \left(\sum_{k=1}^N \frac{|y_1|^2}{\lambda^{2k}}\right)^{1/2}
           +\left(\sum_{k=2}^N \frac{|y_2|^2}{\lambda^{2k-2}}\right)^{1/2}
           +\cdots + \left(\frac{|y_n|^2}{\lambda^2}\right)^{1/2} \right]^2.
\end{split}
$$
Now since $\lambda > 1$ we have a geometric series in each part, so we take each $|y_i|^2$ out of the sum, and then each sum is bounded:
$$\ldots \le \left[K|y_1|+K|y_2|+\cdots+K|y_N|\right]^2
  =\left[K \sum_{k=1}^N|y_k| \right]^2.$$
However, I do not know that $\sum_{k=1}^N|y_k|$ is bounded for any $N$? And the problem with $[\sum_{k=1}^N|y_k|]^2$ is that it contains many cross-terms, it contains $\sum_{k=1}^N |y_k|^2$, but also many more cross-terms.
Any tips?
 A: We need to prove that
$$
\sum_{k=1}^\infty\left|\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}\right|^2\frac{1}{\lambda^2}<+\infty.
$$


*

*Denote $x_k=\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}$, which gives
$$
x_{k+1}=\frac{1}{\lambda}x_k+y_{k+1}, \qquad x_0=0,\ k\ge 0.\tag1
$$

*Do $z$-transform of the equation, i.e. multiply by $z^{k+1}$ and sum up. With the notations 
$$
x(z)=\sum_{k=0}^\infty x_k z^k,\qquad y(z)=\sum_{k=0}^\infty y_k z^k
$$
we get
$$
x(z)=\frac{z}{\lambda}x(z)+y(z)\qquad\Leftrightarrow\qquad
x(z)=\frac{y(z)}{1-\lambda^{-1}z}.
$$

*By Plancherel theorem $y\in\ell^{2+}$ $\Leftrightarrow$ $y(z)\in H^2(\mathbb{T})$ (Hardy space). The function $1/(1-\lambda^{-1}z)\in H^\infty(\mathbb{T})$ due to $\lambda>1$, then $x(z)\in H^2(\mathbb{T})$, so $x\in\ell^{2+}$. 


P.S. The whole point here is to say that the stable ($\lambda>1$) linear difference equation is a linear bounded operator on $\ell^{2+}$. There are alternative ways to show that, for example, by using Riesz-Thorin theorem to the convolution for $x_k$ directly (since $1/\lambda^k\in\ell^{1+}$, it is easy to see that the convolution is a linear bounded operator on $\ell^{\infty+}$ and on $\ell^{1+}$, which interpolates it to a linear bounded on $\ell^{2+}$).
A: Well my solution looks basically the same as that of A.G; perhaps it's a little more elementary. I'll include it since I just spent some time on it!
Suppose
$$\sum_{k=1}^{\infty}|a_k| < \infty, \sum_{k=1}^{\infty}|b_k|^2 < \infty.$$
Let $f(z) = \sum_{k=1}^{\infty}a_kz^k, g(z) = \sum_{k=1}^{\infty}b_kz^k.$ Then in the open unit disc,
$$f(z)g(z) = \sum_{k=2}^{\infty}\left( \sum_{j=1}^{k}a_jb_{k+1-j}\right) z^k.$$
Thus
$$\tag 1 \frac{1}{2\pi }\int_0^{2\pi} |f(re^{it})g(re^{it})|^2  = \sum_{k=2}^{\infty}| \sum_{j=1}^{k}a_jb_{k+1-j} | ^2 r^{2k}.$$
But $|f(re^{it})| \le \sum |a_k| = M,$ so the left side of $(1)$ is $\le M\int_0^{2\pi} |g(re^{it})|^2 \le M \sum |b_k|^2.$ It follows that
$$ \sum_{k=2}^{\infty}|\sum_{j=1}^{k}a_jb_{k+1-j}|^2r^{2k}\le M \sum |b_k|^2$$
for all $r,$ hence
$$\sum_{k=2}^{\infty}| \sum_{j=1}^{k}a_jb_{k+1-j} |^2 \le M \sum |b_k|^2.$$
Now in the problem of the OP, we can take $a_k = 1/\lambda ^k, b_k = y_k.$
