How to prove this limit of derivative

Question

Here is a question that I need help to prove it.

Let $f:\mathbb{R}\to\mathbb{R} \in C^{\infty}$ be periodic of period $1$ and nonnegative. Show that $$ \dfrac{d}{dx}\left(\dfrac{f(x)}{1+cf(x)}\right)\to0 $$ uniformly in $x$ as $c\to\infty$.

My thought: I think it needs uniform convergence of function sequence to prove it. I know the derivative is $$ \frac{f'(x)}{(1+cf(x))^2} $$ but I am not sure how to do it exactly.

Answer 1

Since $f(x)∈C^∞, f(x)⩾0$ and periodic, $f′(x_0)=0$ where $f(x_0)=0$. If $f(x)≡0$, then conclusion is obvious. So assume $f(x)≢0$.

Since $f′(x)$ is continuous and periodic, it is bounded, or $∃M>0, ∀x∈R, |f′(x)|⩽M$. Also for any $ϵ>0, ∃δ,0<δ<1, ∀|x−x_0|<δ$, there is $|f′(x)|<ϵ$, and $$|f(x)|=|f′(t)(x−x_0)+f(x_0)|=|f′(t)(x−x_0)|<ϵδ<ϵ$$ where $|t−x_0|<δ.$

Now let $⋃\limits_{i=1}^{\infty}B(x_i,δ) (f(x_i)=0)$ be a cover of $[0,1]$. Since $[0,1]$ is compact, there is a finite cover $A=⋃\limits_{i=1}^{n}B(x_i,δ)$. And there is $$ \begin{cases} |f′(x)|<ϵ, & \text{ for } x∈A \\ 0\leqslant f(x)<ϵ, & \text{ for } x∈A \end{cases} $$ And $$ \begin{cases} |f′(x)|⩽M, & \text{ for } x∈[0,1]−A \\ f(x)⩾ϵ, & \text{ for } x∈[0,1]−A \end{cases} $$ Now $$ \frac{d}{dx}\left(\frac{f(x)}{1+cf(x)}\right)=\frac{f′(x)}{(1+cf(x))^2} $$ For $x∈A$ $$ \left|\frac{f′(x)}{(1+cf(x))^2}\right|⩽|f′(x)|<ϵ\hspace{5 mm} (c>0) $$ For $x∈[0,1]−A$ $$\left|\frac{f′(x)}{(1+cf(x))^2}\right|⩽\frac{M}{(|cf(x)|−1)^2}⩽\frac{M}{(cϵ−1)^2}⩽ϵ \hspace{5 mm} \left(c>\left(\sqrt{\frac{M}{ϵ}}+1\right)/ϵ\right) $$ So $$\frac{d}{dx}\left(\frac{f(x)}{1+cf(x)}\right)→0$$ uniformly for all $x∈[0,1].$

Since $f(x)$ has period of $1$, it is easily followed that the integral approaches to $0$ uniformly for all $x∈R.$

Answer 2

Denote $g(x)=\frac{f'(x)}{(1+cf(x))^2}$.

1. Since $f$ is periodic it is enough to prove convergence over the period (compact interval $T$).
2. Note that $f(x)=0$ implies $f'(x)=0$ (since $f\ge 0$ and differentiable).

Pick $\epsilon>0$. Any point $x$ with $f(x)=0$ can be covered by an open interval $E_x\subset T$ where $|f'(x)|\le\epsilon$. Denote $E$ the union of all such intervals. Clearly $|g(x)|\le|f'(x)|\le\epsilon$ for $x\in E$.

Now since the complement set $T_c=T\setminus E$ is compact, there exists $M=\max_{x\in T_c}|f'(x)|<+\infty$ and $m=\min_{x\in T_c}|f(x)|>0$ and we can estimate $$ |g(x)|\le\frac{M}{c^2m^2}\le\epsilon\quad\text{for sufficiently large }c. $$ Thus. $g\to 0$ uniformly when $c\to +\infty$.