If $f:E\rightarrow E'$ is a linear embedding of locally convex topological vector spaces, and $A\subseteq E$ open and convex, can we always find $A'\subseteq E'$ open and convex sucht that $f(A)=f(E)\cap A'$? Maybe with the additional requirement that $A$ is balanced?
My first thought was to choose some linear complement $F'\subseteq E'$ of $f(E)$ and considering the set $f(A)+F'$, but I couldn't prove it to be open...
OK, I think I got it. First we can assume w.l.o.g. that $0\in A$, I won't need $A$ to be balanced. Since $f$ is an embedding we can find $B'\subseteq E'$ open with $f(A)=f(E)\cap B'$ but this $B'$ doesn't have to be convex.
Now choose a convex, open neighbourhood $U'\in{\cal U}(0)$ in $E'$ such that $U'\subseteq B'$. For each $x\in A$ we find some $\epsilon>0$ such that $x+\epsilon A\subseteq A$, this holds for all convex, open sets. If we define $U'_{x,\epsilon}:=x+\epsilon U'$ for such $x,\epsilon$, we also have $$x\in U'_{x,\epsilon}\cap f(E)\subseteq f(A).$$
Now set $A'$ as the union over all $U'_{x,\epsilon}$ such that, as above, $x\in A$ and $U'_{x,\epsilon}\cap f(E)\subseteq f(A)$. $A'$ is obviously open and $f(A)=f(E)\cap A'$ but we need to show that $A'$ is indeed convex. But for $U'_{x,\epsilon},U'_{y,\delta}\subseteq A'$ and $0<\lambda<1$ one can easily verify $$\lambda U'_{x,\epsilon}+(1-\lambda)U'_{y,\delta}=U'_{\lambda x+(1-\lambda y),\lambda\epsilon+(1-\lambda)\delta}\subseteq A'$$ which concludes the proof.