Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 24J (Probability and Measure)

For $1\le p<\infty$, the Lp norm is $\Vert f{\Vert }_p=({\int }_E|f{|}^{p}d\mu {)}^{1/p}$; for $p=\infty$ it is the essential supremum of $|f|$. The Lp space consists of measurable functions with finite norm, identifying functions equal almost everywhere. This identification makes the quantity a norm rather than only a seminorm. Convergence in $L^p$ means $\Vert f_n-f{\Vert }_p\to 0$, including the essential-supremum norm when $p=\infty$.

The Minkowski inequality states the triangle inequality for every $L^p$ norm with $1\le p\le \infty$. Apply it to $f_n=(f_n-f)+f$ and then to $f=(f-f_n)+f_n$. The two bounds combine toThus norm convergence implies convergence of the norms themselves.

Put $M=\Vert f{\Vert }_{\infty }>0$ and $M_0=\mu (E)$. There is a positive-measure set on which $|f|>M/2$, so $M_0>0$ and each $M_n$ is positive; boundedness and finite measure make each finite. For $n\ge 1$, apply Cauchy-Schwarz to $|f{|}^{(n-1)/2}$ and $|f{|}^{(n+1)/2}$ to getTherefore the positive ratios $r_n=M_{n+1}/M_n$ are nondecreasing. Telescoping gives $M_n/M_0=r_0\cdots r_{n-1}\le r_n^n$, while $|f{|}^{n+1}\le M|f{|}^n$ almost everywhere gives $r_n\le M$. HenceFor $0<\epsilon <M$, the set $A_{\epsilon }=\{|f|>M-\epsilon \}$ has positive measure by the definition of essential supremum. Thus $\Vert f{\Vert }_n\ge (M-\epsilon )\mu (A_{\epsilon }{)}^{1/n}$, whereas $\Vert f{\Vert }_n\le M\mu (E{)}^{1/n}$. Letting $n\to \infty$ and then $\epsilon \downarrow 0$ shows $\mu (E{)}^{-1/n}\Vert f{\Vert }_n\to M$. The preceding squeeze yields $M_{n+1}/M_n\to \Vert f{\Vert }_{\infty }$. This proves the ratio limit rather than assuming it from the moment-root limit.