# L∞(E) Norm

1. Nov 17, 2013

### SqueeSpleen

Real Analysis, L∞(E) Norm as the limit of a sequence.

$|| f ||_{\infty}$ is the lesser real number $M$ such that $| \{ x \in E / |f(x)| > M \} | = 0$ ($| \cdot |$ used with sets is the Lebesgue measure).

Definition:
For every $1 \leq p < \infty$ and for every $E$ such that $0 < | E | < \infty$ we define:
$N_{p} [f] = (\frac{1}{ | E | } \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}}$

a) $p_{1} < p_{2} \Rightarrow N_{p_{1}} [f] \leq N_{p_{2}} [f]$
b) $N_{p} [f+g] \leq N_{p} [f] + N_{p} [g]$
c) If $\frac{1}{p} + \frac{1}{q} = 1$ $\frac{1}{ | E | } \displaystyle \int_{E} | f g | \leq N_{p} [f] N_{q} [g]$ 
d) $\lim_{p \to \infty} N_{p} [f] = | | f | |_{\infty}$

I need help with the a), I couldn't solve it. I didn't got anything good in my attemps, so I'll show my work in the other three points (Also I'm taking the opportunity to someone find any mistake I could have overlooked).

b) By triangular inequality of the norm $|| \cdot ||_{p}$ and observing that:
$| E |^{\frac{1}{p}} N_{p} = || \cdot ||_{p}$
We have:
$| E |^{\frac{1}{p}} N_{p} [f+g] \leq | E |^{\frac{1}{p}} N_{p} [f] + | E |^{\frac{1}{p}} N_{p} [g]$
Divide both sides by $| E |^{\frac{1}{p}}$ the work is done.

c) By Hölder Inequality we have: $\displaystyle \int_{E} | f g | \leq (\int_{E} | f |^{p} )^{\frac{1}{p}}(\int_{E} | g |^{q})^{\frac{1}{q}}$
We divide both sides by $| E |$ and knowing that as $\frac{1}{p}+\frac{1}{q}=1$ we have:
$\frac{1}{| E |} \displaystyle \int_{E} | f g | \leq \frac{1}{| E |} (\int_{E} | f |^{p} )^{\frac{1}{p}}(\int_{E} | g |^{q})^{\frac{1}{q}} = (\frac{1}{| E |} \int_{E} | f |^{p} )^{\frac{1}{p}}(\frac{1}{| E |} \int_{E} | g |^{q})^{\frac{1}{q}}$

d) Let be the sequence of nonnegative simple functions $\{ \varphi_{n} \}_{n \in \mathbb{N}}$ with $\varphi_{n} \leq \varphi_{n+1} \forall n \in \mathbb{N}$ and:
$\lim_{n \to \infty} \varphi_{n} = | f |$.
It's clear that $\lim_{n \to \infty} \varphi_{n} = | f |$ if and only if $\lim_{n \to \infty} \varphi_{n}^{p} = | f |^{p}$ as:
$\phi_{n} = \displaystyle \sum_{i=1}^{m_{n}} \alpha_{i,n} X_{E_{i,n}} \Longrightarrow \phi_{n}^{p} = \displaystyle \sum_{i=1}^{m_{n}} \alpha_{i,n}^{p} X_{E_{i,n}}$
Let be $\varphi$ a non negative simple function such that:
$\varphi \leq | f |$

$\varphi = \displaystyle \sum_{i=1}^{m} \alpha_{i} X_{E_{i}}$ with $| E_{i} | > 0 \forall i: 1 \leq i \leq m$ and:
$\alpha_{1} < \alpha_{2} < ... < \alpha_{m}$
$\displaystyle \int_{E} \varphi^{p} = \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |$
As $p \to \infty$ $\alpha_{m}^{p} | E_{m} |$ grows faster than $\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |$, to every $\varepsilon > 0 \exists p \in \mathbb{R}$ such that:
$| \frac{\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |}{\alpha_{m}^{p} | E_{m} |} -1 | < \varepsilon$
As all coefficients of the sumatory are positive, we can drop the absolute value bars.
$\frac{\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |}{\alpha_{m}^{p} | E_{m} |} -1 < \varepsilon$
Then
$\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} | < (1+\varepsilon) \alpha_{m}^{p} | E_{m} |$
Then we have:
$\alpha_{m}^{p} | E_{m} | \leq \int_{E} \varphi^{p} < (1 + \varepsilon) \alpha_{m}^{p} | E_{m} |$
We divide by $| E |$ sides and take power to $\frac{1}{p}$ and we have:
$((1 + \varepsilon) \frac{E_{m}}{| E |})^{1/p} \alpha_{m}$
As $p \to \infty$ this converges to $\alpha_{m}$
Now by Beppo-Levi we have:
$\lim_{n \to \infty} \int_{E} \varphi_{n}^p = \int_{E} |f|^{p}$
(We are using the simple functions $\varphi_{n}^p$ to aproximate $| f |^{p}$
We power both sides to $\frac{1}{p}$ and we are almost done.
The only thing left it to prove that $|| f ||_{\infty} = \lim_{n \to \infty} (\lim_{p \to \infty} N_{p} [ \varphi_{n} ])$ which is easy, because to every set with positive measure, taking $n$ big enough we'll have a $\varphi_{n}$ with that set, then the norm infinity is the supremum of that sequence, and as this is an increasing sequence then it's the limit.

Last edited: Nov 18, 2013
2. Nov 18, 2013

### brmath

part a. See the following (the norm he defines here is the $L_p$ norm):

Moreover, if we're on a bounded domain, we also have the relatively standard result that if f∈Lp1 for some p1∈[1,∞), then it is in Lp for every p≤p1 (which can be shown using Hölder's inequality). http://math.stackexchange.com/quest...ble-for-a-function-to-be-in-lp-for-only-one-p

This is more general.
Let M be the set of all measures of subsets of your space. Then, if p<q, then Lp < Lq if M has a positive lower bound, and Lq < Lp if M has a finite upper bound. http://www.johndcook.com/blog/2012/11/11/how-lp-spaces-nest/

3. Nov 18, 2013

### SqueeSpleen

4. Nov 18, 2013

### SqueeSpleen

Isn't the $L^{p}(E)$ norm:
$| | f | | = (\displaystyle \int_{E} | f |^{p} )^{\frac{1}{p}}$
?
This one is different, it has the measure of $E$ dividing before the root.

5. Nov 18, 2013

### brmath

You are right, the E is inside the root, which I didn't quite see. Well, I found you a proof. The finite measure of E is necessary, and shown by the counterexample at the bottom. The division by |E| falls out of his final inequality.

http://math.stackexchange.com/questions/66029/lp-and-lq-space-inclusion

Theorem Let X be a finite measure space. Then, for any 1≤p<q≤+∞
Lq(X,B,m)⊂Lp(X,B,m).
The proof follows from Hölder inequality. Note that 1/p=1/q+1/r, with r>0. Hence
∥f∥Lp≤meas (X)1/r∥f∥Lq.

The case reported on the wikipedia link of commenter answer follows from this, since of course, if X does not contain sets of arbitrary large measure, X itself can't have an arbitrary large measure.

For the counterexample: f(x)=1/x belongs to L2([1,+∞)), but clearly it does not belong to L1([1,+∞)).

Has your prof discussed measure in detail?

6. Nov 18, 2013

### SqueeSpleen

What do you mean by "discussed measure in detail"? (I wouldn't know, I have not any reference course to compare).
I think we did, our first set of practices were enterely of Lebesgue measure and we spend roughly 1/3 of the course (5-6 weeks) doing it.

Elemental sets -> $\sigma$ - elemental sets -> Borelians and a couple of ways of generate them.
External lebesgue measure.
Lebesgue Measure
Internal lebesgue measure, non-measurable sets.

That was more or less what we covered on measure before we moved to measurable functions.

We saw that for sets of finite measure $L^{p}(E) \subset L^{q}(E)$ if $q \leq p$
For infinite measure I know it's false because I already had the oportunity to solve the following exercise:
Let be $E = [0,\infty)$.
Prove that $f(x) = x^{\frac{1}{2}} (1+|ln(x)|)^{-1} \in L^{2} (E)$ but $f \notin L^{p} (E) \forall p \in [0,\infty) - \{ 2 \}$

Last edited: Nov 18, 2013
7. Nov 18, 2013

### SqueeSpleen

I'll transcribe here what I understood from the links.
$| E | < \infty$
Let be $p,q / 1 \leq p \leq q \Longrightarrow 1 < \frac{q}{p}$
Then there exists $s \in \mathbb{R}$ such that:
$1 = \frac{p}{q} + \frac{1}{s}$
There exists a $r \in \mathbb{R}$ such that $s= \frac{r}{p}$
$1 = \frac{p}{q} + \frac{p}{r}$
Now we apply the Hölder inequality:
$\displaystyle \int_{E} | f |^{p} = \displaystyle \int_{E} | f |^{p} \cdot X_{E} \leq | E |^{\frac{p}{r}} ( \displaystyle \int_{E} (| f |^{p})^{\frac{q}{p}} )^{\frac{p}{q}} = | E |^{\frac{p}{r}} ( \displaystyle \int_{E} | f |^{q} )^{\frac{p}{q}}$
We power to $\frac{1}{p}$:
$( \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} \leq | E |^{\frac{1}{r}} ( \displaystyle \int_{E} | f |^{q} )^{\frac{1}{q}}$
$\frac{1}{r} = \frac{1}{p} - \frac{1}{q} \Longrightarrow | E |^{\frac{1}{r}} = \frac{| E |^{\frac{1}{p}}}{| E |^{\frac{1}{q}}}$
$\frac{1}{| E |^{\frac{1}{p}}} ( \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} \leq \frac{1}{| E |^{\frac{1}{q}}} ( \displaystyle \int_{E} | f |^{q} )^{\frac{1}{q}}$
$\therefore ( \frac{1}{| E |} \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} \leq ( \frac{1}{| E |}\displaystyle \int_{E} | f |^{q} )^{\frac{1}{q}}$



8. Nov 19, 2013

### brmath

It sounds like you did indeed cover a lot of measure theory -- certainly more than I ever did. I never took a course in it specifically; covered Lp spaces in a course either on real or functional analysis. So maybe you should be helping me???

It looks to me like your interpretation of the sketchy proof I sent you is correct.

I don't think you need the line about s. If p/q < 1 then there has to be a real number that fills in the gap, and p/r is as good as any (since r can be whatever works). This doesn't invalidate your proof in any way. It's just a little roundabout and where you can simplify that kind of thing it's a good idea.

9. Nov 19, 2013

### SqueeSpleen

The course is of real analysis, both we spend a lot of time in the first chapter.

And I found that you can directly apply the following generalization of Hölder inequality:
$\displaystyle \sum_{i=1}^{m} p_{i} = \frac{1}{r} \wedge f_{i} \in L^{p_{i}} \forall i : 1 \leq i \leq m \Longrightarrow || \displaystyle \prod_{i=1}^{m} f_{i} ||_{r} \leq \displaystyle \prod_{i=1}^{m} | | f_{i} | |_{p_{i}}$

Thank you very much for your help again, I think this is the hardest course I ever toke and I really need help (As I don't speak English very well I really feel I need to say it explicity because I may sound a little rude to people helping me but it's only because my English is quite basic).

Last edited: Nov 19, 2013
10. Nov 19, 2013

### brmath

I wish my real analysis course had spent more time on measures. You have not sounded rude to me, but it is nice to get thanked. I really enjoy helping with problems, and I learn a lot too.