Let $(f_n)_{n\in\Bbb N}:I\to\Bbb R$ be a uniformly convergent on $I\subset\Bbb R$. Is $g_n=\{\frac{f_n}{1+f_n^2}\}_{n\in\Bbb N}$ uniformly convergent?

Question

Let $I\subset\mathbb R$ be an interval and $\left(f_n\right)_{n\in\mathbb N}:I\to\mathbb R$ be a sequence that is uniformly convergent on $I$. Is the sequence $g_n=\left\{\frac{f_n}{1+f_n^2}\right\}_{n\in\mathbb N}$ uniformly convergent?

I dont know where to start with this one :(

Answer 1

If $x\in I$, and if $f$ is the uniform limit of the sequence $(f_n)_{n\in\mathbb N}$, then\begin{align}\left\lvert \frac f{1+f^2}-\frac{f_n}{1+{f_n}^2}\right\rvert&=\left\lvert\frac{f-f_n+{f_n}^2f-f^2f_n}{(1+f^2)(1+{f_n}^2)}\right\rvert\\&=\lvert f-f_n\rvert\left\lvert\frac{1-f_nf}{(1+f^2)(1+{f_n}^2)}\right\rvert\\&\leqslant\lvert f-f_n\rvert,\end{align}since$$(\forall a,b\in\mathbb R):\left\lvert1+ab\right\rvert\leqslant\sqrt{1+a^2}\sqrt{1+b^2}\leqslant(1+a^2)(1+b^2),$$where the first inequality follows from the Cauchy-Schwarz inequality. Therefore, since $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$, $\left(\frac{f_n}{1+{f_n}^2}\right)_{n\in\mathbb N}$ converges uniformly to $\frac f{1+f^2}$.

Answer 2

Let $\varphi(t) = \dfrac{t}{1 + t^2}.$ Then $\varphi'(t) = \dfrac{1 + t^2 - 2t^2}{(1 + t^2)^2}$ is bounded on $\mathbf{R};$ this shows $\varphi$ is uniformly continuous. We conclude the exercise with the following lemma.

Lemma. If $\varphi:\mathrm{J} \to \mathbf{R}$ is uniformly continuous and $f_n:\mathrm{I} \to \mathrm{J}$ is a uniformly convergent sequence to $f,$ then $\varphi \circ f_n$ converges uniformly as well.

Proof. It is rather trivial, given $\varepsilon > 0,$ there exists $\delta > 0$ such that if $x, y \in \mathrm{J}$ are such that $\|x - y\| < \delta$ then $|\varphi(x) - \varphi(y)|<\varepsilon.$ For this $\delta,$ there is an $N$ such that if $n \geq N,$ then $\sup\limits_t \|f_n(t) - f(t)\| < \delta.$ Q.E.D.