Let $X$ be a reflexive Banach space and $T:X\to X$ is a linear operator. Is it true that $$ X = \mathrm{ker}(I-T) \oplus \overline{\mathrm{range}(I-T^\ast)}, $$ where $\oplus$ is the direct sum?
I know this is true when $X$ is a Hilbert space (due to the fact that $Tx=x$ implies $T^\ast x =x$), and I would suspect something like that is true when $T$ is a compact operator. How about the general case?
This fails even in finite dimension. Let $X=\mathbb C^2$, and $$ T=\begin{bmatrix} 1&1\\0&1\end{bmatrix}. $$ Then $(I-T)^2=0$, so $$\overline{\operatorname{ran}(I-T)}\subset\ker(I-T).$$
When $X$ is a Hilbert space, it is true that $$\tag1 X=\ker T\oplus \overline{\operatorname{ran} T^*} $$ for any bounded linear operator $T$, since $\overline{\operatorname{ran} T^*}=(\ker T)^\perp$.
For a general Banach space, the equality $(1)$ makes no sense, as $\ker T\subset X$ and $\operatorname{ran} T^*\subset X^*$.
More dramatically, it is known that any Banach space that is not isomorphic to a Hilbert space has a non-complemented subspace $M$. And if $X$ is separable, it is known that there exists a bounded linear $T:X\to X$ with $\ker T=M$. This impedes $(1)$ in general, and there is nothing you can put on the second summand that will make it work.
