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Let $(\Omega, \mathcal B, \mu)$ be a measure space, and $f:\Omega\rightarrow \mathbb C$ a measurable function. Suppose $M:=\sup\left\{\mu(A)\mid 0<\mu(A)<\infty\right\}<\infty$. Show that for $0


Any hints please?

Chebyshev's inequality does not seem to help here.

1 Answers 1

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Since $\frac{q}p>1$ then the map $\phi: x\mapsto x^{\frac{q}p}$ is convex and therefore by Jenssens inequality(which is convexity) we have

$$\phi\left(\frac{1}{\mu(X)}\int_X|f|^p d\mu\right)\le \frac{1}{\mu(X)}\int_X\phi\circ|f |^pd\mu $$ that is $$\left(\frac{1}{\mu(X)}\int_X|f|^p d\mu\right)^{q/p}\le \frac{1}{\mu(X)}\int_X|f |^qd\mu $$

that is $$\|f\|_p\leq\|f\|_q\mu(X)^{\frac{1}{p}-\frac{1}{q}}$$