# kernel and range of a linear operator in a reflexive space

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 isn't even true in finite dimensional spaces unless you take some further assumptions (such as $T = T^*$) – Omnomnomnom Dec 20 '18 at 0:01
• @Omnomnomnom What if range(I-T) is changed to range(I-T*)? – user58955 Dec 20 '18 at 0:05
• Then that's at least correct in the finite dimensional case – Omnomnomnom Dec 20 '18 at 1:28
• What does it mean to add a subspace of $X$ and a subspace of $X^{*}$? – Kavi Rama Murthy Dec 20 '18 at 5:58

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.