Let f be continuous on [a, b]. For $ 0 < p < q < \infty $, prove that $(\frac{1}{b-a}\int_a^b\lvert f(x)\rvert^p dx) ^{1/p}$ $ \le $ $(\frac{1}{b-a}\int_a^b\lvert f(x)\rvert^q dx) ^{1/q}$ by the following two methods:
(1) Use the inequality $ \int_a^b \lvert f(x)g(x) \rvert dx $ $ \le $ $(\int_a^b \lvert f(x) \rvert ^p dx)^\frac{1}{p}$ $(\int_a^b \lvert g(x) \rvert ^q dx)^\frac{1}{q}$ with 1/p + 1/q = 1
(2) suppose that f is continuous on [a,b] and $ f([a,b]) \subset (c,d) $ If $ \phi $ is convex and differentiable on (c, d) , it satisfies $ \phi(\frac{1}{b-a}\int_a^bf(x)dx) $ $\le $ $\frac{1}{b-a} \int_a^b \phi (f(x)) dx $
In the first ineqaulity, I plug f(x), g(x) into $ f(x)^p $, 1, respectively.
So I got $ \int_a^b \lvert f(x) \rvert dx $ $\le$ $(\int_a^b \lvert f(x) \rvert ^p dx) ^\frac{1}{p} (b-a)^\frac{1}{q} $ and $ \int_a^b \lvert f(x) \rvert dx $ $\le$ $(\int_a^b \lvert f(x) \rvert ^q dx) ^\frac{1}{q} (b-a)^\frac{1}{p} $. But I don't know what to do next.
In the second inequality, I provd that $x^\frac{1}{p}$ is convex, and used it
So I got $ (\frac{1}{b-a}\int_a^bf(x)dx) ^\frac{1}{p} $ $\le $ $\frac{1}{b-a} \int_a^b (f(x))^\frac{1}{p} dx $. But I don't know waht to do next
help me please..
In the first method, use conjugate exponents $q/p$ and $q/(q - p)$ to get $$\int_a^b \lvert f(x)\rvert^p\, dx \le \left(\int_a^b \left(\lvert f(x)\rvert^p\right)^{q/p}\, dx\right)^{\frac{p}{q}} \left(\int_a^b 1^{q/(q-p)} dx\right)^{1-\frac{p}{q}}$$ Simplify this inequality and rearrange to obtain to the result.
In the second method, consider using the convex map $x\mapsto x^{q/p}$ instead. We would get $$\left(\frac{1}{b - a}\int_a^b \lvert f(x)\rvert^p\, dx\right)^{q/p} \le \frac{1}{b - a}\int_a^b \lvert f(x)\rvert^q\, dx$$ and the result follows at once.