Holder’s Inequality Trick

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Holder’s Inequality is a very useful inequality in Functional Analysis, hence many results can be proved by applying Holder’s Inequality.

Suppose that \mu(\Omega)<\infty and 1\leq p_1<p_2\leq\infty. Prove that if f_n\to f in L^{p_2}, then f_n\to f in L^{p_1}.

Proof: Assume f_n\to f in L^{p_2}. Then there exists N such that if n\geq N, then \|f_n-f\|_{p_2}=(\int |f_n-f|^{p_2} d\mu)^{1/p_2}<\epsilon.

Then, \begin{aligned}    \|f_n-f\|_{p_1}^{p_1}&=\int |f_n-f|^{p_1} d\mu\\    &=\||f_n-f|^{p_1}\cdot 1\|_1\\    &\leq\||f_n-f|^{p_1}\|_{p_2/p_1}\|1\|_{(p_2/p_1)'},\ \ \ \text{where } (p_2/p_1)'=\frac{p_2}{p_2-p_1}\ \text{is the Holder conjugate}\\    &=(\int |f_n-f|^{p_2}d\mu)^{p_1/p_2}(\int 1 d\mu)^\frac{p_2-p_1}{p_2}\\    &<\epsilon^{p_1}\cdot \mu(\Omega)^\frac{p_2-p_1}{p_2}    \end{aligned}

Since \epsilon>0 is arbitrary, \|f_n-f\|_{p_1}^{p_1}\to 0 as n\to\infty.

Therefore, \|f_n-f\|_{p_1}\to 0.

Author: mathtuition88


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