Let $X$ be a complex Banach space. Does validity of
$$
\mbox{ker}\left(e^{2\pi \imath \, T} - 1\right) = \overline{\sum\nolimits\_{k\in \mathbb{Z}} \mbox{ker} (T-k) }\quad \forall \, T \in B(X,X)
$$
imply that $\mbox{dim}X<\infty$?



A nullspace identity for operator exponentials
No: every complex Banach space satisfies this condition, and you don't need the closure. 


For any bounded linear operator $T$, $K = \sigma(T) \cap \mathbb Z$ is finite.
Write $$e^{2\pi i z} - 1 = g(z) \prod\_{k \in K} (z - k)$$
where $g(z)$ is analytic and nonzero on a neighbourhood of $\sigma(T)$. Therefore
by the holomorphic functional calculus $g(T)$ is invertible and $$ e^{2\pi i T} - I = g(T) \prod\_{k \in K} (T - k I)$$
so $$\ker(e^{2\pi i T}-I) = \ker\left(\prod\_{k \in K} (T - k I)\right)$$
Now from the partial fraction decomposition
$$ \dfrac{1}{\displaystyle \prod\_{k \in K} (z - k)} = \sum\_{k \in K} \dfrac{c\_k}{z-k}$$
we have for any $v \in \ker \left( \prod\_{k \in K} (T - k I) \right)$
 $$v = \sum\_{k \in K} c\_k \left(\prod\_{j \in K \backslash \{k\}} (T - jI)\right) v \in \sum\_{k \in K} \ker(T - k I)$$


