Let $A$ be a closed operator on some Banach space $X$ and $\lambda \in \sigma_p(A)$. Then the dimension of the eigenspace $\ker(\lambda - A)$ is called the geometric multiplicity of $\lambda$. Moreover, the dimension of the generalized eigenspace $\bigcup_{n \in \mathbb N}\ker(\lambda - A)^n$ is called the algebraic multiplicity of $\lambda$. Clearly, the geometric multiplicity is always less or equal to the algebraic multiplicity.
So if the algebraic multiplicity is finite, then so is the geometric multiplicity. Is the converse also true? I know it is when $\lambda$ is a pole of the resolvent of $A$. But does it hold in general? I would also appreciate a reference if so!
Converse is not true: take $A$ the left-shift on $l^2(\mathbb N)$. Then $\ker(A) = span(e_1)$, $\ker(A^n) = span(e_1 \dots e_n)$. So algebraic multiplicity is infinite, while geometric multiplicity is $1$.