convergence in Lp, convergence for p smaller infinity
Let X=[0,1] , A= ![[0,1]\cap B](http://latex.codecogs.com/png.latex?[0,1]\cap B)
. (B is the Borel- 
algebra), 
the lebesgue measure. In )
explain why these statements are true or false:
1) if 
with 
and 
then 
2) if and 
then
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?
Re: convergence in Lp, convergence for p smaller infinity
Try :=\frac 1n\chi_{(0,1/n)}(x))
.
Re: convergence in Lp, convergence for p smaller infinity
} 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 ^{\frac{1}{p}} =0 )
but why doesn't converge
to 0 ?
![\forall y \in [\frac{1}{3n},\frac{1}{2n}] : |f_n (y)| = |\frac{1}{n}|](http://latex.codecogs.com/png.latex? \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}](http://latex.codecogs.com/png.latex? \mu([\frac{1}{3n},\frac{1}{2n}])=\frac{1}{6n} >0 \forall n \in \mathbb{N})
I'm not sure because for me )
converges to the empty set so why doesn't 
go to 0 ?
Re: convergence in Lp, convergence for p smaller infinity
Sorry, I meant })
for the first example. For the second one, take =\frac{\log x}n)
.
Re: convergence in Lp, convergence for p smaller infinity
Hi
Well, }d\mu=1 \forall n)
but ^{\frac{1}{p}}=\infty \forall p > 1 )
???
And how can I calculate }{n})^p d\mu )
?? It's not a Riemann integral...?
Regards
Re: convergence in Lp, convergence for p smaller infinity
Maybe })
is better for the first question.
For the computation of the integral, use the substitution 
.
Re: convergence in Lp, convergence for p smaller infinity
Hi
What does })
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 ????
so in the end this integral is simply 0 ???
Regards
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 
).
Re: convergence in Lp, convergence for p smaller infinity
Hi
Thank you.
1st:
So isn't a zero set for all n so it isn't one when n goes to infinity as well?
then for all  |f_n(x)|=n^{\frac{1}{2}})
and this is not bounded.
This is the correct reason for that yes?
2nd:
=\int \frac{1}{n^p}*\log(x)^p d\mu(x)= \frac{1}{n^p}*\int t^p*e^t d\mu(t))
(e-(p-2)(e-...)))) )
(*)
But what is (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 (Nod)
Re: convergence in Lp, convergence for p smaller infinity
Yes asb 
is convergent. But I cheated a little as I took an unbounded function.
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 
for n to infinity?
I don't understand ??
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, 
is not essentially bounded (otherwise 
on )
would be so).
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?
Re: convergence in Lp, convergence for p smaller infinity
Yes, as 
if 
, we should have 
for each integer 
, which is not possible.
Re: convergence in Lp, convergence for p smaller infinity
So the reason here is formally said for the 2nd part:
but
)=\frac{1}{n}>0 \forall n)
and
 \Rightarrow f_n (x) \ge \frac{\log{\frac{1}{n}}}{n} = 0)
---doesn't work
