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Considering $\ell^1$ with the norm $\lVert·\rVert_1$, we define the following map $ f:\ell^1 \rightarrow \mathbb{R}$ with: \begin{equation} g((x_j)_{j=1}^\infty)=\sum_{j=1}^\infty (-1)^j x_j.\end{equation} We have to show that it is well defined and continuous.

Taking into account that absolute convergence implies convergence, I think we can say that each sequence maps to only one point in $\mathbb{R}$ and thus say it is well defined. But for continuity I've been trying with the triangle inequality and the reverse triangle inequality to try to use the definition with $\epsilon$ and $\delta$.

Alex A.G.
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1 Answers1

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Hint

Use the sequential equivalence of continuity

You have already seen that $f : \ell^1 \to \mathbb{R}$ where

$$f((x_j)) := \sum\limits_{j \in \mathbb{N}} (-1)^jx_j$$

is well defined.

To prove continuity through sequence, take a sequence in $\ell^1$ (a sequence of sequences!) $(x^k_j)$ for $k,j \in \mathbb{N}$, such that $x^k_j \to (a_j)$ in $\ell^1$ for some fixed $(a_j) \in \ell^1$.

What can you say about the sequence $f((x^k_j))$, $k \in \mathbb{N}$?, does it converges to $f(a_j)$?. If it does, you can say that $f$ is continuous!

César VB
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  • But how could I prove (for the well defined part) that the function maps only onto $\mathbb{R}$ and not onto $\mathbb{C}$? Or is it not necessary? – Alex A.G. Jan 19 '24 at 14:34
  • @AlexA.G. In my answer I'm assuming $\ell^1$ is the set of all absolutely convergent sequences in $\mathbb{R}$! if you take $\mathbb{C}$ I'm not sure... it may not be well defined. Are you taking sequences in $\mathbb{C}$ or in $\mathbb{R}$? http://mathonline.wikidot.com/the-1-sequences-normed-linear-space – César VB Jan 19 '24 at 21:12