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Let for $n\geq 1$: $f_n(x):=\dfrac{x}{1+nx^2}$ and $\mathcal{F}:=\{f_n:n=1,2,3,\ldots\}$

I'd like to know if $\mathcal{F}$ is equicontinuous.

This is what I have done:

$\left|\frac{x}{1+nx^2}-\frac{x_0}{1+nx_0^2}\right|<\varepsilon$

this becames $|x-x_0|\left|\frac{1-nxx_0}{(1+nx^2)(1+nx_0^2)}\right|<\varepsilon$. Now I thought to define this function $g(n)=\frac{1-nxx_0}{(1+nx^2)(1+nx_0^2)}$ where we think $n\in\mathbb{R}$ and $x,x_0$ as parameters. Now $f(n)\rightarrow0$ if $n\rightarrow\infty$ and it is continuous, so there will be an interval $[-M,M]$ such that this function will be less than a certain $\nu$ ($\nu$ and $M$ depend on $x$ and $x_0$). And on $[-M,M]$ $g(n)$ will have a maximum $N$ (depending on $x,x_0$). So we have

$|x-x_0|\left|\frac{1-nxx_0}{(1+nx^2)(1+nx_0^2)}\right|<\delta\cdot\mathrm{max}\{{N,\nu}\}$

But now I don't know how to continue, please could you help me?

1 Answers 1

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We have f_n'(x)=\frac{1+nx^2-x(2nx)}{(1+nx^2)^2}=\frac{1-nx^2}{(1+nx^2)^2}, therefore \sup_{x\in\mathbb R}|f_n'(x)|=\sup_{x\in\mathbb R}\left|\frac{1-nx^2}{(1+nx^2)^2}\right|\leq \sup_{x\in\mathbb R}\frac{1+nx^2}{(1+nx^2)^2}=\sup_{x\in\mathbb R}\frac 1{1+nx^2}=1. Now, for $x,y\in\mathbb R$ and $n\in\mathbb N$ we have |f_n(x)-f_n(y)|\leq \left|\int_x^y|f_n'(t)|dt\right|\leq |x-y|, which shows that $\mathcal F$ is uniformly equi-continuous (and equi-continuous at each point). Here, it's a particular case of a family of Lipschitz continuous functions such that the Lipschitz constants are uniformly bounded.

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    @t.b. Ha! I already noticed that there are over 50 posts with "continous", but I plan only to correct new ones. I don't want to annoy people *too* much.2011-10-13