description/proof of that for bounded interval on \(\mathbb{R}\) with Borel \(\sigma\)-algebra and Lebesgue measure and \(L^2\) space, integral of absolute function is equal to or smaller than square-root of measure of interval times square-root of integral of squared absolute function
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of \(L^p\) over measure space.
- The reader admits the proposition that \(L^2\) over any measure space with the canonical inner product is a Hilbert space.
- The reader admits the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
Target Context
- The reader will have a description and a proof of the proposition that for any bounded interval on \(\mathbb{R}\) with the Borel \(\sigma\)-algebra of the Euclidean subspace topology and the Lebesgue measure and the \(L^2\) space over the measure space, the integral of any absolute function is equal to or smaller than the square-root of the measure of the interval times the square-root of the integral of the squared absolute function.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\((I, A, \lambda)\): \(\in \{\text{ the bounded intervals of } \mathbb{R}\}\), \(= (l, h)\), \((l, h]\), \([l, h)\), or \([l, h]\), with the subspace topology, with the Borel \(\sigma\)-algebra and the Lebesgue measure
\(\mathbb{C}\): \(= \text{ the complex Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(F\): \(\in \{\mathbb{C}, \mathbb{R}\}\)
\(L^2 (I, A, \lambda, F)\):
\(f\): \(\in L^2 (I, A, \lambda, F)\)
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Statements:
\(\int_I \vert f \vert d \lambda \le \sqrt{h - l} \sqrt{\int_I \vert f \vert^2 d \lambda}\)
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2: Proof
Whole Strategy: Step 1: see that \(L^2 (I, A, \lambda, F)\) is a Hilbert space; Step 2: apply the Cauchy-Schwarz inequality to \(1, \vert f \vert \in L^2 (I, A, \lambda, F)\).
Step 1:
\(L^2 (I, A, \lambda, F)\) is a Hilbert space with the inner product, \(\langle f, g \rangle = \int_I f \overline{g} d \lambda\), by the proposition that \(L^2\) over any measure space with the canonical inner product is a Hilbert space.
Step 2:
\(1 \in L^2 (I, A, \lambda, F)\), where \(1\) is the constantly-1 function, because \(1\) is measurable (the preimage of any measurable subset of \(F\) is \(I\) or \(\emptyset\)) and \(\int_I \vert 1 \vert^2 d \lambda = h - l \lt \infty\).
\(\vert f \vert \in L^2 (I, A, \lambda, F)\) is a well-known fact.
Let us apply the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space to \(1, \vert f \vert \in L^2 (I, A, \lambda, F)\): \(\vert \langle 1, \vert f \vert \rangle \vert \le \sqrt{\langle 1, 1 \rangle} \sqrt{\langle \vert f \vert, \vert f \vert \rangle}\).
\(\vert \langle 1, \vert f \vert \rangle \vert = \int_I 1 \vert f \vert d \lambda = \int_I \vert f \vert d \lambda\).
\(\sqrt{\langle 1, 1 \rangle} \sqrt{\langle \vert f \vert, \vert f \vert \rangle} = \sqrt{\int_I 1 1 d \lambda} \sqrt{\int_I \vert f \vert^2 d \lambda} = \sqrt{h - l} \sqrt{\int_I \vert f \vert^2 d \lambda}\).
So, \(\int_I \vert f \vert d \lambda \le \sqrt{h - l} \sqrt{\int_I \vert f \vert^2 d \lambda}\).
