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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?

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We need to prove that $$ \sum_{k=1}^\infty\left|\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}\right|^2\frac{1}{\lambda^2}<+\infty. $$

  1. Denote $x_k=\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}$, which gives $$ x_{k+1}=\frac{1}{\lambda}x_k+y_{k+1}, \qquad x_0=0,\ k\ge 0.\tag1 $$
  2. 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 $$ x(z)=\frac{z}{\lambda}x(z)+y(z)\qquad\Leftrightarrow\qquad x(z)=\frac{y(z)}{1-\lambda^{-1}z}. $$
  3. 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 $\ell^{2+}$).

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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 $a_k = 1/\lambda ^k, b_k = y_k.$

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