1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]
    \varphi_g(h) := \int^1_0 g(t) h(t) dt, \qquad h \in X
    Define [tex]W := \{ \varphi_g | g \in X \}[/tex].

    Prove that [tex]\varphi_g \in X^*[/tex] and calculate

    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]
    \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
    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
    d(h,h') < \delta \implies d(\varphi_g(h), \varphi_g(h')) <
    \varepsilon \text{.}
    The distance function is [tex]\displaystyle d(h,h') = ||h - h'|| =
    \sup_{x \in [0,1]} |h(t) - h'(t)|[/tex].
    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
    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
    ||\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\}
    How can I find the value of this?
    Last edited: May 26, 2010
  2. jcsd
  3. May 26, 2010 #2


    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]
    [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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook