Let $E$ be a vector space. A topology $\tau$ on $E$ is called (linearly) submetrizable if there is a (linear) metrizable topology $\pi$ on $E$ which is weaker than $\tau$, i.e. $\pi\subset\tau$.
Is it true that if $\tau$ is linear and submetrizable, it is linearly submetrizable?
Essentially, the question asks whether it is true that if there is a metric which is continuous with respect to a linear topology $\tau$, then there is a translation invariant metric which is continuous.
Yes. Let $(X,\tau)$ be a toplogical vector space and $d$ a metric on $X$ generatin a coarser topology $\pi$. Recursively, one finds balanced $\tau$-neighbourhoods $U_n$ of the origin with $$ U_{n+1}+U_{n+1}\subseteq U_n \subseteq B_d(0,1/n)$$ (the open $d$-ball around $0$ with radius $1/n$. In the terminolgy of Adasch, Ernst, and Keim the sequence $(U_n)_{n\in\mathbb N}$ is a topological string which generates a so-called $F$-norm and hence a translation invariant metric which is coarser than $\tau$. This construction can also be seen on this wikipedia page about metrizable topological vector spaces.
