The task is to show that if $f \in L^{\infty}\left(\mathbb{R}\right)$, then

$$\lim\limits_{n\to \infty}\left(\int\limits_{\mathbb{R}}\frac{|f(x)|^n}{1+x^2}dx\right)^{\frac{1}{n}}=\left\|f\right\|_{L^{\infty}\left(\mathbb{R}\right)} $$

I want to show it with sqeeze theorem. From one hand,

$$\left(\int\limits_{\mathbb{R}}\frac{|f(x)|^n}{1+x^2}dx\right)^{\frac{1}{n}}\le \left(\int\limits_{\mathbb{R}}\frac{\left\|f\right\|_{L^{\infty}}^n}{1+x^2}dx\right)^{\frac{1}{n}}=\left\|f \right\|_{L^{\infty}\left(\mathbb{R}\right)}\left(\int\limits_{\mathbb{R}}\frac{1}{1+x^2}dx\right)^{\frac{1}{n}}=\left\|f \right\|_{L^{\infty} } \pi^{\frac{1}{n}}$$

So, everything looks good, but I can't find an estimate from another hand. Have you any ideas how to do it, or, perhaps, just how to prove this statement?


Given $M<\|f\|_\infty$, let $A_M=\{x\in\mathbb R : f(x) \geq M\}$. By the definition of $\|f\|_\infty$, each such set $A_M$ has positive Lebesgue measure. Then for each $M<\|f\|_\infty$, \begin{align*} \left(\int_{\mathbb R} \frac{|f(x)|^n}{1+x^2}\ dx\right)^{1/n} &\geq \left(\int_{A_M} \frac{M^n}{1+x^2}\ dx \right)^{1/n} \\ &= M\left(\int_{A_M} \frac{1}{1+x^2}\ dx \right)^{1/n}\\ &\to M \end{align*} since $\int_{A_M}\frac{1}{1+x^2}\ dx>0$, and this gives the desired bound.


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