Check the uniform convergence of $f_n=\frac{n}{nx^{2}+1}$ on $[\frac12,\infty)$ and $(0,\infty)$

Question

I think its clear to see that it converges to $\frac{1}{x^{2}}$ and its domain is all real numbers except for zero. The exercise asks to analyse the uniform convergence of this sequence in the intervals $[\frac{1}{2},\infty]$ and in $]0,\infty]$. So it would suffice to show that $\forall \epsilon \gt 0 \forall x$ in the domain $\exists N $ such that $\forall n \gt N$ $|f_n-\frac{1}{x^{2}}|\lt \epsilon$.

I am not sure on how to find a way to make $N$ depend only on $\epsilon$. Can someone give me a hint or point some mistake in my thought? Thanks in advance. Can also accept linked answers (really didnt found duplicates)

Answer 1

Building on @Didier's comment, when $x \in (0, \infty)$, we can consider the sequence $f_n(1/\sqrt{n})$ to show that the convergence is not uniform. In this case

$$ |f_n(1/\sqrt{x}) - \frac{1}{x^2}| = \left \lvert \frac{n}{n(\frac{1}{\sqrt{n}^2}) + 1} - n\right \lvert = n/2. $$

Clearly the convergence here cannot be uniform, since for $\epsilon > 0$, for any proposed $N$ we can take $x = 1/\sqrt{N}$ to show that the requirement for uniform convergence is not satisfied.

As for the case where $x \in [\frac{1}{2}, \infty)$, there is the inequality

\begin{align*} \left \lvert \frac{n}{nx^2+1} - \frac{1}{x^2} \right \lvert &= \left \lvert \frac{1}{x^2(nx^2+1)} \right \lvert \\ &\leq \left \lvert \frac{1}{(\frac{1}{2})^2((\frac{1}{2})^2n+1)}\right \lvert \\ & = \frac{4}{n/4+1}. \end{align*}

Can you continue from here?

Answer 2

For $x\in[\frac12,\infty)$, we have

$$\left|f_n(x)-\frac1{x^2}\right|=\left|\frac{1}{(nx^{2}+1)x^2}\right|\le\frac1{|nx^4|}\le\frac{16}n<\epsilon$$

hence, it is uniformly convergent and you can choose $N=\frac{16}\epsilon$.

For $x\in(0,\infty)$, let $\epsilon=\frac17$, for arbitrary $N$, we can choose $n=N, m=2N$, $x=\frac1{\sqrt N}\in(0,\infty)$

$$|f_n(x)-f_m(x)|=\left|\frac N{N(\frac{1}{\sqrt N})^2+1}-\frac{2N}{2N(\frac1{\sqrt N})^2+1} \right|=\left|\frac N2-\frac{2N}3\right|=\frac N6\ge\frac16>\frac17=\epsilon$$

hence, it is NOT uniformly convergent on $(0,\infty)$.

Answer 3

$f_n(x)$ converges pointwise on $\Bbb R^+$ to $g(x)=1/x^2.$ Each $f_n$ is a bounded function but $g$ is unbounded.

Without any detailed calculation we know that a sequence $(f_n)_n$ of bounded functions cannot converge uniformly to an unbounded function $g,$ as follows:

Let $y_n=\sup_x |f_n(x)|.$ For each $n$ there exists $x_n$ with $|g(x_n)|>1+y_n,$ so $$|g(x_n)-f_n(x_n)|\ge |g(x_n)|-|f_n(x_n)|>1+y_n-|f_n(x_n)|\ge 1+y_n-y_n=1.$$

Answer 4

A fun way to disprove uniform convergence is to consider $(0,b)\subset (0,\infty)$. Then

$$\lim_{n\to\infty}\int_0^b \frac{n}{nx^2+1}dx = \lim_{n\to\infty}\sqrt{n}\arctan(b\sqrt{n}) \to +\infty$$

but

$$\int_0^b\lim_{n\to\infty}\frac{n}{nx^2+1}dx = \int_0^bdx = b$$

which are not equivalent. Therefore the sequence is not uniformly convergent on the set, and subsequently not uniformly convergent on the superset either.