I am trying to solve the following problem.
Let $H$ be a Hilbert space and $T \in B(H)$ where $B(H)$ means the set of bounded linear functions from $H$ to $H$. Prove that $H = \overline{Im~ T} \oplus Ker ~T^*$.
Here's what I've done so far. We need to show that:
1) $Y:=\overline{Im~ T}$ is a closed subspace of $H$ so that we can apply the Orthogonal Decomposition Theorem.
2) $Y^\perp = Ker ~T^*$.
I'll start from Part 2. To show $Y^\perp \subset Ker ~T^*$, take $y_\perp\in Y^\perp$. We have $\langle y,y_\perp \rangle =0$ for every $y\in Y$. Since $y \in \overline{Im~ T}$, we have $y=\overline{T(y_0)}$ for some $y_0\in H$. Therefore, $\langle \overline{T(y_0)}, y_\perp \rangle=0$ for every $y_0\in H$. If there was no complex conjugate in the inner-product, then I could have done the following:
$$ 0 = \langle {T(y_0)},y_\perp \rangle = \langle y_0, T^*(y_\perp)\rangle, \ \ \textrm{for all } y_0 \in H$$
Then take $y_0 = T^*(y_\perp)$ and conclude that $y_\perp \in Ker ~T^*$.
But how do I deal with the complex conjugate? Also, is there a short way to prove Part 1 (I know the image of an operator is a subspace.)?
Best Answer
You have confused your symbolisms: $\overline{Im(T)}$ refers to the closure of the set $Im(T)$ in the norm topology and not to something that has to do with complex conjugation? I don't see how this makes sense actually.
For part 1, show that $Im(T)$ is a subspace and prove that the closure of any subspace is again a subspace. That's quite easy. Therefore it will be a closed subspace, since it is a subspace and the closure of a set.
For part 2: If $x\in\overline{Im(T)}$ then let $x=\lim T(x_n)$ for some sequence $(x_n)$. For $z\in\ker(T^*)$ we have $\langle x,z\rangle=\lim\langle Tx_n, z\rangle=\lim\langle x_n, T^*z\rangle=0$ (prove that we can interchange inner product with existent limits: you will need Cauchy-Schwarz) therefore $\overline{Im(T)}\subset\ker(T^*)^\bot$ and that yields $\ker(T^*)\subset\overline{Im(T)}^\bot$ (taking orthogonal complements reverses inclusions and the double orthogonal complement of a closed subspace -such as the kernel of a bounded operator- is the space itself: prove these as exercises!). For the reverse inclusion, let $z\bot\overline{Im(T)}$. Then for any $x\in H$ it is $\langle T^*z,x\rangle=\langle z, Tx\rangle=0$, hence $T^*z=0$ (it is vertical to the whole space) and therefore $z\in\ker(T^*)$.