4
$\begingroup$

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!

$\endgroup$
5
$\begingroup$

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. $E = H$.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.