8 — Hölder’s Inequality Refinements, $L^p$ Norms, Triangle Inequality, Bounded Convergence, Convergence in Measure
Lecture 8 covers:
Refinements of Hölder’s inequality including the 𝑝 = 1 ,
𝑞
∞ p=1, q=∞ case,
Definition and properties of 𝐿 𝑝 L p spaces,
Triangle inequality for 𝐿 𝑝 L p ,
The Bounded Convergence Theorem,
Convergence in measure vs almost everywhere,
Examples illustrating why convergence in measure does not imply convergence of integrals unless additional assumptions hold.
We work on a σ-finite measure space $(\Omega, \mathcal{F}, \mu)$.
1. The $L^p$ Norm and Measurability
For $1 \le p < \infty$, \(\vert f\vert _p = \left( \int_\Omega \vert f(\omega)\vert ^p\, d\mu(\omega) \right)^{1/p}.\)
A function $f : \Omega \to \mathbb{R}$ is measurable if: \(f^{-1}(B) \in \mathcal{F}, \quad \forall B \subset \mathbb{R} \text{ Borel}.\)
2. Hölder’s Inequality, Including the Case $p=1,\, q=\infty$
For $1 \le p < \infty$ and $q$ such that $1/p + 1/q = 1$: \(\int_\Omega \vert f g\vert \, d\mu \le \vert f\vert _p \cdot \vert g\vert _q.\)
Case $p=1,\ q=\infty$
Here $\vert g\vert _\infty$ is the essential supremum: \(\vert g\vert _\infty = \inf\{ a \ge 0 : \mu(\{ \vert g\vert > a \}) = 0 \}.\)
Then: \(\int_\Omega \vert f g\vert \, d\mu \le \vert f\vert _1 \, \vert g\vert _\infty.\)
Reason (page 1):
On the set where $\vert g\vert \le \vert g\vert _\infty$ (ignoring measure-zero sets):
\(\vert fg\vert \le \vert f\vert \cdot \vert g\vert _\infty,\)
hence integrate and use monotonicity.
3. $L^p$ Spaces
\[L^p(\Omega, \mathcal{F}, \mu) = \{f\text{ measurable}: \vert f\vert _p < \infty\}.\]The full notation in the notes is: \(L^p = L^p(\Omega, \mathcal{F}, \mu).\)
4. Triangle Inequality for $L^p$ ($1 \le p \le \infty$)
Case $1 \le p < \infty$
Goal: \(\vert f+g\vert _p \le \vert f\vert _p + \vert g\vert _p.\)
Proof (page 2)
Start from: \(\vert f+g\vert ^p = \vert f+g\vert \cdot \vert f+g\vert ^{p-1} \le \vert f\vert \, \vert f+g\vert ^{p-1} + \vert g\vert \, \vert f+g\vert ^{p-1}.\)
Integrate:
\[\int \vert f+g\vert ^p \le \int \vert f\vert \; \vert f+g\vert ^{p-1} + \int \vert g\vert \; \vert f+g\vert ^{p-1}.\]Apply Hölder:
- Exponent $p$ for $\vert f\vert $ or $\vert g\vert $,
- Exponent $q = \frac{p}{p-1}$ for $\vert f+g\vert ^{p-1}$.
We get: \(\int \vert f\vert \; \vert f+g\vert ^{p-1} \le \vert f\vert _p \, \vert f+g\vert _p^{p-1},\) and similarly with $g$.
Thus: \(\vert f+g\vert _p^p \le (\vert f\vert _p + \vert g\vert _p)\, \vert f+g\vert _p^{p-1}.\)
Divide both sides by $\vert f+g\vert _p^{p-1}$ (nonzero unless the inequality is trivial):
\[\boxed{ \vert f+g\vert _p \le \vert f\vert _p + \vert g\vert _p. }\]Case $p=\infty$
\[\vert f+g\vert _\infty = \operatorname{esssup} \vert f+g\vert \le \operatorname{esssup} \vert f\vert + \operatorname{esssup} \vert g\vert = \vert f\vert _\infty + \vert g\vert _\infty.\]5. Bounded Convergence Theorem (Book Version)
Assume:
- $\mu(\Omega) < \infty$,
- $\vert f_n\vert \le M < \infty$ for all $n$,
- $f_n \to f$ almost everywhere.
Then: \(f_n \to f \quad \text{in measure}.\)
And: \(\int f_n \, d\mu \to \int f \, d\mu.\)
This is weaker than Dominated Convergence (because the bound $M$ does not need to be integrable on infinite measure spaces).
6. Convergence in Measure
Definition: \(f_n \xrightarrow{\mu} f \quad \text{means} \quad \mu\{ \vert f_n - f\vert > \varepsilon \} \to 0 \quad \forall \varepsilon > 0.\)
Relation to almost everywhere convergence:
\[f_n \xrightarrow{\text{a.e.}} f \quad \Longrightarrow \quad f_n \xrightarrow{\mu} f.\]Proof idea (page 3):
Let $A_{n,\varepsilon} = {\vert f_n - f\vert > \varepsilon}$.
If $f_n \to f$ a.e., then:
\(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_{k,\varepsilon}\)
has measure 0.
Since $\mu(\Omega) < \infty$, this shows $\mu(A_{n,\varepsilon}) \to 0$.
7. Examples (Lebesgue Measure on $\mathbb{R}$)
Let $\mu = \lambda$ be Lebesgue measure.
Example 1: Convergence a.e. and in measure, but $\int f_n$ does not converge to 0.
Define: \(f_n(\omega) = \frac{1}{n} \mathbf{1}_{(n,2n]}(\omega),\quad n=1,2,\dots\)
- $f_n \to 0$ a.e.
- $f_n \to 0$ in measure.
- But: \(\int_{\mathbb{R}} f_n\, d\lambda = 1,\quad \forall n.\)
So the integrals do not converge to 0.
Example 2: Diverging integral despite convergence in measure
Define: \(f_n(\omega) = \mathbf{1}_{(n,2n]}(\omega).\)
Then:
- $f_n \to 0$ a.e.
- $f_n \to 0$ in measure.
- But: \(\int_{\mathbb{R}} f_n\, d\lambda = n \to \infty.\)
This shows that convergence in measure does not control integrals unless there is additional uniform integrability or a dominating integrable function.
8. Convergence Relationships
Almost everywhere ⇒ In measure
(always true if $\mu(\Omega) < \infty$)
In measure ⇏ Almost everywhere
(need subsequences to get a.e. convergence)
In measure or a.e. does NOT imply convergence of integrals
(see examples above)
This finishes Lecture 8: a key transition from integration theory into convergence modes, preparing for the Dominated Convergence Theorem and Fatou’s lemma in the next lectures.
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