# Math Help - convergence in Lp, convergence for p smaller infinity

1. ## convergence in Lp, convergence for p smaller infinity

Let X=[0,1] , A= $[0,1]\cap B$. (B is the Borel- $\sigma$ algebra), $\lambda$ the lebesgue measure. In $(X,A,\lambda)$ explain why these statements are true or false:

1) if $\exists 1 \le p < \infty$ with $f_n \in L_p$ and $||f_n||_p \rightarrow 0$ then $||f_n||_\infty \rightarrow 0$

2) if $f_n \in L_p$ and $||f_n||_p \rightarrow 0 \forall 1 \le p < \infty$ then $||f_n||_\infty \rightarrow 0$

Well I honestly do not have much ideas how to solve this. It's clear 1) isn't true because 2) wouldn't make sense if 1) was true. But a formal explanation? A prove? I don't know.

All I tried was finding functions that are for example in L_1 but not in L_2 . But I couldn't find a contradiction. Also there exist Lebesgue-integrable functions that are not bounded...??

Could please someone give me a hint?

2. ## Re: convergence in Lp, convergence for p smaller infinity

Try $f_n(x):=\frac 1n\chi_{(0,1/n)}(x)$.

3. ## Re: convergence in Lp, convergence for p smaller infinity

$\int \frac{1}{n^p}* \chi_{(0,\frac{1}{n})} d\mu=\frac{1}{n^{p+1}} < \infty \forall n \in \mathbb{N} \Rightarrow f_n \in L_p \forall 1 \le p < \infty$
and $lim_{n \to \infty} (\frac{1}{n^{p+1}})^{\frac{1}{p}} =0$
but why doesn't converge

$||f_n||_{\infty}$ to 0 ?

$\forall y \in [\frac{1}{3n},\frac{1}{2n}] : |f_n (y)| = |\frac{1}{n}|$and
$\mu([\frac{1}{3n},\frac{1}{2n}])=\frac{1}{6n} >0 \forall n \in \mathbb{N}$

I'm not sure because for me $(0,\frac{1}{n})$ converges to the empty set so why doesn't $||f_n||_{\infty}$ go to 0 ?

4. ## Re: convergence in Lp, convergence for p smaller infinity

Sorry, I meant $n\chi_{(0,n^{-1})}$ for the first example. For the second one, take $f_n(x)=\frac{\log x}n$.

5. ## Re: convergence in Lp, convergence for p smaller infinity

Hi

Well, $\int n \chi_{(0,\frac{1}{n})}d\mu=1 \forall n$ but $\lim_{n\rightarrow \infty} (n^{p-1})^{\frac{1}{p}}=\infty \forall p > 1$ ???

And how can I calculate $\int (\frac{log(x)}{n})^p d\mu$ ?? It's not a Riemann integral...?

Regards

6. ## Re: convergence in Lp, convergence for p smaller infinity

Maybe $f_n=\sqrt n\chi_{(0,n^{-1})}$ is better for the first question.

For the computation of the integral, use the substitution $t=\log x$.

7. ## Re: convergence in Lp, convergence for p smaller infinity

Hi

What does $\sqrt{(0,\frac{1}{n})}$ mean? I've never seen this before. The square root of an interval???

The substitution didn't help here either. I don't know what to do because this is not a Riemann integral ????

$\frac{dt}{d\mu}=0$ so in the end this integral is simply 0 ???

Regards

8. ## Re: convergence in Lp, convergence for p smaller infinity

I've edited for the first; for the second is just an improper integral (I "cheated" a little as the function is not in $L^{\infty}$).

9. ## Re: convergence in Lp, convergence for p smaller infinity

Hi

Thank you.
1st:
So $(0,\frac{1}{n})$ isn't a zero set for all n so it isn't one when n goes to infinity as well?

then for all $x\in (0,\frac{1}{n}) |f_n(x)|=n^{\frac{1}{2}}$ and this is not bounded.

This is the correct reason for that yes?

2nd:
$\int |f_n|^p d\mu(x)=\int \frac{1}{n^p}*\log(x)^p d\mu(x)= \frac{1}{n^p}*\int t^p*e^t d\mu(t)$
$= \frac{1}{n^p} (e-p(e-(p-1)(e-(p-2)(e-...))))$ (*)

But what is $(e-p(e-(p-1)(e-(p-2)(e-...))))$ ? Is it smaller than infinity?

But here when n goes to infinity (*) goes to zero and so does the pth root of (*) or what did I do wrong here?

Regards

10. ## Re: convergence in Lp, convergence for p smaller infinity

Yes asb $\int_0^{+\infty}e^{-t}t^pdt$ is convergent. But I cheated a little as I took an unbounded function.

11. ## Re: convergence in Lp, convergence for p smaller infinity

Hi

1st: You just didn't answer my question. What about my arguments concerning the first part?

2nd: Why isn't $||f_n||_{\infty}\to 0$ for n to infinity?

I don't understand ??

12. ## Re: convergence in Lp, convergence for p smaller infinity

You are right for the first question (I forget answering you before, sorry...).

For the second one, $f_n$ is not essentially bounded (otherwise $t\mapsto \log t$ on $(0,1)$ would be so).

13. ## Re: convergence in Lp, convergence for p smaller infinity

Hi

But your "cheating" isn't a problem at all. The function has to be in L_p for all p only... it can be unbounded?

14. ## Re: convergence in Lp, convergence for p smaller infinity

Yes, as $\log t\geq n$ if $0, we should have $\lVert \log \cdot\rVert_{\infty}\geq n$ for each integer $n$, which is not possible.

15. ## Re: convergence in Lp, convergence for p smaller infinity

So the reason here is formally said for the 2nd part:

$f_n\in L_p \forall 1 \le p < \infty$ but

$\mu((0,\frac{1}{n}))=\frac{1}{n}>0 \forall n$ and
$\forall x \in (0,\frac{1}{n}) \Rightarrow f_n (x) \ge \frac{\log{\frac{1}{n}}}{n} = 0$

---doesn't work

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