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!

Homework Help: Continuity of the identity function on function spaces.

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that if [tex]p\in (1,\infty)[/tex] the identity functions [tex]id:C^{0}_{1}[a,b]\longrightarrow C^{0}_{p}[a,b][/tex] and [tex]id:C^{0}_{p}[a,b]\longrightarrow C^{0}_{\infty}[a,b][/tex] are not continuous.

    2. Relevant equations

    [tex]C^{0}_{p}[a,b][/tex] is the space of continuous functions on the [a,b] with the p-norm [tex]\left\vert\left\vert f\right\vert\right\vert_{p}=\int_{a}^{b}\vert f\vert^{p}\,dx[/tex]

    3. The attempt at a solution

    It is sufficient to prove that for [tex]C^{0}[0,1][/tex] because I can map the interval [tex][0,1][/tex] to the interval [tex][a,b][/tex] via [tex]x=(b-a)t+a[/tex]. In fact, since [tex]C^{0}[0,1][/tex] is a vector space is sufficient to prove that for [tex]f_{0}\equiv 0[/tex] the identity is discontinuous.

    To prove discontinuity I have to prove that [tex]\exists \epsilon>0 \mid \forall\delta>0[/tex] I can show that [tex]\left\vert\left\vert f\right\vert\right\vert_{1}<\delta \longrightarrow \left\vert\left\vert f\right\vert\right\vert_{p}>\epsilon[/tex] or what's the same, there is a sequence of functions that the area below their absolute value is less than delta but the area below their absolute value elevated to p is not bounded by epsilon. For the second case I must prove that there exists a sequence of functions for that the maximum of each is not bounded by epsilon but the area below their absolute value elevated to p is bounded by delta.

    In other words I have to prove that the [tex]p[/tex]-norm and the [tex]\infty[/tex]-norm (or the [tex]1[/tex]-norm) are not equivalent.

    My problem is that to prove vía sequences I can't figure out a function sequence that elevated to p can help me to prove that.

    I have already proven something similar: that [tex]id:C^{0}_{1}[0,1]\longrightarrow C^{0}_{\infty}[0,1][/tex] is not continuous. I have done it via the functions [tex]g_{\delta}(x)=1-\frac{1}{\delta}x[/tex] for [tex]0\leq x\leq \delta[/tex] and 0 for [tex]\delta\leq x\leq 1[/tex] and it's straightforward. However this functions don't help me in my present problem. Please Help!
  2. jcsd
  3. Sep 6, 2009 #2
    Try [tex]f_n(x)=\exp{(x/n)}[/tex].

    Btw, I'm sure you just forgot to put this in, but the p-norm is
    [tex]\|f\|_p = \left(\int_a^b |f|^p\,dx\right)^{1/p}.[/tex]​
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook