Let $x\in U$ as interior for the set $U$ in topological vector space.
Can we show $\lambda \ne 0$ then $\lambda x\in \lambda U$ is interior for $\lambda U$?
I do as follows:
first since $x\in U$ is interior.i.e. exist $U_x = x +U_o\subset U$
Now $\lambda x + \lambda U_o \subset \lambda U$ now we know for TVS $\lambda U_o$ is also neiborhood of origin,hence exist some neiborhood of $\lambda x $ contains in $\lambda U$ hence it's interior.
Is my proof correct?
One Corollary is if zero is interior for $U$ then zero is interior for all $nU$ correct?