Let $\mathfrak{g}$ be a solvable, non zero subalgebra of $gl(V)$, with $V$ a finite dimensional vector space over an algebraically closed field $\mathbb{F}$ of characteristic zero. Then $V$ contains a common eigenvector for all the endomorphisms in $\mathfrak{g}$.
Fulton and Humphreys states this theorem. Proof works on the key idea that $V$ is of finite dimension, mainly on mathematical induction.
My doubt is can we extend this result to infinite dimensional vector spaces ? Then how will the proof works ? If not possible does there exist a counter example ?
We already know that another well known theorem on nilpotent (Engel's Theorem) does not hold in the infinite dimensional case.
Take your vector space over $\mathbb{C}$ to be $\mathbb{C}[x]$. Consider the linear transformation $T:\mathbb{C}[x]\rightarrow\mathbb{C}[x]$ that sends $P(x)$ to $xP(x)$. Take your subalgebra ${g}$ to be the lie subalgebra generated by $T$. $g$ is 1 dimensional (abelian), so solvable. However, $g$ does not have any common eigenvector because $T$ itself has no eigenvectors.
The thing is the fact "any endomorphism of a finite dimensional vector space over an algebraically closed field must have an eigenvector" does not hold in infinite dimensions. Let alone Lie's theorem.