Let \(f\in L^\infty([0,1])\).
Prove that for \(1<p<\infty\), \[ \|f\|_p\le\|f\|_\infty. \]
Prove that \[ \lim_{p\to\infty}\|f\|_p=\|f\|_\infty. \]
Solution
1. For every \(1<p<\infty\),\[ \norm{f}_p\le\norm{f}_\infty. \]
Let \(M=\norm{f}_\infty\). Since \([0,1]\) has measure \(1\), \[ \norm{f}_p^p=\int_0^1\abs{f}^p\le M^p, \] and taking \(p\)th roots gives the claim. In particular, \[ \limsup_{p\to\infty}\norm{f}_p\le M. \]
2. If \(M>0\), then\[ \liminf_{p\to\infty}\norm{f}_p\ge M. \]
Fix \(0<\varepsilon<M\). By the definition of essential supremum, \[ A_\varepsilon=\{x:\abs{f(x)}>M-\varepsilon\} \] has positive measure. Therefore \[ \norm{f}_p^p\ge (M-\varepsilon)^p m(A_\varepsilon), \] so \[ \norm{f}_p\ge (M-\varepsilon)m(A_\varepsilon)^{1/p}. \] Letting \(p\to\infty\) gives \[ \liminf_{p\to\infty}\norm{f}_p\ge M-\varepsilon. \] Since \(\varepsilon>0\) is arbitrary, \[ \liminf_{p\to\infty}\norm{f}_p\ge M. \]
3. Therefore\[ \boxed{ \lim_{p\to\infty}\norm{f}_p=\norm{f}_\infty. } \]
If \(M=0\), then \(f=0\) almost everywhere and the result is immediate. If \(M>0\), combine the limsup bound from step 1 with the liminf bound from step 2.
4. Q.E.D.
Step 1 proves part (a), and step 3 proves part (b).