Show that $||e^{-\lambda k}x-x||\to0$ as $\lambda\to0$ when $x = \{x_k\}_k \in \ell^2$

Question

Let be $x=\{x_k\}_k\in \ell^2(\mathbb{R})$ , consider $\lambda>0$ and show that $||e^{-\lambda k}x-x||\to0$ as $\lambda\to0$.

I tried to solve this exercise by this way:

$$||e^{-\lambda k}x-x||^2=||(e^{-\lambda k}-1)x||^2\leqslant\sum_k |e^{-\lambda k}-1|^2|x_k|^2 \leqslant ||x||^2 \sum_k |e^{-\lambda k}-1|^2\ $$

... how can I show the convergence?

Answer 1

Wiyh your assumptions, $e^{-\lambda k}x_k-x_k\in{\mathbb R}$. Hence $$\left| e^{-\lambda k}x_k-x_k \right| =(1-e^{-\lambda k}) x_k\to (bounded)\cdot 0=0$$.

BEcause $x\in\ell^2$, one has that $\sum|x_k|^2$ is convergente,hence, $|x_k|^2\to0$ and then $|x_k|\to0$.

EDIT With the new question (edited by author), $$\|e^{-\lambda k}x-x\|=(1-e^{-\lambda k})\|x\|\to 0$$ because $e^{-\lambda k}\to 1$ as $k\to\infty$.

In my opinion, the former question was more interesting.

Answer 2

You could show that for all $N\in \mathbb N,~\sum_{k=0}^N |e^{-\lambda k}-1|^2 |x_k|^2$ goes to zero while $$\sum_{k=N+1}^{+\infty} |e^{-\lambda k}-1| |x_k| \leq \sum_{k=N+1}^{+\infty} 2 |x_k|$$ We just have to formalize the argument as follows.

Let $\varepsilon>0$,

$\quad$ Since the sum $\sum_{n\in \mathbb N}{|x_k|^2}$ converges, there exists $N\in \mathbb N$ such that $\sum_{k=0}^{+\infty} 2|x_k|\leq \varepsilon/2$.

$\quad$ Let $N\in \mathbb N$ as above, since $\lim_{\lambda\rightarrow 0} \sum_{k=0}^N |e^{-\lambda k}-1|^2|x_k|^2=0$, there exists $\eta>0$ such that for all $\lambda\in \,]0,\eta[$, ~$\sum_{k=0}^N |e^{-\lambda k}-1|^2 |x_k|^2 \leq \varepsilon/2$.

Finally, Let $\eta>0$ as above, forall $\lambda \in \,]0,\eta[$ : \begin{eqnarray} \sum_{k=0}^{+\infty} |e^{-\lambda k}-1|^2|x_k|^2&=&\sum_{k=0}^N |e^{-\lambda k}-1|^2|x_k|^2 + \sum_{k=N+1}^{+\infty}|e^{-\lambda k}-1|^2|x_k|^2\\ &\leq&\varepsilon/2+\varepsilon/2\\ &=& \varepsilon \end{eqnarray}