I have a question please :
Let $f:[0,2\pi]\times \mathbb{R} \rightarrow \mathbb{R}$ a differential function satisfying : $\displaystyle k^2\leq \liminf_{|x|\rightarrow \infty} \frac{f(t,x)}{x}\leq \limsup_{|x|\rightarrow \infty}\frac{f(t,x)}{x} \leq (k+1)^2$
Let $(x_n)\subset H^1([0,2\pi],\mathbb{R})=\lbrace x\in L^2([0,2\pi],\mathbb{R}),x'\in L^2([0,2\pi],\mathbb{R}),x(0)=x(2\pi)\rbrace$
sucht that $\|x_n\|\rightarrow \infty$ when $n \rightarrow \infty$
Why : the sequence $\left(\displaystyle\frac{f(t,x_n)-k^2 x_n}{\|x_n\|}\right)$ is bounded ?
Is this answer given by :@TZakrevskiy true ?
Well, we write
$\limsup\limits_{|x|\to\infty} \frac{f(x)-k^2x}{x}\le 2k+1,$
$\liminf\limits_{|x|\to\infty} \frac{f(x)-k^2x}{x}\ge 0.$
Plus, $\frac{f(x)-k^2x}{x}$ and $\frac{f(x)-k^2x}{|x|}$ have the same absolute value. Then, if $\limsup y_n\le C$ for a sequence $y_n$, then $y_n$ has a finite upper bound, same goes for $\liminf$ and a finite lower bound. Thus, for positive $x_n$ the sequence $\frac{f(x_n)-k^2x_n}{|x_n|}$ is bounded.
For $x_n<0$ we study
$\limsup\limits_{|x|\to\infty} \frac{k^2x-f(x)}{-x}=\limsup\limits_{|x|\to\infty} \frac{k^2x-f(x)}{|x|}\le 2k+1,$
$\liminf\limits_{|x|\to\infty} \frac{k^2x-f(x)}{-x}=\liminf\limits_{|x|\to\infty} \frac{k^2x-f(x)}{|x|}\ge 0,$ which gives us for negative $x_n$ the sequence $\frac{k^2x-f(x)}{|x|}$ is bounded; and, clearly, so is the sequence $-\frac{k^2x-f(x)}{|x|}$. This conludes the proof.
Please
Thank you .