- #1
SqueeSpleen
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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.
|| 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.
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