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I'm familiar with the result that $$\lim_{p \to \infty} ||f||_p=||f||_\infty$$ when $f \in L^p([0,1])$, but I've come a cross a variation of this fact that I'm having trouble showing.

The assertion is that given $f \in L^\infty(\mathbb{R})$ $$\lim_{n \to \infty}\left(\int \frac{|f(x)|^n}{1+x^2} \, dx\right)^\frac{1}{n}=||f||_\infty$$

The function $\frac{1}{1+x^2}$ inside the integrand is what is tripping me up. I'm not sure how to deal with it in order to run the typical argument.

Gabrielek
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Newman
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    A "computational" way to go is to see that $\frac{|f(x)|^n}{2} \leq \frac{|f(x)|^n}{1+x^2} \leq \frac{|f(x)|^n}{1}$ everywhere, so the same inequalities hold for the integrals, and then the squeeze theorem and the result you already have finishes the job. You see now that what matters is just that $\frac{1}{1+x^2}$ is measurable and bounded away from zero and infinity, nothing else really. – Ian Jul 28 '20 at 16:44
  • @Ian This method works even though the integral is over all of $\mathbb{R}$? I was concerned about trying this method because of the bounds of integration. – Newman Jul 28 '20 at 16:51
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    Sorry, I missed that you changed focus to the domain of all of $\mathbb{R}$. In this case you can adapt this approach with some work (by approximating by functions with bounded support), but it is easier to go with the more abstract approach adopted by the answers (based on the idea that the function $| \cdot |_{L^\infty(\mu)}$ is determined entirely by the null sets of $\mu$). – Ian Jul 28 '20 at 16:54

2 Answers2

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Apply the result you are familiar with to the measure $\mu:=\frac 1{1+x^2}dx$: the quantity $\left(\int \frac{|f(x)|^n}{1+x^2} \, dx\right)^\frac{1}{n}$ is the $L^n$ norm with respect to this measure. Moreover, $\mu(A)=0$ if and only if the Lebesgue measure of $A$ is zero hence the $L^\infty$-norm with respect to $\mu$ is the same as the $L^\infty$-norm with respect to the Lebesgue measure.

Davide Giraudo
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Let $\mu(dx)=\frac{1}{1+x^2}\,dx$. This measure is equivalent to Lebesgue measure $\lambda(dx)=dx$ in the sense that $\mu\ll \lambda$ and $\lambda\ll \mu$.

It is a general result that $\|f\|_{L_n(\mu)}\xrightarrow{n\rightarrow\infty}\|f\|_{L_\infty(\mu)}$.

Since $\mu$ and $\lambda$ are equivalent, $\|f\|_{L_\infty(\mu)}=\|f\|_{L_\infty(\lambda)}$.

Mittens
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