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Functional Analysis problems need checking

  1. Jul 28, 2005 #1
    Question 1

    Prove that if [itex](V, \|\cdot\|)[/itex] is a normed vector space, then

    [tex] \left| \|x\| - \|y\| \right| \leq \|x-y\|[/tex]

    for every [itex]x,y \in V[/itex]. Then deduce that the norm is a continuous function from [itex]V[/itex] to [itex]\mathbb{R}[/itex].
     
  2. jcsd
  3. Jul 28, 2005 #2
    Solution

    Let [itex]x = x-y+y[/itex]. Taking norms on both sides gives

    [tex]\|x\| = \|x-y+y\| \leq \|x-y\| + \|y\|[/tex]

    This implies that

    [tex]\|x\| - \|y\| \leq \|x-y\|[/tex] (1)

    Now let [itex]y = y -x + x[/itex]. Taking norms on both sides gives

    [tex]\|y\| = \|y - x + x\| \leq \|y - x\| + \|x\|[/tex]

    Which implies that

    [tex]\|y\| - \|x\| \leq \|y-x\|[/tex]

    Which is also

    [tex]-\left(\|x\|-\|y\|\right) \leq \|x-y\|[/tex] (2)

    By considering both (1) and (2) we have

    [tex]\left| \|x\| - \|y\| \right| \leq \|x - y\|[/tex]

    For continuity, suppose we fix [itex]y = x_0 \in V[/itex]. Then for every [itex]x \in V[/itex] the triangle inequality gives us

    [tex]\left| \|x\| - \|x_0\| \right| \leq \|x - x_0\|[/tex]

    Which implies

    [tex]\left| \|x\| - \|x_0\| \right| < \epsilon[/tex]

    Satisfying

    [tex]\|x-x_0\| < \epsilon[/tex]
     
    Last edited: Jul 28, 2005
  4. Jul 28, 2005 #3

    Galileo

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    Looks good. If it's homework I suggest to write down the continuity proof more clearly and completely like you did with the first question.
     
  5. Jul 29, 2005 #4
    Thankyou for replying Galileo. Here is my next question/solution...


    Question 2

    Let [itex]\{a_n\}[/itex] be a fixed bounded sequence of complex numbers, and define [itex]T:l^2 \rightarrow l^2[/itex] by

    [tex]T(\{x_n\}) = \{a_nx_n\}[/tex]

    Prove that [itex]T[/itex] is linear and bounded with

    [tex]\|T\| = \sup |a_n|[/tex]
     
  6. Jul 29, 2005 #5
    The first thing I did was assume that the norm is the usual one

    [tex] \|x\| = \left(\sum_{n=1}^{\infty} |x_n|^2 \right)^{1/2}[/tex]

    Then to show that [itex]T[/itex] is linear I showed two things:

    (1) [tex]T(\{x_n+y_n\}) = \{a_n(x_n+y_n)\} = \{a_nx_n + a_ny_n\} = \{a_nx_n\} + \{a_ny_n\} = T(\{x_n\}) + T(\{y_n\})[/tex]

    (2) [tex]T(\lambda\{x_n\}) = \{\lambda a_nx_n\} = \lambda \{a_n x_n\} = \lambda T(\{x_n\})[/tex]

    For every [itex]x_n,y_n \in l^{\infty}[/itex]

    Hence linear.


    Further, for every [itex]x_n \in l^{\infty}[/itex] we have

    [tex]\|T(\{x_n\})\|^2 = \sum_{i=1}^{\infty} |a_nx_n|^2 \leq \left(\sup_{n\in\mathbb{N}}|a_n|\right)^2\sum_{n=1}^{\infty}|x_n|^2 = \left(\sup_{n\in\mathbb{N}}|a_n|\right)^2\|x\|^2[/tex]

    Therefore [itex]T:l^2\rightarrow l^2[/itex] is bounded, with

    [tex]\|T(\{x_n\})\| \leq M \|x\|[/tex]

    For all [itex]\{x_n\} \in l^{\infty}[/itex] where

    [tex]M = \sup_{n\in\mathbb{N}}|a_n| = \|a_n\|_{\infty}[/tex]

    ie, the supremum norm of [itex]\{a_n\} \in l^{\infty}[/itex].
     
    Last edited: Jul 29, 2005
  7. Jul 29, 2005 #6
    Your first proof works for me just fine, but you have to use the given norm that tells you that
    [tex] \| T(\{x_n\}) \| = \textrm{sup}(a_n) [/tex]
    This is a function mapping sequences to sequences, and you've been given a norm differing from the Euclidean norm, so you have to use that one.
     
  8. Jul 29, 2005 #7
    Thankyou for replying MS.

    If the norm I am actually meant to be using is

    [tex]\|T\| = \sup |a_n|[/tex]

    Then isn't [itex]T[/itex] obviously bounded by the supremum of the [itex]a_n's[/itex]? This is probably wrong, since Im used to seeing an inequality instead of an equality - which is alarming.
     
    Last edited: Jul 29, 2005
  9. Jul 29, 2005 #8
    Yeah, the norm should be bounded in the case of bounded coefficients.
     
  10. Aug 1, 2005 #9
    How would I formalize this statement? Or is it enough to simply say that the norm is bounded by the supremum of the coefficients?
     
  11. Aug 4, 2005 #10
    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.
     
  12. Aug 4, 2005 #11
    1) (To prove that [itex]T^*[/itex] is unique I'll be referring to Riesz's Theorem.)

    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]
     
    Last edited: Aug 4, 2005
  13. Aug 4, 2005 #12
    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]

    Hence

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

    [itex]T^*[/itex] is linear. [itex]\square[/itex]
     
    Last edited: Aug 5, 2005
  14. Aug 5, 2005 #13
    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]
     
    Last edited: Aug 5, 2005
  15. Aug 8, 2005 #14
    Does anyone know if my 3 solutions are correct? ...anyone?
     
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