MHB Show Inclusion of Measures: Hölder's Inequality

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In the discussion on Hölder's Inequality, the main focus is on demonstrating that for measurable sets E with finite measure, the space L^{p_2} is strictly contained in L^{p_1} when 1 ≤ p_1 < p_2 ≤ +∞. The user correctly applies Hölder's inequality to establish that the norm ||f||_{p_1} is bounded by ||f||_{p_2} multiplied by a measure-dependent term. However, the challenge lies in proving that ||f||_{p_1} is strictly less than ||f||_{p_2} for non-zero functions, which is necessary to confirm the strict inclusion. The discussion also raises questions about alternative methods to establish this result. Ultimately, the goal is to clarify the relationship between these two function spaces under the given conditions.
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)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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