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Functional Analysis Problems

  1. Aug 5, 2005 #1
    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 [itex]T^*:K\rightarrow H[/itex]
    2) That [itex]T^*[/itex] is bounded and linear.
    3) That [itex]T:H\rightarrow K[/itex] is isometric if and only if [itex]T^*T = I[/itex].
    4) Deduce that if [itex]T[/itex] is an isometry, then [itex]T[/itex] has closed range.
    5) If [itex]S \in B(K,H)[/itex], then [itex](TS)^* = S^*T^*[/itex], and that [itex]T^*^* = T[/itex].
    6) Deduce that if [itex]T[/itex] is an isometry, then [itex]TT^*[/itex] is the projection onto the range of [itex]T[/itex].

    Note that [itex]H,K[/itex] 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.
  2. jcsd
  3. Aug 5, 2005 #2
    I want to prove that there exists a unique mapping [itex]T^*:K\rightarrow H[/itex] such that

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

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

    [tex]\langle Th,k \rangle_K = \langle h,z \rangle_H \quad \forall \, h \in H[/tex].

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

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

    Therefore there exists a unique [itex]T^*[/itex]. [itex]\square[/itex]
  4. Aug 5, 2005 #3
    2a) To see that [itex]T^*[/itex] is linear, take [itex]k_1, k_2 \in K[/itex] and [itex]\lambda \in \mathbb{F}[/itex], then for any [itex]h \in H[/itex] we have

    [tex]\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


    [tex]T^*(k_1+\lambda k_2) = T^*(k_1) + \lambda T^*(k_2)[/tex]

    [itex]T^*[/itex] is linear. [itex]\square[/itex]
  5. Aug 5, 2005 #4
    2b) To prove that [itex]T^*[/itex] is bounded note first that

    [tex]\|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 [/tex]

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

    [tex]\|T^*k\| \leq \|T\|\|k\| \quad \forall \, k \in K[/tex]

    Note that this is trivial if [itex]\|T^*k\| = 0[/itex].

    Therefore [itex]T^*[/itex] is bounded. [itex]\square[/itex]
  6. Aug 5, 2005 #5
    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?

    [tex]\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 [/tex]

    Does this prove that [itex]T^*^* = T[/itex]?

    [tex]\langle (TS)^*x,y \rangle = \langle y, (TS)x \rangle = \langle T^*x, Sy \rangle = \langle S^*T^*x,y \rangle [/tex]

    And does this prove that [itex](TS)^* = S^*T^*[/itex]?
  7. Aug 5, 2005 #6
    3) We have to prove the following if and only if statement:

    [tex]\|Th\| = \|h\| \Leftrightarrow T^*T = I[/tex]


    Suppose [itex]T^*T = I[/itex] is true, then

    [tex]\|Th\|^2 = \langle Th,Th \rangle = \langle h,T^*Th \rangle = \langle h,h \rangle = \|h\|^2[/tex]

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


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

    [tex]\|Th\|^2 = \|h\|[/tex]

    That is

    [tex]\langle Th,Th \rangle = \langle h,h \rangle[/tex]

    This implies that

    [tex]\langle h,T^*Th \rangle = \langle h,h \rangle[/tex]

    Which implies that

    [tex]T^*T = I[/tex]

    Therefore [itex]\|Th\| = \|h\| \Leftrightarrow T^*T = I[/itex]. [itex]\square[/tex]
    Last edited: Aug 5, 2005
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