Integral Version of Hölder Inequality

Statement: Let f1,f2,,fnf_1,f_2,\cdots, f_n are nn functions who are positive in the region [a,b][a,b] . And p1,p2,,pnp_1, p_2, \cdots,p_n be positive rational numbers such that p1+p2++pn=1.p_1+p_2+\cdots+p_n=1 . Thenk=1n[(abfk(x)dx)pk]ab[k=1n(fk(x))pk]dx\prod_{k=1}^n \left[\left(\int_a^b f_k (x)dx \right)^{p_k}\right]\geq \int_a^b\left[ \prod_{k=1}^n\left(f_k (x)\right)^{p_k}\right]dx

Proof:
k=1n[pkfkabfk(x)dx]k=1n[(fk(x)abfk(x)dx)pk]\sum_{k=1}^n \left[\frac{p_kf_k}{\displaystyle{\int_a^b f_k (x)dx}}\right]\geq \prod_{k=1}^n \left[\left(\frac{f_k (x)}{\displaystyle{\int_a^b f_k (x)dx}} \right)^{p_k}\right]     ab[k=1n[pkfk(x)abfk(x)dx]]dxab[k=1n[(fk(x)abfk(x)dx)pk]]dx\implies \int_a^b \left[\sum_{k=1}^n \left[\frac{p_kf_k(x)}{\displaystyle{\int_a^b f_k (x)dx}}\right] \right]dx \geq \int_a^b \left[ \prod_{k=1}^n \left[\left(\frac{f_k (x)}{\displaystyle{\int_a^b f_k (x)dx}} \right)^{p_k}\right]\right]dx     k=1n[pkabfk(x)dxabfk(x)dx]ab[k=1n(fk(x))pk]dxk=1n[(abfk(x)dx)pk] \implies \sum_{k=1}^n \left[\frac{\displaystyle{p_k\int_a^bf_k(x)dx}}{\displaystyle{\int_a^b f_k (x)dx}}\right]\geq \frac{\displaystyle{\int_a^b \left[ \prod\limits_{k=1}^n \left( f_k (x)\right)^{p_k}\right]dx }}{\displaystyle{\prod\limits_{k=1}^n \left[\left(\int_a^b f_k (x)dx \right)^{p_k}\right]}}     k=1n[(abfk(x)dx)pk]ab[k=1n(fk(x))pk]dx\implies \prod\limits_{k=1}^n \left[\left(\int_a^b f_k (x)dx \right)^{p_k}\right] \geq \int_a^b\left[ \prod_{k=1}^n\left(f_k (x)\right)^{p_k}\right]dx

Note by Soham Chatterjee
1 month, 3 weeks ago

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Wonderful I really love the argument

Aruna Yumlembam - 1 month, 3 weeks ago

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Nice!

Yajat Shamji - 1 month, 3 weeks ago

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