I am struggling with the following problem:
Let $M >1$ and $\lambda \in (0,1)$, $\mathbf{z} \in \mathcal{l}_p$. If $|v_t|^p < (M\sum_{s=t}^\infty \lambda^{s-t+1} |z_s| )^p $, then $\mathbf{v} \in \mathcal{l}_p$.
Proof: We have to show that $(\sum_{t=0}^\infty |v_t|^p)^{1/p} < \infty$ using (somehow) the fact that $$\biggl(\sum_{s=0}^\infty |z_s|^p\biggr)^{1/p} < \infty.$$
Hence we start with this inequality: $$\biggl(\sum_{t=0}^\infty |v_t|^p\biggr)^{1/p} < \biggl(\sum_{t=0}^\infty \biggl(M\sum_{s=t}^\infty \lambda^{s-t+1} |z_s| \biggr)^p\biggr)^{1/p} = M \biggl(\sum_{t=0}^\infty \biggl(\sum_{s=t}^\infty \lambda^{s-t+1} |z_s| \biggr)^p\biggr)^{1/p}.$$
And here I am lost. I tried to apply Holder inequality, but it was too much, it didn't work. Any ideas? Inequalities that might be useful?
Thanks!
PS: Using Holder inequility, I obtained
$$\biggl(\sum_{t=0}^\infty |v_t|^p\biggr)^{1/p} < M \biggl(\sum_{t=0}^\infty \biggl(\sum_{s=t}^\infty \lambda^{s-t+1} |z_s| \biggr)^p\biggr)^{1/p} < M \biggl(\sum_{t=0}^\infty \biggl(\sum_{s=t}^\infty \lambda^{q(s-t+1)}\biggr)^{p/q} \biggl(\sum_{s=t}^\infty |z_s|^p \biggr)\biggr)^{1/p} = M \biggl(\frac{\lambda^q}{1-\lambda^q}\biggr)^{1/q}\biggl(\sum_{t=0}^\infty\sum_{s=t}^\infty |z_s|^p \biggr)^{1/p} = M \biggl(\frac{\lambda^q}{1-\lambda^q}\biggr)^{1/q}\biggl(\sum_{t=0}^\infty (t+1) |z_t|^p \biggr)^{1/p}.$$
Since $p\geqslant 1$, it follows from an application of Jensen's inequality that $$|v_t|^p\leqslant \left(\sum_{j\geqslant 1}\lambda^j \right)^{p-1}\cdot\sum_{s\geqslant t}\lambda^{s-t +1} |z_s|^p=:C_p\sum_{s\geqslant t}\lambda^{s-t +1} |z_s|^p$$ (the constant $M$ can be cancelled if we consider $Mz_s$ instead of $z_s$). Therefore, switching the sums, we have $$\sum_{t=0}^\infty|v_t|^p\leqslant C_p \sum_{t=0}^\infty\sum_{s\geqslant t}\lambda^{s-t +1} |z_s|^p= C_p \sum_{s=0}^{\infty}|z_s|^p\sum_{t=0}^s\lambda^{s-t +1}.$$ We conclude noticing that $$\sum_{t=0}^s\lambda^{s-t +1}=\sum_{u=1}^{s +1} \lambda^{u} ,$$ which can be bounded uniformly in $s$ by $ 1/(1-\lambda)$.