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Let $X$ be the normed linear spaceof sequences of reals that have only finitely many non-zero terms. Given $x = \{x_n\} \in X$, define $$f(x) = \displaystyle \sum_{n=1}^{\infty} x_n$$ I think that it is quite easy to prove that $f$ is linear (please correct me if you think otherwise). My question is whether $f$ is continuous if we give $X$ the topology induced by the $l^{\infty}$ norm.

Thank you very much in advance!

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2 Answers 2

up vote 2 down vote accepted

Observe that for $(x^k)_k=((x_n^k)_n)_k \subset X$, where $x_n^k = 1$ for $n \leq k$ and $x_n^k = 0$ otherwise, we have $\lVert x^k \rVert_{\ell^\infty} = 1$ and $$ f(x^k) = k. $$ Hence $\lim_{k \to \infty} f(x^k) = \infty$ which proves that $f$ cannot be continuous.

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No. Consider $B(0,2\varepsilon) \subset X$ and let

$$y_n=\sum_{k=1}^n \varepsilon e_k$$

where $e_k$ is the usual basis for $X$. Then $y_n \in B(0,2\varepsilon)$ and $f(y_n)=\varepsilon n$ in particular the image of any neighborhood around $0$ is unbounded. It follows that we can't find an open set containing $0$ in $f^{-1}((-\varepsilon,\varepsilon))$ whose image is contained in $(-\varepsilon,\varepsilon)$.

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