Let \(\phi\in L^\infty(\RR)\). Show that the following limit exists and satisfies the equality \[ \lim _{n \to \infty} \left(\int _{\mathbb{R}} \frac{|\phi(x)|^{n}}{1+x^{2}} \, dx \right) ^ {\frac{1}{n}} = \norm{\phi}_\infty. \]
Solution
Put \[ M:=\|\phi\|_\infty, \qquad A_n:=\left(\int_{\mathbb R}\frac{|\phi(x)|^n}{1+x^2}\,dx\right)^{1/n}. \] If \(M=0\), then \(\phi=0\) a.e. and \(A_n=0\) for every \(n\). Assume \(M>0\).
Since \(|\phi|\le M\) a.e., \[ A_n^n\le M^n\int_{\mathbb R}\frac{dx}{1+x^2}=\pi M^n, \] so \[ \limsup_{n\to\infty}A_n\le M. \]
Now fix \(0<\varepsilon<M\) and let \[ E_\varepsilon:=\{x:|\phi(x)|>M-\varepsilon\}. \] By the definition of essential supremum, \(m(E_\varepsilon)>0\). Since \((1+x^2)^{-1}>0\) everywhere, \[ c_\varepsilon:=\int_{E_\varepsilon}\frac{dx}{1+x^2}>0. \] Therefore \[ A_n^n\ge (M-\varepsilon)^n c_\varepsilon, \] and hence \[ A_n\ge (M-\varepsilon)c_\varepsilon^{1/n}. \] Letting \(n\to\infty\) gives \[ \liminf_{n\to\infty}A_n\ge M-\varepsilon. \] Since \(\varepsilon>0\) is arbitrary, \[ \liminf_{n\to\infty}A_n\ge M. \] Together with the upper bound, \[ \boxed{\lim_{n\to\infty}A_n=\|\phi\|_\infty.} \]