# Hölder inequality from Jensen inequality

I'm taking a course in Analysis in which the following exercise was given.

Exercise Let $(\Omega, \mathcal{F}, \mu)$ be a probability space. Let $f\ge 0$ be a measurable function.

1. Using Jensen's inequality, prove that for $1\le p < q$, $$\left(\int f^p\, d\mu\right)^{\frac{1}{p}}\le \left(\int f^q\, d\mu\right)^{\frac{1}{q}}.$$
2. Deduce Hölder's inequality from Jensen's inequality and discuss the cases of equality.

The first part is standard and I had no problems with it. On the contrary, the second part is somewhat unclear. The standard proof of Hölder's inequality uses Young's inequality which may be proved by means of the convexity of the exponential function. So, strictly speaking, this is a way of "deducing Hölder's inequality from Jensen's", but I don't think this is what the examiner had in mind. More likely, one is supposed to look for a proof employing Jensen's inequality in $(\Omega, \mathcal{F}, \mu)$, or perhaps applying directly the first point. But I have no idea on how to do this.

Thank you.

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Could show Jensen's inequality ? –  Nikita Evseev Oct 12 '12 at 12:55
Per chance your examiner has in mind this version of Jensen's inequality? –  Willie Wong Oct 12 '12 at 13:17
@WillieWong: Yes, exactly. –  Giuseppe Negro Oct 12 '12 at 14:22
@nikita2: By "Jensen's inequality in $(\Omega, \mathcal{F}, \mu)$" I mean the following: $$\varphi\left(\int f\, d\mu\right)\le \int\varphi(f)\, d\mu,$$where $\varphi\colon \mathbb{R}\to\mathbb{R}$ is a convex function. See previous comment. –  Giuseppe Negro Oct 12 '12 at 14:24
Suppose f,g > 0 and look at jensen applied to probability measure $\frac {g^q d \mu}{\int g^q d \mu}$ and $h = \frac f {g^{q-1}}$ –  mike Oct 12 '12 at 14:51

As Mike suggests, take the measure $\nu:=\frac{g^q}{\int g^qd\mu}\cdot \mu$ (a probability measure) and $h:=\frac f{g^{q-1}}$. Then $$\int fgd\mu=\int hg^qd\mu=\int g^qd\mu\cdot\int hd\nu\leqslant \int g^qd\mu \left(\int h^pd\nu\right)^{1/p}=\left(\int g^q\right)^{1/q}\left(\int f^pd\mu\right)^{1/p}.$$