Show Inclusion of Measures: Hölder's Inequality

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SUMMARY

The discussion centers on demonstrating the inclusion of Lebesgue spaces, specifically that \( L^{p_2} \subsetneq L^{p_1} \) for \( 1 \leq p_1 < p_2 \leq +\infty \) in a Lebesgue measurable set \( E \subset \mathbb{R}^d \) with finite measure \( m(E) \). The user correctly applies Hölder's inequality, stating that \( ||f||_{p_1} \leq ||f||_{p_2} \mu(E)^{1/p_1 \cdot q} \), and seeks clarification on how to establish that \( ||f||_{p_1} < ||f||_{p_2} \). The discussion indicates a need for further exploration of the properties of these spaces to solidify the proof.

PREREQUISITES
  • Understanding of Lebesgue spaces \( L^p \) and their properties
  • Familiarity with Hölder's inequality and its applications
  • Knowledge of measure theory, particularly Lebesgue measure
  • Basic concepts of functional analysis
NEXT STEPS
  • Study the proof of the inclusion \( L^{p_2} \subsetneq L^{p_1} \) using specific examples
  • Explore the implications of Hölder's inequality in different contexts
  • Investigate the relationship between \( L^p \) spaces and convergence of functions
  • Learn about the completeness of \( L^p \) spaces and their duals
USEFUL FOR

Mathematicians, students of analysis, and anyone interested in functional analysis and measure theory will benefit from this discussion, particularly those studying properties of Lebesgue spaces.

mathmari
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Hey! :o

Let $1 \leq p_1 \leq p_2 \leq +\infty$. Show that in a Lebesgue measurable $E\subset R^d$ with $0<m(E)<+\infty$ we have that $L^{p_2} \subsetneq L^{p_1}$.

Using Hölder's inequality I got that $||f||_{p_1} \leq ||f||_{p_2} \mu (E)^{1/p_1 \cdot q}$.

Is this correct so far?? How could I continue to show that $||f||_{p_1} < ||f||_{p_2}$ ?? (Wondering)

Or is there an other way to show this?? (Wondering)
 
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mathmari said:
How could I continue to show that $||f||_{p_1} < ||f||_{p_2}$ ?? (Wondering)

Or isn't this that we want to show so that $L^{p_2} \subsetneq L^{p_1}$ ?? (Wondering)
 

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