Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help! Stuck in proof of Riesz Representation Theorem

  1. Oct 2, 2011 #1
    I am lecturing out of R.Bartle, The Elements of Integration and Lebesgue Measure, for the first time. In the most recent lecture I got stuck not being able to argue an inequality on page 107. I cannot post the text here, sorry. But if anyone has the book, can you also explain how to derive the inequality in line 6:

    g(t + n[itex]^{-2}[/itex]) [itex]\leq[/itex] g(t) + 2[itex]\epsilon[/itex] ?

    The inequality G([itex]\psi[/itex][itex]_{n}[/itex]) [itex]\leq[/itex] g(t) + 2[itex]\epsilon[/itex] from the previous line does not seem to help much, since
    G([itex]\psi[/itex][itex]_{n}[/itex]) [itex]\leq[/itex] G([itex]\varphi[/itex][itex]_{t + n^{-2}, n}[/itex])
    follows from G being positive, and this is the opposite inequality of the desired.
    Or what am I missing here? Thanks for any enlightenment!

    Kind regards, Tommy
  2. jcsd
  3. Oct 2, 2011 #2


    User Avatar
    Science Advisor

    This looks like a result of g being continuous. Since I don't have the book, I can't comment on the rest.
  4. Oct 3, 2011 #3
    Thanks for the reply!

    Actually, in the part of the proof which I am asking about, the aim is the prove that the function g is continuous.

    There appears to be a mistake in the proof though. So for those who do have the book, here is a fix for later reference.

    Recall that J = [a,b] and C(J) is the set of real-valued continuous functions on J. A linear, bounded and positive functional G on C(J) is assumed given. Fix any t [itex]\in[/itex] [a,b) and a sufficiently large integer n. Let [itex]\varphi[/itex][itex]_{t,n}[/itex] be the function that maps x to 1 if a ≤ x ≤ t, maps x to 0 if t+[itex]\frac{1}{n}[/itex] ≤ x ≤ b, and maps x to 1 - n(x-t) if t < x < t+[itex]\frac{1}{n}[/itex]. Then [itex]\varphi[/itex][itex]_{t,n}[/itex] [itex]\in[/itex] C(J), and G([itex]\varphi[/itex][itex]_{t,n}[/itex]) is non-negative and non-increasing as a function of n.

    Let g(t) = lim[itex]_{n → ∞}[/itex] G([itex]\varphi[/itex][itex]_{t,n}[/itex]). Let g(s)=0 if s < a, and g(s)=G(1) if s ≥ b. Then g is monotone increasing. The aim is to show that g is everywhere continuous from the right, then to extend the Borel-Stieltjes measure generated by g to a measure defined on the Borel algebra, using the Hahn Extension Theorem.

    I cannot make the proof in the book work to prove continuity of g from the right. But the fix is easier than in the suggested proof.

    Let ε > 0 and assume that n is large enough to satisfy n > 2, n > [itex]\frac{1}{ε}[/itex]||G||, and g(t) ≤ G([itex]\varphi[/itex][itex]_{t,n}[/itex]) ≤ g(t) + ε.
    Then ||[itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex] - [itex]\varphi[/itex][itex]_{t,n}[/itex]|| = [itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}(t+\frac{1}{n})[/itex] = [itex]\frac{1}{n}[/itex] implies ||G|| ≥ |G(n([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex] - [itex]\varphi[/itex][itex]_{t,n}[/itex]))| = n(G([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex]) - G([itex]\varphi[/itex][itex]_{t,n}[/itex])), hence
    G([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex]) ≤ G([itex]\varphi[/itex][itex]_{t,n}[/itex]) + [itex]\frac{1}{n}[/itex]||G|| ≤ g(t) + 2ε.
    The right continuity of g at t follows by noting that
    g(t) ≤ g(t+[itex]\frac{1}{n^2}[/itex]) = lim[itex]_{k → ∞}[/itex] G([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},k}[/itex]) ≤ G([itex]\varphi[/itex][itex]_{t+\frac{1}{n^2},n}[/itex]) ≤ g(t) + 2ε.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook