## Holder’s Inequality Trick

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