- #1
SqueeSpleen
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Real Analysis, L∞(E) Norm as the limit of a sequence.
[itex]|| f ||_{\infty}[/itex] is the lesser real number [itex] M [/itex] such that [itex]| \{ x \in E / |f(x)| > M \} | = 0 [/itex] ([itex] | \cdot | [/itex] used with sets is the Lebesgue measure).
Definition:
For every [itex]1 \leq p < \infty[/itex] and for every [itex]E[/itex] such that [itex] 0 < | E | < \infty [/itex] we define:
[itex]N_{p} [f] = (\frac{1}{ | E | } \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} [/itex]
a) [itex]p_{1} < p_{2} \Rightarrow N_{p_{1}} [f] \leq N_{p_{2}} [f][/itex]
b) [itex]N_{p} [f+g] \leq N_{p} [f] + N_{p} [g][/itex]
c) If [itex]\frac{1}{p} + \frac{1}{q} = 1[/itex] [itex]\frac{1}{ | E | } \displaystyle \int_{E} | f g | \leq N_{p} [f] N_{q} [g][/itex] [itex][/itex]
d) [itex]\lim_{p \to \infty} N_{p} [f] = | | f | |_{\infty}[/itex]
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 [itex]|| \cdot ||_{p}[/itex] and observing that:
[itex]| E |^{\frac{1}{p}} N_{p} = || \cdot ||_{p}[/itex]
We have:
[itex]| E |^{\frac{1}{p}} N_{p} [f+g] \leq | E |^{\frac{1}{p}} N_{p} [f] + | E |^{\frac{1}{p}} N_{p} [g][/itex]
Divide both sides by [itex]| E |^{\frac{1}{p}}[/itex] the work is done.c) By Hölder Inequality we have: [itex]\displaystyle \int_{E} | f g | \leq (\int_{E} | f |^{p} )^{\frac{1}{p}}(\int_{E} | g |^{q})^{\frac{1}{q}}[/itex]
We divide both sides by [itex]| E |[/itex] and knowing that as [itex]\frac{1}{p}+\frac{1}{q}=1[/itex] we have:
[itex]\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}} [/itex]d) Let be the sequence of nonnegative simple functions [itex]\{ \varphi_{n} \}_{n \in \mathbb{N}}[/itex] with [itex]\varphi_{n} \leq \varphi_{n+1} \forall n \in \mathbb{N}[/itex] and:
[itex]\lim_{n \to \infty} \varphi_{n} = | f |[/itex].
It's clear that [itex]\lim_{n \to \infty} \varphi_{n} = | f |[/itex] if and only if [itex]\lim_{n \to \infty} \varphi_{n}^{p} = | f |^{p}[/itex] as:
[itex]\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}}[/itex]
Let be [itex]\varphi[/itex] a non negative simple function such that:
[itex]\varphi \leq | f |[/itex]
[itex]\varphi = \displaystyle \sum_{i=1}^{m} \alpha_{i} X_{E_{i}}[/itex] with [itex] | E_{i} | > 0 \forall i: 1 \leq i \leq m[/itex] and:
[itex]\alpha_{1} < \alpha_{2} < ... < \alpha_{m} [/itex]
[itex]\displaystyle \int_{E} \varphi^{p} = \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |[/itex]
As [itex]p \to \infty[/itex] [itex]\alpha_{m}^{p} | E_{m} |[/itex] grows faster than [itex] \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |[/itex], to every [itex] \varepsilon > 0 \exists p \in \mathbb{R}[/itex] such that:
[itex]| \frac{\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |}{\alpha_{m}^{p} | E_{m} |} -1 | < \varepsilon[/itex]
As all coefficients of the sumatory are positive, we can drop the absolute value bars.
[itex] \frac{\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |}{\alpha_{m}^{p} | E_{m} |} -1 < \varepsilon[/itex]
Then
[itex] \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} | < (1+\varepsilon) \alpha_{m}^{p} | E_{m} | [/itex]
Then we have:
[itex]\alpha_{m}^{p} | E_{m} | \leq \int_{E} \varphi^{p} < (1 + \varepsilon) \alpha_{m}^{p} | E_{m} |[/itex]
We divide by [itex]| E |[/itex] sides and take power to [itex]\frac{1}{p}[/itex] and we have:
[itex]((1 + \varepsilon) \frac{E_{m}}{| E |})^{1/p} \alpha_{m}[/itex]
As [itex]p \to \infty[/itex] this converges to [itex]\alpha_{m}[/itex]
Now by Beppo-Levi we have:
[itex]\lim_{n \to \infty} \int_{E} \varphi_{n}^p = \int_{E} |f|^{p}[/itex]
(We are using the simple functions [itex]\varphi_{n}^p[/itex] to aproximate [itex]| f |^{p}[/itex]
We power both sides to [itex]\frac{1}{p}[/itex] and we are almost done.
The only thing left it to prove that [itex]|| f ||_{\infty} = \lim_{n \to \infty} (\lim_{p \to \infty} N_{p} [ \varphi_{n} ])[/itex] which is easy, because to every set with positive measure, taking [itex]n[/itex] big enough we'll have a [itex]\varphi_{n}[/itex] 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.
[itex]|| f ||_{\infty}[/itex] is the lesser real number [itex] M [/itex] such that [itex]| \{ x \in E / |f(x)| > M \} | = 0 [/itex] ([itex] | \cdot | [/itex] used with sets is the Lebesgue measure).
Definition:
For every [itex]1 \leq p < \infty[/itex] and for every [itex]E[/itex] such that [itex] 0 < | E | < \infty [/itex] we define:
[itex]N_{p} [f] = (\frac{1}{ | E | } \displaystyle \int_{E} | f |^{p})^{\frac{1}{p}} [/itex]
a) [itex]p_{1} < p_{2} \Rightarrow N_{p_{1}} [f] \leq N_{p_{2}} [f][/itex]
b) [itex]N_{p} [f+g] \leq N_{p} [f] + N_{p} [g][/itex]
c) If [itex]\frac{1}{p} + \frac{1}{q} = 1[/itex] [itex]\frac{1}{ | E | } \displaystyle \int_{E} | f g | \leq N_{p} [f] N_{q} [g][/itex] [itex][/itex]
d) [itex]\lim_{p \to \infty} N_{p} [f] = | | f | |_{\infty}[/itex]
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 [itex]|| \cdot ||_{p}[/itex] and observing that:
[itex]| E |^{\frac{1}{p}} N_{p} = || \cdot ||_{p}[/itex]
We have:
[itex]| E |^{\frac{1}{p}} N_{p} [f+g] \leq | E |^{\frac{1}{p}} N_{p} [f] + | E |^{\frac{1}{p}} N_{p} [g][/itex]
Divide both sides by [itex]| E |^{\frac{1}{p}}[/itex] the work is done.c) By Hölder Inequality we have: [itex]\displaystyle \int_{E} | f g | \leq (\int_{E} | f |^{p} )^{\frac{1}{p}}(\int_{E} | g |^{q})^{\frac{1}{q}}[/itex]
We divide both sides by [itex]| E |[/itex] and knowing that as [itex]\frac{1}{p}+\frac{1}{q}=1[/itex] we have:
[itex]\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}} [/itex]d) Let be the sequence of nonnegative simple functions [itex]\{ \varphi_{n} \}_{n \in \mathbb{N}}[/itex] with [itex]\varphi_{n} \leq \varphi_{n+1} \forall n \in \mathbb{N}[/itex] and:
[itex]\lim_{n \to \infty} \varphi_{n} = | f |[/itex].
It's clear that [itex]\lim_{n \to \infty} \varphi_{n} = | f |[/itex] if and only if [itex]\lim_{n \to \infty} \varphi_{n}^{p} = | f |^{p}[/itex] as:
[itex]\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}}[/itex]
Let be [itex]\varphi[/itex] a non negative simple function such that:
[itex]\varphi \leq | f |[/itex]
[itex]\varphi = \displaystyle \sum_{i=1}^{m} \alpha_{i} X_{E_{i}}[/itex] with [itex] | E_{i} | > 0 \forall i: 1 \leq i \leq m[/itex] and:
[itex]\alpha_{1} < \alpha_{2} < ... < \alpha_{m} [/itex]
[itex]\displaystyle \int_{E} \varphi^{p} = \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |[/itex]
As [itex]p \to \infty[/itex] [itex]\alpha_{m}^{p} | E_{m} |[/itex] grows faster than [itex] \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |[/itex], to every [itex] \varepsilon > 0 \exists p \in \mathbb{R}[/itex] such that:
[itex]| \frac{\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |}{\alpha_{m}^{p} | E_{m} |} -1 | < \varepsilon[/itex]
As all coefficients of the sumatory are positive, we can drop the absolute value bars.
[itex] \frac{\displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} |}{\alpha_{m}^{p} | E_{m} |} -1 < \varepsilon[/itex]
Then
[itex] \displaystyle \sum_{i=1}^{m} \alpha_{i}^{p} | E_{i} | < (1+\varepsilon) \alpha_{m}^{p} | E_{m} | [/itex]
Then we have:
[itex]\alpha_{m}^{p} | E_{m} | \leq \int_{E} \varphi^{p} < (1 + \varepsilon) \alpha_{m}^{p} | E_{m} |[/itex]
We divide by [itex]| E |[/itex] sides and take power to [itex]\frac{1}{p}[/itex] and we have:
[itex]((1 + \varepsilon) \frac{E_{m}}{| E |})^{1/p} \alpha_{m}[/itex]
As [itex]p \to \infty[/itex] this converges to [itex]\alpha_{m}[/itex]
Now by Beppo-Levi we have:
[itex]\lim_{n \to \infty} \int_{E} \varphi_{n}^p = \int_{E} |f|^{p}[/itex]
(We are using the simple functions [itex]\varphi_{n}^p[/itex] to aproximate [itex]| f |^{p}[/itex]
We power both sides to [itex]\frac{1}{p}[/itex] and we are almost done.
The only thing left it to prove that [itex]|| f ||_{\infty} = \lim_{n \to \infty} (\lim_{p \to \infty} N_{p} [ \varphi_{n} ])[/itex] which is easy, because to every set with positive measure, taking [itex]n[/itex] big enough we'll have a [itex]\varphi_{n}[/itex] 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.
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