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Let $(X,\tau)$ be a complex metrizable topological vector space with the metric $d$. Does the following hold: $$d(0,\alpha x)=d(0,x),\ \forall \alpha \in \mathbb C, |\alpha|=1 \ ?$$ In general, the following holds: $$d(0,\alpha x)\neq |\alpha|d(0,x), \ \alpha \in \mathbb C.$$

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  • $\begingroup$ Which examples did you check? $\endgroup$ – Did Jan 20 '19 at 11:05
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Absolutely not. One can easily find counterexamples, but it seems more important to understand why one should not expect such an equality. "Metrizable topological vector space" just means that the topology generated by the metric satisfies certain conditions (namely making the vector space operations continuous). This leaves plenty of room to break your proposed equality without changing the topology.

For a concrete example take $\mathbb{C}$ and let $$\phi\colon\mathbb{C}\to\mathbb{C},\,z\mapsto \operatorname{Re}z+4 \mathrm{i}\operatorname{Im} z.$$ This map is clearly continuous with continuous inverse, thus $$ d\colon\mathbb{C}\times \mathbb{C}\to [0,\infty),\,d(z,w)=|\phi(z)-\phi(w)| $$ induces the Eucliean topology on $\mathbb{C}$ (which is of course a vector space topology). However $d(0,2)=2$, while $d(0,2i)=8$.

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