L1 Space: Difference between revisions
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<math> 3\implies 2: </math> Since <math> f=g</math> a.e., <math>|f-g|=0</math> a.e. Take a simple function, <math> \phi</math>, such that <math> 0\leq \phi\leq |f-g|</math>, such <math>\phi</math> must be <math>0</math> a.e. Therefore, <math> \int |f-g|=\sup\left\{\int \phi: 0\leq\phi\leq|f-g|,\phi\text{ simple}\right\}=0</math> | <math> 3\implies 2: </math> Since <math> f=g</math> a.e., <math>|f-g|=0</math> a.e. Take a simple function, <math> \phi</math>, such that <math> 0\leq \phi\leq |f-g|</math>, such <math>\phi</math> must be <math>0</math> a.e. Therefore, <math> \int |f-g|=\sup\left\{\int \phi: 0\leq\phi\leq|f-g|,\phi\text{ simple}\right\}=0</math> | ||
<math> 1\implies 3: </math> Suppose the set <math>\{ x\in X: f(x)\neq g(x)\} </math> does not have measure zero. Then either <math> E_+=\{x\in X: (f-g)_+(x)\neq 0\} </math> or <math>E_-=\{x\in X: (f-g)_-(x)\neq0\}</math> has nonzero measure, where <math> (f-g)_+</math> denotes <math> \max\{f-g,0\}</math> and <math> (f-g)_ | <math> 1\implies 3: </math> Suppose the set <math>\{ x\in X: f(x)\neq g(x)\} </math> does not have measure zero. Then either <math> E_+=\{x\in X: (f-g)_+(x)\neq 0\} </math> or <math>E_-=\{x\in X: (f-g)_-(x)\neq0\}</math> has nonzero measure, where <math> (f-g)_+</math> denotes <math> \max\{f-g,0\}</math> and <math> (f-g)_-</math> denotes <math> \max\{-(f-g),0\}</math>. WLOG, assume <math>E_+</math> has nonzero measure. Define the following sets <math> E_{+,n}=\{x\in X: (f-g)_+>1/n\}</math>, then from continuity from below, <math>\mu(E_+)=\mu\left(\cup_i^\infty E_{+,i}\right)=\lim_{i\to\infty}\mu(E_{+,i})>0</math>. This shows that there exists some <math> i\in\mathbb{N}</math> such that <math> \mu(E_{+,i})>0</math>, which implies that <math>\int_E (f-g)\geq\int_E(f-g)_+\geq \int_{E_{+,i}}(f-g)_+\geq \mu(E_{+,i})/i>0</math>, contradicting 1. | ||
With the proposition, we define our norm on <math>L^1(\mu)</math> to be <math>\lVert f\rVert=\int |f|</math>. This is indeed a norm since: | With the proposition, we define our norm on <math>L^1(\mu)</math> to be <math>\lVert f\rVert=\int |f|</math>. This is indeed a norm since: |
Revision as of 02:38, 18 December 2020
Very much not finished
Introduction
Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (X,\mathcal{M},\mu)} be a measure space. From our study of integration, we know that if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f,g} are integrable functions, the following functions are also integrable:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f+g}
- , for
This shows that the set of integrable functions on any measurable space is a vector space. Furthermore, integration is a linear functional on this vector space, ie a linear function sending elements in our vector space to , one would like to use integration to define a norm on our vector space. However, if one were to check the axioms for a norm, one finds integration fails to be a norm by taking almost everywhere, then . In other words, there are non zero functions which has a zero integral. This motivates our definition of to be the set of integrable functions up to equivalence to sets of measure zero.
Space
In this section, we will construct .
Definition
Let denote the set of integrable functions on , ie . Define an equivalence relation: if a.e. Then .
To make sense of the definition, we need the following proposition:
Proposition: Let , then the following are equivalent:
- for all
- a.e.
Proof
Since a.e., a.e. Take a simple function, , such that , such must be a.e. Therefore,
Suppose the set does not have measure zero. Then either or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_-=\{x\in X: (f-g)_-(x)\neq0\}} has nonzero measure, where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f-g)_+} denotes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \max\{f-g,0\}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f-g)_-} denotes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \max\{-(f-g),0\}} . WLOG, assume Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_+} has nonzero measure. Define the following sets Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{+,n}=\{x\in X: (f-g)_+>1/n\}} , then from continuity from below, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(E_+)=\mu\left(\cup_i^\infty E_{+,i}\right)=\lim_{i\to\infty}\mu(E_{+,i})>0} . This shows that there exists some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\in\mathbb{N}} such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(E_{+,i})>0} , which implies that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_E (f-g)\geq\int_E(f-g)_+\geq \int_{E_{+,i}}(f-g)_+\geq \mu(E_{+,i})/i>0} , contradicting 1.
With the proposition, we define our norm on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^1(\mu)} to be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lVert f\rVert=\int |f|} . This is indeed a norm since:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int |f+g| \leq \int |f|+\int|g|}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int |cf|=c\int |c||f|, c\in \mathbb{R}}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int |f|=0\iff f=0} a.e
Convergence in
With our norm defined, we can the metric to be . With a metric, one can talk about convergence in . This gives us a fourth mode of convergence for a sequence of functions. It is useful to compare these mode of convergence: Uniform Convergence Pointwise Convergence Pointwise a.e. Convergence
However, convergence in does not imply pointwise a.e. convergence and vice versa. To see that, we look at the following examples: