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Does $g(v_n) longrightarrow g(0)$ for all $v_n text{s.t.} ||v_{n+1}|| leq ||v_n||$ imply $g$ continuos at $0$?

The question is pretty much summed up in the title:

Let $V,W$ be normed vector spaces and $$ g: V to W$$.

Suppose g fulfills $$g(v_n) longrightarrow g(0) $$ for all sequences $v_n$ such that $$v_n longrightarrow 0$$ and $$||v_{n+1}|| leq ||v_n||.$$

Does this imply continuity of $g$ at $0$?

Intuititively, this should hold true, but I am looking for a proof (or counterexample). Of course, for any sequence $v’_n longrightarrow 0$ we can select a subsequence that fulfills the above requirement and will have the limit $g(0)$. But is this enough to ensure the convergence? Do we need additional restrictions on the spaces $V,W$ for this to hold?

Mathematics Asked by a_student on November 12, 2021

2 Answers

2 Answers

Assume that $v_n rightarrow 0$ and $g(v_n) nrightarrow g(0)$. Then there exists a subsequence $v_{k_n}$ s.t. $||g(v_{k_n}) - g(0)|| ge varepsilon$ for some $varepsilon > 0$. Let $u_n$ denote the sequence $v_{k_n}$. There exists a subsequence $u_{m_n}$ s.t. $||u_{m_{n+1}}||le ||u_{m_n}||$. Then $g(u_{m_n}) rightarrow g(0)$. But it is already known that $||g(u_{m_n}) - g(0)|| ge varepsilon$. This contradiction shows that $g(v_n) rightarrow g(0)$ for all $v_n rightarrow 0$. Thus, $g$ is continuous at $0$.

Answered by Matsmir on November 12, 2021

Take an arbitrary subsequence ${g(v_{n_k})}_k$ and then choose a subsequence ${v_{n_{k_l}}}_l$ such that $| v_{n_{k_{l+1}}} | leq | v_{n_{k_{l}}} |$. Using the assumptions we can deduce $g(v_{n_{k_l}}) to g(0)$. Why does this imply $g(v_n) to g(0)$ already?

Answered by neca on November 12, 2021

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