# 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

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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