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

Question

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?

Answer 1

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 $0$.

Answer 2

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?