I have the normed space $ \ell^1 =\{(x_n) : \sum_{n=1}^\infty |x_n| < \infty\}$ with the norm defined by $\|x\|_1 = \sum_{n=1}^\infty|x_n| $, and a function $$f(x):=\sum_{n=1}^\infty x_n\sin(n) .$$ I wanna prove that it is continuous from $( \ell^1, \|\cdot\|)$ to $\mathbb{R}$ with usual metric.

Here's what I've done: let $x,y \in \ell^1$ $$\begin{align} |f(y) - f(x)|&= |\sum_{n=1}^\infty y_n\sin(n) - \sum_{n=1}^\infty x_n\sin(n) | \\ &= |\sum_{n=1}^\infty (y_n -x_n) \sin(n) | \\ &\leq | \sum_{n=1}^\infty (y_n -x_n)|\\ &= |\sum_{n=1}^\infty y_n - \sum_{n=1}^\infty x_n |\\ & = |\|y\| - \|x\|| \\ & \leq\|x-y\|. \end{align}$$ So I got $ d_R \leq d_{\ell^1} $.

I think the main work is done. I just don't know how to fit it together? Using the definition of continuous functions, should I choose, $\epsilon \leq \delta $ ?


2 Answers 2


A few notes:

  1. Forget $d_R$. You should have started with $\bigl|f(x)-f(y)\bigr|$.
  2. After $\displaystyle\left|\sum_{n=1}^\infty(y_n-x_n)\sin(n)\right|$, what you should have is$$\leqslant\sum_{n=1}^\infty|y_n-x_n|=\|y-x\|.$$
  3. The conclusion is that you can take $\delta=\varepsilon$.
  4. Since your map is linear, it would be simpler just to have proved that it is continuous at $0$. That would have been enough.

When checking continuity of linear maps between normed spaces, there is a characterization which is much easier to verify:$$f\text{ continuous } \iff \exists C>0: |f(x)|\leq C\|x\|\quad \forall x $$ Now, $$|f(x)|\leq \sum_{n=0}^{\infty}\left|x_n \sin (n)\right|\leq \sum_{n=0}^{\infty}|x_n|=\|x\| $$ So the inequality $|f(x)|\leq C\|x\|$ works with $C=1$.


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