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Under these conditions, $f$ is $C^1(\mathbb{R},\mathbb{R})$

$$\lim_{|t|\rightarrow0}\frac{f(t)}{a(|t|)|t|}=0, \lim_{|t|\rightarrow+\infty}\frac{f(t)}{a_*(|t|)|t|)}=c$$

Where $a,a_{*}:[0,+\infty)\rightarrow [0,+\infty)$ is $C^1$

How to obtain that : "given $\varepsilon>0$, there exist $C_{\varepsilon}>0$ such that $$ f(t)t\leq \varepsilon a(|t|)|t|^2+C_{\varepsilon} a_*(|t|)|t|^2, \forall t\in \mathbb{R} $$

1 Answers 1

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I will use the extra hypothesis $a_*(|t|)>0$ on $(0,\infty)$.

Given $\epsilon>0$ take $\delta>0$ such that $$ \Bigl|\frac{f(t)}{a(|t|)|t|}\Bigr|\le\epsilon\quad,|t|\le\delta. $$ Then $$ |t\,f(t)|\le\epsilon\,a(|t|)\,t^2,\quad |t|\le\delta. $$ From the second limit we deduce that there exists $R>0$ such that $$ \Bigl|\frac{f(t)}{a_*(|t|)|t|}-c\Bigr|\le1,\quad|t|>R. $$ Then $$ |t\,f(t)|\le(|c|+1)\,a_*(|t|)\,t^2,\quad |t|>R. $$ Let $$ M=\max_{\delta\le|t|\le R}{|t\,f(t)|}\text{ and }m=\min_{\delta\le|t|\le R}a_*(|t|)\,t^2>0. $$ Then $$ |t\,f(t)|\le\frac{M}{m}\,a_*(|t|)\,t^2,\quad\delta\le|t|\le R. $$ Finally, let $$ C_\epsilon=\max\Bigl(\frac{M}{m},|c|+1\Bigr). $$

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    Why $|t f(t)|\leq \frac{M}{m} a_*(|t|)t^2$ and not $|t f(t)|\leq \frac{M}{m} a(|t|)t^2$? – 2017-01-11
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    Because $m$ is defined to be the minimum of $a_*(|t|)\,t^2$. – 2017-01-11
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    We can efined as the minimum of a ? and please why we take $C_{varepsilon}=\max(\frac{M}{m},c+\varepsilon)$ and not $\frac{M}{m}+c+\varepsilon$ ? – 2017-01-11
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    In the inequality you want to prove the second term is $C_\epsilon\,a_*(|t|)\,t^2$; that's why I defined $m$ as I did. As for the second question it as a matter of taste. It is clear that $a+b\ge\max(a,b)$ if $a,b\ge0$. – 2017-01-11
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    Ok, for the second question, i write like this $$|t\,f(t)|\leq \varepsilon a(|t|)t^2+(|c|+\varepsilon)\,a_*(|t|)\,t^2+ \frac{M}{m}a_*(|t|)t^2, ~\forall t $$ – 2017-01-11