1
$\begingroup$

Let the sequence space $ s = \{ $ all sequences of complex numbers $\}$ with distance $$ d(x,y) = \sum_{j=1}^{\infty} \frac{1}{2^j} \frac{ | \xi_j - \eta_j| }{ 1 + |\xi_j - \eta_j|}.$$ Let $ x_n = (\xi_j^{(n)})=(\xi_1^{(n)}, \xi_2^{(n)},...)$, $x= (\xi_1, \xi_2, ...) $.

I want to show that $x_n \to x$ iff $\xi_j^{(n)} \to \xi_j \quad \forall j $

For the first implication, suppose $x_n \to x$. Fix $\epsilon > 0.$ There exists $N$ s.t. $n\geq N$ implies $d(x_n, x) < \epsilon$, i.e. $$ d(x_n,x) = \sum_{j=1}^{\infty} \frac{1}{2^j} \frac{ | \xi_j^{(n)} - \xi_j| }{ 1 + |\xi_j^{(n)} - \xi_j|} < \epsilon$$

Fix $j$ . I want to show that $ | \xi_j^{(n)} - \xi_j| < \epsilon $. I know that : $$ \frac{1}{2^j} \frac{ | \xi_j^{(n)} - \xi_j| }{ 1 + |\xi_j^{(n)} - \xi_j|} < \epsilon $$

What should I do next? Thanks!

$\endgroup$
0
$\begingroup$

For proving the convergence $\xi_j^{(n)}\to\xi_j$, let's better use the definition of $x_n\to x$ with $\frac{1}{2^{j+2}}>\frac{\epsilon}{2^{j+1}}>0$ then, as you did

$$\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}<\frac{\epsilon}{2^{j+1}}$$

from where $\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}<\frac{\epsilon}{2}$, i.e.

$$|\xi_j^{(n)}-\xi_j|<\frac{\epsilon/2}{1-\epsilon/2}\leq\epsilon$$


Now we need the other direction. Assume each $\xi_j^{(n)}\to\xi_j$ and let $\epsilon>0$.

Observe that the quotients $\frac{|X|}{1+|X|}<1$. Therefore there is $J\in\mathbb{N}$ such that $\sum_{j>J}\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)-\xi_j}|}<\frac{\epsilon}{2}$.

Now choose $N\in\mathbb{N}$ such that $|\xi_j^{(n)}-\xi_j|<\min\{\frac{\epsilon}{2},1\}$ for $0<j\leq J$ and $n>N$.

then $$\begin{align}d(x_n,x)&=\sum_{i\leq j}\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}+\sum_{j>J}\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}\\&\leq\sum_{i\leq j}\frac{1}{2^{j+1}}|\xi_j^{(n)}-\xi_j|+\sum_{j>J}\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}\\&\leq\frac{\epsilon}{2}\sum_{j\leq J}\frac{1}{2^{j+1}}+\frac{\epsilon}{2}\\&\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\end{align}$$

$\endgroup$
  • 1
    $\begingroup$ the inequality $\frac{\epsilon}{1-\epsilon} \leq \epsilon$ is false if $\epsilon < 1$, in which case it would be equivalent to $\epsilon \leq \epsilon - \epsilon ^2$ $\endgroup$ – user159517 Jan 21 '15 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.