description/proof of that \(L^2\) over measure space with canonical inner product is Hilbert space
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 knows a definition of Lebesgue integral of integrable complex function over measurable subset of measure space.
- The reader knows a definition of Hilbert space.
- The reader admits the proposition that for any map from any measure space into the \(1\)-dimensional complex Euclidean topological space with the Borel \(\sigma\)-algebra, if and only if the map is measurable, the real and imaginary parts maps are measurable.
- The reader admits the proposition that for any real number equal to or larger than \(1\) or \(\infty\), \(p\), its exponent conjugate, \(q\), any measure space, any element of \(\mathcal{L}^p\) over the measure space, and any element of \(\mathcal{L}^q\) over the measure space, the Lebesgue integral of the absolute product of the \(2\) elements is equal to or smaller than the product of the seminorms of the elements (Hölder's inequality).
- The reader admits the parallelogram law on any vectors space normed induced by any inner product.
- The reader admits the proposition that for any normed vectors space, if the norm satisfies the parallelogram law, the norm is induced by the unique inner product as this.
- The reader admits the proposition that \(L^p\) over any measure space is complete.
Target Context
- The reader will have a description and a proof of the proposition that \(L^2\) over any measure space with the canonical inner product is a Hilbert space.
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:
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\(\mathbb{C}\): \(= \text{ the complex Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(F\): \(\in \{\mathbb{C}, \mathbb{R}\}\)
\(L^2 (M, A, \mu, F)\): with the inner product, \(\langle [f_1], [f_2] \rangle = \int_M f_1 \overline{f_2} d \mu\)
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Statements:
\(L^2 (M, A, \mu, F) \in \{\text{ the Hilbert spaces }\}\)
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2: Proof
Whole Strategy: Step 1: see that the inner product is well-defined; Step 2: see that the norm is induced by the inner product; Step 3: conclude the proposition.
Step 1:
Let us see that the inner product is well-defined.
Let \([f_1], [f_2], [f_3] \in L^2 (M, A, \mu, F)\) and \(r_1, r_2 \in F\) be any.
\(\overline{f_2}\) is measurable, by the proposition that for any map from any measure space into the \(1\)-dimensional complex Euclidean topological space with the Borel \(\sigma\)-algebra, if and only if the map is measurable, the real and imaginary parts maps are measurable.
\(f_1 \overline{f_2}\) is measurable, as a well-known fact: as we accept that additions and multiplications of real measurable functions are measurable, the real part and the imaginary part of \(f_1 \overline{f_2}\) are measurable, and the proposition that for any map from any measure space into the \(1\)-dimensional complex Euclidean topological space with the Borel \(\sigma\)-algebra, if and only if the map is measurable, the real and imaginary parts maps are measurable applies.
Let us see that \(f_1 \overline{f_2}\) is integrable.
It is so if and only if \(\vert f_1 \overline{f_2} \vert\) is integrable, as is well known.
\(\int_M \vert f_1 \overline{f_2} \vert d \mu \le \Vert f_1 \Vert \Vert \overline{f_2} \Vert\), by the proposition that for any real number equal to or larger than \(1\) or \(\infty\), \(p\), its exponent conjugate, \(q\), any measure space, any element of \(\mathcal{L}^p\) over the measure space, and any element of \(\mathcal{L}^q\) over the measure space, the Lebesgue integral of the absolute product of the \(2\) elements is equal to or smaller than the product of the seminorms of the elements (Hölder's inequality): \(2\) is the exponent conjugate of \(2\), \(= \Vert f_1 \Vert \Vert f_2 \Vert \lt \infty\).
So, \(f_1 \overline{f_2}\) is integrable.
So, \(\int_M f_1 \overline{f_2} d \mu\) is well-defined.
Let us see that \(\int_M f_1 \overline{f_2} d \mu\) does not depend on the representatives, \(f_1, f_2\).
Let \(f'_1, f'_2 \in \mathcal{L}^2 (M, A, \mu, F)\) be any such that \([f'_j] = [f_j]\).
\(\int_M f'_1 \overline{f'_2} d \mu = \int_M (f'_1 - f_1 + f_1) (\overline{f'_2} - \overline{f_2} + \overline{f_2}) d \mu = \int_M (f'_1 - f_1) (\overline{f'_2} - \overline{f_2}) d \mu + \int_M (f'_1 - f_1) \overline{f_2} d \mu + \int_M f_1 (\overline{f'_2} - \overline{f_2}) + \int_M f_1 \overline{f_2} d \mu\).
But \(\vert \int_M (f'_1 - f_1) (\overline{f'_2} - \overline{f_2}) d \mu \vert \le \int_M \vert (f'_1 - f_1) (\overline{f'_2} - \overline{f_2}) \vert d \mu\), as a well known fact, \(\le \Vert f'_1 - f_1 \Vert \Vert \overline{f'_2} - \overline{f_2} \Vert\), by the proposition that for any real number equal to or larger than \(1\) or \(\infty\), \(p\), its exponent conjugate, \(q\), any measure space, any element of \(\mathcal{L}^p\) over the measure space, and any element of \(\mathcal{L}^q\) over the measure space, the Lebesgue integral of the absolute product of the \(2\) elements is equal to or smaller than the product of the seminorms of the elements (Hölder's inequality).
But \(\Vert f'_1 - f_1 \Vert = \Vert [f'_1 - f_1] \Vert = \Vert [0] \Vert = 0\), so, \(\int_M (f'_1 - f_1) (\overline{f'_2} - \overline{f_2}) d \mu = 0\).
Likewise, \(\vert \int_M (f'_1 - f_1) \overline{f_2} d \mu \vert \le \int_M \vert (f'_1 - f_1) \overline{f_2} \vert d \mu \le \Vert f'_1 - f_1 \Vert \Vert \overline{f_2} \Vert = 0\), so, \(\int_M (f'_1 - f_1) \overline{f_2} d \mu = 0\).
Likewise, \(\vert \int_M f_1 (\overline{f'_2} - \overline{f_2}) d \mu \vert \le \int_M \vert f_1 (\overline{f'_2} - \overline{f_2}) \vert d \mu \le \Vert f_1 \Vert \Vert \overline{f'_2} - \overline{f_2} \Vert = \Vert f_1 \Vert \Vert \overline{f'_2 - f_2} \Vert = \Vert f_1 \Vert \Vert f'_2 - f_2 \Vert = 0\), so, \(\int_M f_1 (\overline{f'_2} - \overline{f_2}) d \mu = 0\).
So, \(\int_M f'_1 \overline{f'_2} d \mu = \int_M f_1 \overline{f_2} d \mu\).
Let us see that \(\langle [f_1], [f_2] \rangle\) is indeed an inner product.
1) \((0 \le \langle [f_1], [f_1] \rangle)\) \(\land\) \((0 = \langle [f_1], [f_1] \rangle \iff [f_1] = 0)\): \(0 \le \int_M f_1 \overline{f_1} d \mu\); if \(0 = \int_M f_1 \overline{f_1} d \mu\), \(0 = \Vert [f_1] \Vert\), so, \([f_1] = 0\); if \([f_1] = 0\), \(0 = \inf_M 0 \overline{0} d \mu\).
2) \(\langle [f_1], [f_2] \rangle = \overline{\langle [f_2], [f_1] \rangle}\), where the over-line denotes the complex conjugate: \(\langle [f_1], [f_2] \rangle = \int_M f_1 \overline{f_2} d \mu = \int_M \overline{\overline{f_1} f_2} d \mu = \overline{\int_M f_2 \overline{f_1} d \mu} = \overline{\langle [f_2], [f_1] \rangle}\).
3) \(\langle r_1 [f_1] + r_2 [f_2], [f_3] \rangle = r_1 \langle [f_1], [f_3] \rangle + r_2 \langle [f_2], [f_3] \rangle\): \(\langle r_1 [f_1] + r_2 [f_2], [f_3] \rangle = \langle [r_1 f_1 + r_2 f_2], [f_3] \rangle = \int_M (r_1 f_1 + r_2 f_2) \overline{f_3} d \mu = \int_M r_1 f_1 \overline{f_3} + r_2 f_2 \overline{f_3} d \mu = r_1 \int_M f_1 \overline{f_3} d \mu + r_2 \int_M f_2 \overline{f_3} d \mu = r_1 \langle [f_1], [f_3] \rangle + r_2 \langle [f_2], [f_3] \rangle\).
So, \(\langle [f_1], [f_2] \rangle\) is an inner product.
Step 2:
The norm of \(L^2 (M, A, \mu, F)\) is induced by the inner product, because \(\Vert [f] \Vert = (\int_M \vert f \vert^2 d \mu)^{1 / 2} = \langle [f], [f] \rangle^{1 / 2}\).
In fact, this inner product is the unique possible inner product from which the norm is induced, by the parallelogram law on any vectors space normed induced by any inner product and the proposition that for any normed vectors space, if the norm satisfies the parallelogram law, the norm is induced by the unique inner product as this.
Step 3:
\(L^2 (M, A, \mu, F)\) is complete with the metric induced by the norm, by the proposition that \(L^p\) over any measure space is complete.
So, \(L^2 (M, A, \mu, F)\) is a Hilbert space.
