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Coming up with counterexamples in Real Analysis

  1. Aug 17, 2007 #1
    Coming up with counterexamples is hard. So to prove or not to prove, that depends if there exists a counterexample.

    Question 1 has been ANSWERED!: If [itex]f[/itex] has a bounded variation on [itex] [a,b] [/itex], then is it true that [itex]f[/itex] is of Riemann integration on [itex][a,b][/itex]?


    Question 2 has been ANSWERED!: Is it true that [itex]L^1(\mathbb{R}) \cap L^3(\mathbb{R}) \subseteq L^2(\mathbb{R}) [/itex]?


    Question 3. Is it true that
    [itex]\cap_{1 \leq p<\infty} \: L^{p}(\mathbb{R},m) \subseteq L^{\infty}(\mathbb{R},m) [/itex] where [itex]m[/itex] denotes Lebesgue measure on [itex]\mathbb{R}[/itex].




    Thank you.
     
    Last edited: Aug 17, 2007
  2. jcsd
  3. Aug 17, 2007 #2
    Question 2:

    Proposition. If [itex]0<p<q<r \le \infty[/itex], then [itex]L^p\cap L^r \subset L^q[/itex] and [itex]\|f\|_q\le \|f\|_p^\lambda \|f\|_r^{1-\lambda}[/itex], where [itex]\lambda \in (0,1)[/itex] is defined by

    [tex]\frac{1}{q}=\frac{\lambda}{p}+\frac{1-\lambda}{r}.[/tex]

    Proof. Use Hölder's inequality.
     
  4. Aug 17, 2007 #3
    Thanks! I think I have seen your proposition before in some book!
     
  5. Aug 17, 2007 #4
    For Question 1 I believe its false.

    Eg. f(x)=1 if x is rational 0 o.w.
     
  6. Aug 17, 2007 #5
    Hi ansrivas, you might be right, as long as the bounded variation is for a finite partition of the interval [a,b]. That is,

    [itex]\sum_{i=1,..., N} |f(x_i)-f(x_i-1)| \leq M [/itex] for some M.

    It's because for your function f, the total variation of f is infinite, isn't it?
     
  7. Aug 17, 2007 #6
    As for my own answer to Question 3, I think if f is in [itex] L^1 \cap L^\infty[/itex], then [itex]f\in L^p[/itex] for every [itex]p\geq 1[/itex].

    So the converse of Problem 3 is certainly true! But I don't think this is true...
     
  8. Aug 17, 2007 #7
    Does this work?

    By definition, [itex]ess \sup f(x) = \inf \{M : m\{x: f(x)> M\}=0 \} [/itex].

    So suppose such finite M does not exist. Then [itex]m\{x: f(x)> n\} >0 [/itex] for all n.

    Then by Tchevbychev, [itex]\int_{\mathbb{R}} |f|^p \geq n m(E) >0 [/itex] where [itex]E = \{x: f(x)> n \} [/itex].

    So as [itex]n \rightarrow \infty[/itex], [itex]\int_{\mathbb{R}}|f|^p \rightarrow \infty[/itex]?

    Contradiction?
     
  9. Aug 17, 2007 #8

    mathwonk

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    all possible counterexamples are in the book of gelbaum and olmstead.
     
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