Let XX be a measure space, with positive measure mu\mu. Assume f inL^(r)(mu)f\in L^r(\mu ) for some real rr, and ||f||_(oo) < oo\|f\|_\infty <\infty. We then assert that ||f||_(p)rarr||f||_(oo)\|f\|_p\to \|f\|_\infty as p rarr oop\to \infty.
Proof
If ||f||_(oo)=0\|f\|_\infty=0, then f=0f=0 a.e., and consequently, ||f||_(p)=0\|f\|_p=0 for all real positive pp. The result then follows trivially.
Otherwise, put F=(f)/(||f||_(oo))F=\frac{f}{\|f\|_\infty}, and let 0 < epsi < 10<\varepsilon<1 be given. Choose 0 < delta < epsi0<\delta <\varepsilon. Since ||F||_(oo)=1\|F\|_\infty=1, it follows that for sufficiently large p inR^(+)p\in\mathbb{R}^+:((1-epsi)/(1-delta))^(p) < mu(|F|^(-1)((1-delta,oo))),\left(\frac{1-\varepsilon}{1-\delta}\right)^p<\mu \bigg( |F|^{-1}\Big( (1-\delta,\infty ) \Big)\bigg),so that1-epsi < ||(1-delta)chi_(|F|^(-1)((1-delta,oo)))||_(p) <= ||F||_(p).1-\varepsilon<\left\|(1-\delta ) \chi_{|F|^{-1}\big( (1-\delta,\infty ) \big)}\right\|_p\leq \left\|F\right\|_p.Next: since |F| <= 1|F|\leq 1 a.e., then p > rp>r implies |F|^(p) <= |F|^(r)|F|^p\leq |F|^r a.e. and hence ||F||_(p) <= ||F||_(r)^(r//p)\|F\|_p\leq \|F\|_r^{r/p}.
As 0 < ||F||_(r)^(r) < oo0<\|F\|_r^{r}<\infty, it follows that ||F||_(r)^(r//p)rarr1\|F\|_r^{r/p}\to 1 as p rarr oop\to\infty. Thus, ||F||_(p)rarr1\|F\|_p\to 1 as p rarr oop\to \infty. It is then trivial to deduce ||f||_(p)rarr||f||_(oo)\|f\|_p\to \|f\|_\infty as p rarr oop\to \infty. ////////////
Additionally, this relation still holds when ||f||_(oo)=oo\|f\|_\infty=\infty, but this verification is easier, and is thus left as an exercise for the reader.
References
Rudin, Walter. Real and Complex Analysis. 3rd ed, McGraw-Hill, 1987.
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