What is the $\lim_{n\to\infty} n^a\int_0^1 \frac{f(x)dx}{1+n^2x^2}$ depending on $a\in\mathbb{R}$, if $f\in L^1(0,1)$? By Banach-Steinhaus theorem I deduced that the limit is zero for $a\leq 0$, but I cannot decide what happens when $a>0$. Any help is welcome. Thanks in advance.
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1$\begingroup$ There is not really a definitive answer to this in my opinion. If you choose $f(x) = c$ for any $c \in \mathbb{R}$, then $\int^1_0 \tfrac{f(x)}{1+n^2x^2}~\mathrm{d}x = cn^a\tfrac{\arctan(n)}{n}$, which, by variyng $a$ and $c$ can be pretty much anything. $\endgroup$– Hyperbolic PDE friendCommented Jun 16 at 20:41
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1$\begingroup$ well the integrand is clearly bigger than $\tfrac{f(x)}{1+n^2}$, at least for non-negative $f$, so there is a lower bound of $\|f\|_1 \tfrac{n^a}{1+n^2}$, which means that if $a> 2$ this is unbounded. $\endgroup$– AnCarCommented Jun 16 at 20:41
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$\begingroup$ @HyperbolicPDEfriend if put in terms of $f$, there are three straightforward cases. $\endgroup$– Ninad MunshiCommented Jun 16 at 21:25
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2 Answers
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Use the substitution $y=nx$:
$$\lim_{n\to\infty}n^{a-1}\int_0^n\frac{f\left(\frac{y}{n}\right)}{1+y^2}dy$$
The term on right of the product converges to
$$\lim_{n\to\infty}\int_0^n\frac{f\left(\frac{y}{n}\right)}{1+y^2}dy \longrightarrow \int_0^\infty\frac{f(0)}{1+y^2}dy = \frac{\pi}{2}f(0)$$
by dominated convergence.
Therefore the value of the limit depends on the value of $a$ relative to $a=1$.
answered Jun 16 at 20:59
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1$\begingroup$ $f(0)$ makes no sense if $f \in L^1$ $\endgroup$ Commented Jun 16 at 22:20
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$\begingroup$ @HyperbolicPDEfriend I think we can safely plug the hole with the exceptions that arise from locally integrable functions which have a singularity at $0$. If it is a nonlogarithmic locally integrable singularity, we can find an $a\in(0,1)$ where limit switches from $0$ to nonzero. If it is logarithmic, the answer will always be $0$ until $a\geq 1$ becomes a joint case. $\endgroup$ Commented Jun 16 at 23:33
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$\begingroup$ I see... But the problem is finding $a$ I guess. Moreover, do we have pointwise convergence of $f\left( \tfrac{y}{n}\right)$ that you implicitly use? $\endgroup$ Commented Jun 17 at 12:31
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For $a\in (0,2)$ we have $$ n^a \frac{f(x) }{1+n^2x^2} = \frac{f(x) }{1/n^a+n^{2-a}x^2} = \frac{1}{n^{2-a}} \frac{f(x) }{1/n^{2}+x^2} $$
Therefore we have $$ \lim_{n\to\infty} \int_0^1 n^a\frac{f(x)}{1+n^2x^2} dx = (\lim_{n\to\infty}\frac{1}{n^{2-a}})(\lim_{n\to\infty} \int_0^1 \frac{f(x) }{1/n^{2}+x^2} dx) $$
if both parts converge. The left bracket converges to $0$. For the right bracket we have that $(\frac{f(x) }{1/n^{2}+x^2})_{n\in ℕ}$ is monotonously ascending.
So if $\int_0^1 \frac{f(x) }{x^2} dx<\infty$ , we have that $\lim_{n\to\infty} \int_0^1 n^a\frac{f(x)}{1+n^2x^2} dx = 0$ (via dominated convergence).
But for $f(x)=1$ we have $\lim_{n\to\infty} \int_0^1 n^a\frac{f(x)}{1+n^2x^2} dx =\lim_{n\to\infty}\arctan(n)n^{a-1}$ which goes to $\infty$ for $a> 1$. So the function $I(a,f):= \lim_{n\to\infty} n^a\int_0^1 \frac{f(x)dx}{1+n^2x^2}$ has to depend on $f$ for at least $a\in (1,2)$.
