# 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 $T\rightarrow K$ 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 I am 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]

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