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Prove the function $T: l_2 \rightarrow l_1 $defined by $T(\{x_n\})=\{\frac{x_n}{n}\}$ is uniformly continuous.

$l_2=\{\{x_n\}:\sum\limits_{n=1}^{\infty} |x_n|^2 < \infty \}$ and $l_1=\{\{x_n\}:\sum\limits_{n=1}^{\infty} |x_n| < \infty \}$ where the metric of each one is $d_2(x,y)=(\sum\limits_{n=1}^{\infty} |x_n-y_n|^2)^{1/2}$ and $d_1(x,y)=\sum\limits_{n=1}^{\infty} |x_n-y_n|$, respectively.

What I tried is

Let $\epsilon>0$ and $\{x_n\}, \{y_n\} \in l_2$ then

$d_1(T(x),T(y))=\sum\limits_{n=1}^{\infty} \frac{|x_n-y_n|}{n}\leq \sum\limits_{n=1}^{\infty} |x_n-y_n| $ but I have no idea how to relate that sum with $(\sum\limits_{n=1}^{\infty} |x_n-y_n|^2)^{1/2}$. Could you help me?

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HINT

$$(\sum_n\frac{|x_y-y_n|}{n} )^2 \leq( \sum_n \frac{1}{n^2})(\sum_n|x_n-y_n|^2)$$

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  • $\begingroup$ thank you so much $\endgroup$ – lepstein Oct 28 '19 at 22:46

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