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

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?

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