Define $g:(l_2,d_1) \to (\Bbb R,d_2)$ by $\displaystyle g(x)=\sum_{n=1}^{\infty}\frac{x_n}{n}$. Test whether $g$ is continuous or NOT.

Attempt : Take $\displaystyle x=\left (\frac{1}{n},\cdots ,\frac{1}{n} ,\cdots \right)\in l_2$ and $y=(0,0, \cdots , 0,\cdots)\in l_2$.

Now , $$d_2(g(x),g(y))=|g(x)-g(y)|=\left|\sum\frac{1}{n^2}-0\right|=\frac{\pi^2}{6}.$$ and $$d_1(x,y)=\left(\sum \frac{1}{n^2}\right)^{1/2}=\frac{\pi}{\sqrt 6}.$$

So , $d_2(g(x),g(y))>d_1(x,y)$. So $g$ is NOT continuous.

Check whether it is correct or wrong.

  • $\begingroup$ How is that checking continuity? $\endgroup$ – Jacky Chong Apr 16 '17 at 6:57

Observe $x = (x_n), y=(y_n) \in \ell^2$, then we see \begin{align} d_2(g(x), g(y))=|g(x)-g(y)| = \left|\sum^\infty_{n=1}\frac{x_n-y_n}{n}\right| \leq \sqrt{\sum^\infty_{n=1}\frac{1}{n^2}}\sqrt{\sum^\infty_{n=1}|x_n-y_n|^2}= \frac{\pi}{\sqrt{6}}d_1(x, y) \end{align} which is Lipschitz continuous.

  • $\begingroup$ Where my fault in my answer ? $\endgroup$ – Empty Apr 16 '17 at 7:08
  • $\begingroup$ @S717717 You weren't checking continuity. What you claimed is similar to saying $y=2x$ is not continuous. $\endgroup$ – Jacky Chong Apr 16 '17 at 7:09

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