# Coming up with counterexamples in Real Analysis

1. Aug 17, 2007

### 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.

Last edited: Aug 17, 2007
2. Aug 17, 2007

### 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.

3. Aug 17, 2007

### bham10246

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

4. Aug 17, 2007

### ansrivas

For Question 1 I believe its false.

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

5. Aug 17, 2007

### 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?

6. Aug 17, 2007

### 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...

7. Aug 17, 2007

### 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 $n \rightarrow \infty$, $\int_{\mathbb{R}}|f|^p \rightarrow \infty$?