Functional Analysis Problems

[Oxymoron]

Im having some difficulties proving some basic properties of the adjoint operator. I want to prove the following things:

1) There exists a unique map T^*:K\rightarrow H
2) That T^* is bounded and linear.
3) That T:H\rightarrow K is isometric if and only if T^*T = I.
4) Deduce that if T is an isometry, then T has closed range.
5) If S \in B(K,H), then (TS)^* = S^*T^*, and that T^*^* = T.
6) Deduce that if T is an isometry, then TT^* is the projection onto the range of T.

Note that H,K are Hilbert Spaces.

There are quite a few questions, and I am hoping that by proving each one I will get a much better understanding of these adjoint operators. Now I think I have made a fairly good start with these proofs, so I'd like someone to check them please.

We'll begin with the first one.

[Oxymoron]

I want to prove that there exists a unique mapping T^*:K\rightarrow H such that

\langle Th,k \rangle = \langle h, T^*k \rangle \quad \forall \, h \in H, \, k\in K

For each k \in K, the mapping h \rightarrow \langle Th, k\rangle_K is in H^*. Hence by Riesz's theorem, there exists a unique z \in H such that

\langle Th,k \rangle_K = \langle h,z \rangle_H \quad \forall \, h \in H.

Therefore there exists a unique map T^*: K \rightarrow H such that

\langle Th, k\rangle_K = \langle h,T^*k\rangle_H \quad \forall \, h \in H, \, k \in K .

Therefore there exists a unique T^*. \square

[Oxymoron]

2a) To see that T^* is linear, take k_1, k_2 \in K and \lambda \in \mathbb{F}, then for any h \in H we have

\langle Th,k_1+\lambda k_2 \rangle_K &=& \langle Th,k_1\rangle_K + \overline{\lambda}\langle Th, k_2 \rangle_K \\
&=& \langle h,T^*k_1\rangle_K + \overline{\lambda}\langle h,T^*k_2\rangle_K \\
&=& \langle h, T^*k_1 + \lambda T^*k_2\rangle_K


Hence

T^*(k_1+\lambda k_2) = T^*(k_1) + \lambda T^*(k_2)

T^* is linear. \square

[Oxymoron]

2b) To prove that T^* is bounded note first that

\|T^*k\|^2 = \langle T^* k, T^*k \rangle_K = \langle T(T^*k), k \rangle \leq \|T(T^*k)\|\|k\| \leq \|T\|\|T^*k\|\|k\| \quad \forall \, k \in K

Now suppose that \|T^*k\| > 0. Then dividing the above by \|T^*k\| we have

\|T^*k\| \leq \|T\|\|k\| \quad \forall \, k \in K

Note that this is trivial if \|T^*k\| = 0.

Therefore T^* is bounded. \square

[Oxymoron]

Im not sure how to begin part 3 and 4 so I'll skip it for now.

5) Now Im not sure if what I have done here proves anything?

\langle T^*^* h,k \rangle = \langle x, T^* y\rangle = \overline{\langle T^*y, x\rangle} = \overline{\langle y, Tx \rangle} = \langle Tx,y \rangle

Does this prove that T^*^* = T?

\langle (TS)^*x,y \rangle = \langle y, (TS)x \rangle = \langle T^*x, Sy \rangle = \langle S^*T^*x,y \rangle

And does this prove that (TS)^* = S^*T^*?

[Oxymoron]

3) We have to prove the following if and only if statement:

\|Th\| = \|h\| \Leftrightarrow T^*T = I

(\Leftarrow)

Suppose T^*T = I is true, then

\|Th\|^2 = \langle Th,Th \rangle = \langle h,T^*Th \rangle = \langle h,h \rangle = \|h\|^2

Hence \|Th\| = \|h\|[/tex] after taking square roots of both sides.

(\Rightarrow)

Suppose [itex]\|Th\| = \|h\| is true, then we have

\|Th\|^2 = \|h\|

That is

\langle Th,Th \rangle = \langle h,h \rangle

This implies that

\langle h,T^*Th \rangle = \langle h,h \rangle

Which implies that

T^*T = I

Therefore \|Th\| = \|h\| \Leftrightarrow T^*T = I. [itex]\square[/tex]