Coming up with counterexamples in Real Analysis

[bham10246]

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 f has a bounded variation on [a,b], then is it true that f is of Riemann integration on [a,b]?


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


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




Thank you.

 

[AiRAVATA]

Question 2:

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

\frac{1}{q}=\frac{\lambda}{p}+\frac{1-\lambda}{r}.

Proof. Use Hölder's inequality.

 

[bham10246]

Thanks! I think I have seen your proposition before in some book!

 

[ansrivas]

For Question 1 I believe its false.

Eg. f(x)=1 if x is rational 0 o.w.

 

[bham10246]

Hi ansrivas, you might be right, as long as the bounded variation is for a finite partition of the interval [a,b]. That is,

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

It's because for your function f, the total variation of f is infinite, isn't it?

 

[bham10246]

As for my own answer to Question 3, I think if f is in L^1 \cap L^\infty, then f\in L^p for every p\geq 1.

So the converse of Problem 3 is certainly true! But I don't think this is true...

 

[bham10246]

Does this work?

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

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

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

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

Contradiction?

 

[mathwonk]

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