description/proof of that for real number equal to or larger than \(1\) or \(\infty\), \(p\), its exponent conjugate, \(q\), element of \(\mathcal{L}^p\), and element of \(\mathcal{L}^q\), Lebesgue integral of absolute product of \(2\) elements is equal to or smaller than product of seminorms (Hölder's inequality)
Topics
About: measure space
The table of contents of this article
Starting Context
- The reader knows a definition of exponent conjugate of exponent.
- The reader knows a definition of \(\mathcal{L}^p\) over measure space.
- The reader admits the proposition that for any real number larger than \(1\), \(p\), its exponent conjugate, \(q\), and any non-negative real numbers, \(r_1\) and \(r_2\), \(r_1 r_2\) is equal to or smaller than \({r_1}^p / p\) plus \({r_2}^q / q\).
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
Target Context
- The reader will have a description and a proof of 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).
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:
\(\mathbb{C}\): \(= \text{ the complex Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(F\): \(\in \{\mathbb{C}, \mathbb{R}\}\)
\(p\): \(\in \mathbb{R} \cup \{- \infty, \infty\}\), such that \(1 \le p\)
\(q\): \(= \text{ the exponent conjugate of } p\), \(\in \mathbb{R} \cup \{- \infty, \infty\}\)
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\(\mathcal{L}^p (M, A, \mu, F)\):
\(\mathcal{L}^q (M, A, \mu, F)\):
\(f_1\): \(\in \mathcal{L}^p (M, A, \mu, F)\)
\(f_2\): \(\in \mathcal{L}^q (M, A, \mu, F)\)
//
Statements:
\(\int \vert f_1 f_2 \vert d \mu \le \Vert f_1 \Vert_p \Vert f_2 \Vert_q\)
//
Especially, \(f_1 f_2 \in \mathcal{L}^1 (M, A, \mu, F)\).
2: Note
In fact, this proposition is used for proving that \(\Vert \bullet \Vert_p\) is indeed a seminorm for \(\mathcal{L}^p (M, A, \mu, F)\), which is not any problem, because Proof does not depend on \(\Vert \bullet \Vert_p\)'s being a seminorm (especially, satisfying the triangle inequality) but just depends on the definition of \(\Vert \bullet \Vert_p\).
3: Proof
Whole Strategy: Step 1: suppose that \(1 \lt p \lt \infty\); Step 2: conclude the proposition using the proposition that for any real number larger than \(1\), \(p\), its exponent conjugate, \(q\), and any non-negative real numbers, \(r_1\) and \(r_2\), \(r_1 r_2\) is equal to or smaller than \({r_1}^p / p\) plus \({r_2}^q / q\); Step 3: suppose that \(p = 1\); Step 4: conclude the proposition by seeing that \(\vert f_1 f_2 \vert \le \vert f_1 \vert \Vert f_2 \Vert_{\infty}\) holds almost everywhere; Step 5: suppose that \(p = \infty\); Step 6: conclude the proposition.
Step 1:
Let us suppose that \(1 \lt p \lt \infty\).
Step 2:
Let us suppose that \(\Vert f_1 \Vert_p = 0\).
\(\vert f_1 \vert =_{a.e} 0\), as is well known.
So, \(\vert f_1 f_2 \vert =_{a.e} 0\).
So, \(\int \vert f_1 (m) f_2 (m) \vert d \mu = 0\).
So, \(\int \vert f_1 f_2 \vert d \mu = 0 \le \Vert f_1 \Vert_p \Vert f_2 \Vert_q\).
Likewise when \(\Vert f_2 \Vert_q = 0\).
Let us suppose that \(\Vert f_1 \Vert_p \neq 0\) and \(\Vert f_2 \Vert_q \neq 0\).
Let \(f'_1 := 1 / \Vert f_1 \Vert_p f_1\) and \(f'_2 := 1 / \Vert f_2 \Vert_q f_2\).
\(\Vert f'_1 \Vert_p = \Vert 1 / \Vert f_1 \Vert_p f_1 \Vert_p = 1 / \Vert f_1 \Vert_p \Vert f_1 \Vert_p = 1\); likewise, \(\Vert f'_2 \Vert_q = 1\).
For each \(m \in M\), \(\vert f'_1 (m) f'_2 (m) \vert \le 1 / p \vert f'_1 (m) \vert^p + 1 / q \vert f'_2 (m) \vert^q\), by the proposition that for any real number larger than \(1\), \(p\), its exponent conjugate, \(q\), and any non-negative real numbers, \(r_1\) and \(r_2\), \(r_1 r_2\) is equal to or smaller than \({r_1}^p / p\) plus \({r_2}^q / q\).
So, \(\int \vert f'_1 (m) f'_2 (m) \vert d \mu \le \int (1 / p \vert f'_1 (m) \vert^p + 1 / q \vert f'_2 (m) \vert^q) d \mu = 1 / p \int \vert f'_1 (m) \vert^p d \mu + 1 / q \int \vert f'_2 (m) \vert^q d \mu = r {\Vert f'_1 \Vert_p}^p + 1 / q {\Vert f'_2 \Vert_q}^q = 1 / p + 1 / q = 1\).
So, \(\int \vert f'_1 (m) f'_2 (m) \vert d \mu = \int \vert 1 / \Vert f_1 \Vert_p f_1 1 / \Vert f_2 \Vert_q f_2 \vert d \mu = 1 / \Vert f_1 \Vert_p 1 / \Vert f_2 \Vert_q \int \vert f_1 f_2 \vert d \mu \le 1\).
So, \(\int \vert f_1 f_2 \vert d \mu \le \Vert f_1 \Vert_p \Vert f_2 \Vert_q\).
Step 3:
Let us suppose that \(p = 1\).
\(q = \infty\).
Step 4:
Let us think of \(S := \{m \in M \vert f_1 (m) \neq 0\}\).
Let \(S_0 = \{m \in M \vert 1 \le \vert f_1 (m) \vert\}\) and for each \(j \in \mathbb{N} \setminus \{0\}\), \(S_j = \{m \in M \vert 1 / (j + 1) \le \vert f_1 (m) \vert \lt 1 / j\}\).
\(S = \cup_{j \in \mathbb{N}} S_j\), because for each \(s \in S\), \(1 \le \vert f_1 (s) \vert\) or \(1 / (j + 1) \le \vert f_1 (s) \vert \lt 1 / j\), so, \(s \in S_j\) for a \(j \in \mathbb{N}\); on the other hand, for each \(s \in \cup_{j \in \mathbb{N}} S_j\), \(s \in S_j\) for a \(j \in \mathbb{N}\), and \(\vert f_1 (s) \vert \neq 0\), so, \(s \in S\).
\(0 \le 1 / (j + 1) \chi_{S_j} \le \vert f_1 \vert \chi_{S_j} \le \vert f_1 \vert\).
So, \(\int 1 / (j + 1) \chi_{S_j} d \mu \le \int \vert f_1 \vert d \mu\).
But the left hand side is \(1 / (j + 1) \mu (S_j)\) while the right hand side is finite.
So, \(\mu (S_j) = (j + 1) \int \vert f_1 \vert d \mu\), finite.
Let \(S' := \{m \in M \vert \Vert f_2 \Vert_{\infty} \lt \vert f_2 \vert\}\), which is locally negligible.
\(S \cap S' = (\cup_{j \in \mathbb{N}} S_j) \cap S' = \cup_{j \in \mathbb{N}} (S_j \cap S')\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, but as \(\mu (S_j) \lt \infty\), \(S_j \cap S'\) is negligible, so, there is a \(N_j \in A\) such that \(S_j \cap S' \subseteq N_j\) and \(\mu (N_j) = 0\), so, \(S \cap S' \subseteq \cup_{j \in \mathbb{N}} N_j\), and \(\mu (\cup_{j \in \mathbb{N}} N_j) \le \sum_{j \in \mathbb{N}} \mu (N_j) = 0\), which means that \(S \cap S'\) is negligible.
So, \(\vert f_1 (m) f_2 (m) \vert \le_{a.e} \vert f_1 (m) \vert \Vert f_2 \Vert_{\infty}\).
So, \(\int \vert f_1 f_2 \vert d \mu \le \int \vert f_1 \vert \Vert f_2 \Vert_{\infty} d \mu = \Vert f_2 \Vert_{\infty} \int \vert f_1 \vert d \mu = \Vert f_1 \Vert_1 \Vert f_2 \Vert_{\infty}\).
Step 5:
Let us suppose that \(p = \infty\).
Then, \(q = 1\).
Step 6:
As this proposition is symmetric with respect to \(p\) and \(q\), by Step 4, the conclusion holds.
