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Let $H$ be a separable Hilbert space and $T:=I-A$, where $A:H\to H$ is a compact operator. If $T^\ast$ is the adjoint operator of $T$ it can be proved that $\ker T$ and $\text{im } T^\ast:=T^\ast (H)$ are closed thanks to the compactness of $A$ (pp. 469-70 here; I can translate for anybody interested). It is clear, from the very definition of adjoint operator, that $\ker T\perp \text{im } T^\ast$.

Therefore, as Kolmgorov-Fomin's says, in order to prove that $H=\ker T\oplus \text{im } T^\ast$, it is enough to show that no non-null vector can be orthogonal to both $\ker T$ and $\text{im } T^\ast$. Why? I suppose it is a trivial thing, but the book does not explain such a fact in the chapter on linear operators and I know nothing analogous from finite dimension linear algebra cases. Thank you so much!

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Since $\ker T$ and $\operatorname{im} T^\ast$ are closed and orthogonal to each other, the subspace

$$E := \ker T \oplus \operatorname{im} T^\ast$$

of $H$ is closed. Thus we have $E = H$ if an only if $E$ is dense. But a subspace $F$ of $H$ is dense, if and only if $F^\perp = \{0\}$. So showing that

$$(\ker T \oplus \operatorname{im} T^\ast)^\perp = (\ker T)^\perp \cap (\operatorname{im} T^\ast)^\perp = \{0\}$$

shows that $E$ is a dense closed subspace of $H$, i.e. $E = H$.

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    $\begingroup$ Thank you! Excuse me: it isn't clear to me why $\ker T\oplus \text{im}T^\ast$ is closed and why $F$ is dense iff $F^\perp=\{0\}$... Thank you so much again!!! $\endgroup$ – Self-teaching worker Dec 4 '14 at 19:54
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    $\begingroup$ Take a Cauchy sequence $(x_n)$ in $\ker T \oplus \operatorname{im} T^\ast$. We can write $x_n = a_n + b_n$ in a unique way with $a_n \in \ker T$ and $b_n\in \operatorname{im} T^\ast$, since the sum is direct. Since the spaces are orthogonal, we have $\lVert x_n - x_m\rVert^2 = \lVert a_n - a_m\rVert^2 + \lVert b_n - b_m\rVert^2$, so $\lVert a_n-a_m\rVert\leqslant \lVert x_n-x_m\rVert$ and ditto for $\lVert b_n-b_m\rVert$, hence $(a_n)$, $(b_n)$ are Cauchy sequences. Since the two subspaces are closed, they are complete, and $a_n \to a \in \ker T$, $b_n\to b\in\operatorname{im} T^\ast$. $\endgroup$ – Daniel Fischer♦ Dec 4 '14 at 20:05
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    $\begingroup$ Thus $x_n\to a+b\in E$, and $E$ is complete. Hence $E$ is closed. We could also obtain that $E$ is closed by other means, since $\ker T$ is finite-dimensional, and the sum (direct or not) of a closed subspace and a finite-dimensional subspace is closed. The equivalence $\overline{F} = H \iff F^\perp = \{0\}$ is an application of the Hahn-Banach theorem. If $F$ isn't dense, there is a nonzero continuous linear functional vanishing on $\overline{F}$. By the Riesz representation theorem, that means $F^\perp \neq \{0\}$. $\endgroup$ – Daniel Fischer♦ Dec 4 '14 at 20:06
  • $\begingroup$ I see that the things you say in the comments also holds without the separability of $H$. Very interesting. I heartily thank you! $\endgroup$ – Self-teaching worker Dec 4 '14 at 21:02

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