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Homework Help: Dual space isometry

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data

    For the Banach space [tex]X = C[0,1][/tex] with the supremum norm, fix
    an element [tex]g \in X[/tex] and define a map [tex]\varphi_g : X \to \mathbb{C}[/tex]
    by
    [tex]
    \begin{align*}
    \varphi_g(h) := \int^1_0 g(t) h(t) dt, \qquad h \in X
    \end{align*}
    [/tex]
    Define [tex]W := \{ \varphi_g | g \in X \}[/tex].


    Prove that [tex]\varphi_g \in X^*[/tex] and calculate
    [tex]||\varphi_g||_{X^*}[/tex].




    2. Relevant equations



    3. The attempt at a solution

    The supremum norm for a function [tex]f \in C[0,1][/tex] is [tex]\displaystyle
    ||f|| = \sup_{x \in [0,1]} |f(x)|[/tex].

    For [tex]h_1,h_2 \in X[/tex] and [tex]\lambda \in \mathbb{C}[/tex]
    [tex]
    \begin{align*}
    \varphi_g(h_1 + h_2) =& \int^1_0 g(t) (h_1(t) + h_2(t)) dt =
    \int^1_0 g(t) h_1(t) dt + \int^1_0 g(t) h_2(t) dt = \varphi_g(h_1) +
    \varphi_g(h_2) \\
    \varphi_g(\lambda h_1) =& \int^1_0 \lambda g(t) h_1(t) dt = \lambda
    \int^1_0 g(t) h_1(t) dt
    \end{align*}
    [/tex]
    So [tex]\varphi_g[/tex] is linear functional.

    For any [tex]h \in X[/tex], [tex]\varphi_g[/tex] is continuous if [tex]\forall \varepsilon
    > 0[/tex] [tex]\exists \delta > 0[/tex] such that any [tex]h' \in X[/tex] satisfies
    [tex]
    \begin{align*}
    d(h,h') < \delta \implies d(\varphi_g(h), \varphi_g(h')) <
    \varepsilon \text{.}
    \end{align*}
    [/tex]
    The distance function is [tex]\displaystyle d(h,h') = ||h - h'|| =
    \sup_{x \in [0,1]} |h(t) - h'(t)|[/tex].
    [tex]
    \begin{align*}
    d(\varphi_g(h), \varphi_g(h')) =& |\int^1_0 g(t) h(t) dt -
    \int^1_0 g(t) h'(t) dt | = |\int^1_0 g(t) h(t) - g(t) h'(t) dt | \\
    =& |\int^1_0 g(t) (h(t) - h'(t)) dt| \leq \int^1_0 |g(t) (h(t) - h'(t))| dt \\
    \leq & \int^1_0 |g(t)| |(h(t) - h'(t))| dt \leq \sup_{t \in
    [0,1]} |(h(t) - h'(t))| \int^1_0 |g(t)| dt \\
    =& d(h,h') \int^1_0 |g(t)| dt
    \end{align*}
    [/tex]
    For any [tex]\varepsilon > 0[/tex], we get [tex]\displaystyle d(h,h') \int^1_0
    |g(t)| dt < \varepsilon[/tex] when [tex]\displaystyle d(h,h') <
    \frac{\varepsilon}{\displaystyle \int^1_0 |g(t)| dt} = \delta[/tex].
    Thus [tex]\varphi_g[/tex] is continuous. Are the above proofs correct?

    The formula for [tex]||\varphi_g||_{X^*}[/tex] is
    [tex]
    \begin{align*}
    ||\varphi_g||_{X^*} :=& \sup \{ |\varphi_g(h)| | \norm{h} = 1
    \} \\
    =& \sup \left\{ |\int^1_0 g(t)h(t) dt| \bigg| \sup_{t \in [0,1]}
    |h| = 1 \right\}
    \end{align*}
    [/tex]
    How can I find the value of this?
     
    Last edited: May 26, 2010
  2. jcsd
  3. May 26, 2010 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For the supremum, maybe some sort of triangle inequality application can give you a starting upper bound
     
  4. May 27, 2010 #3
    As far as I can say, they are.

    You have
    [tex]\left|\int^1_0 g(t)h(t) dt\right| \le \int^1_0 |g(t)| |h(t)| dt \le \int^1_0 |g(t)| dt[/tex]
    therefore
    [tex]||\varphi_g|| \le \int^1_0 |g(t)| dt [/tex].

    I think that [tex]||\varphi_g|| = \int^1_0 |g(t)| dt [/tex] holds. A naive approach to show this would be trying to approximate the step function [tex]h(t)=g(t)/|g(t)|[/tex] by continuous functions; but I guess there is a simpler solution.

    BTW the same question was asked here: http://www.mathhelpforum.com/math-h...tial-geometry/146627-dual-space-isometry.html
     
  5. May 27, 2010 #4
    Is this correct?

    [PLAIN]http://9ya7ng.blu.livefilestore.com/y1pFP94kdkanTL7oWHQXuecgsG7MYfNfM3fHYVt7AE01cgDtbQY8VkjQk94V8H5WceDMp8kOlh-X1WSs79GZtIUTEWTFMCBtceU/q3.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
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