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I'm looking for the example of a continuous function $f$, which is:
1. positive and strictly increasing in $(0,a)$ for some $a>0$,
2. $f(0)=0$,
3. $\lim\limits_{x\rightarrow0^{+}}\frac{x}{f(x)}=0$
and for which there doesn't exist an interval $(0,b)$, $b>0$, such that the map $x\rightarrow\frac{x}{f(x)}$ is strictly increasing for all $x\in(0,b)$.
Thank you in advance for your help!

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2 Answers 2

Choose a positive exponent $u$ and a positive scalar $z$ and consider the function $f$ defined by $$ \color{red}{f(x)=x^u+zx\cos(x^{u-1})}. $$ Then $f(x)>0$ for every $x$ in a neighborhood of $0^+$ if $u<1$ because $f(x)\geqslant x^u-zx$, $f(0)=0$ because $u>0$, and $f$ is increasing near $0^+$ for suitable values of $u$ because $f'(x)$ has the sign of $$ u+z(1-u)\sin(x^{u-1})+zx^{1-u}\cos(x^{u-1}), $$ whose liminf when $x\to0^+$ is $u-z(1-u)$. Thus, every $u>z/(z+1)$ makes $f$ increasing in a neighborhood of $0^+$. Furthermore, $f(x)/x$ is equivalent to $x^{u-1}$ because $u<1$ hence $x/f(x)\to0$ when $x\to0^+$. And $f(x)/x=g(x^{u-1})$ with $g(t)=t+z\cos(t)$ hence the sense of variation of $f(x)/x$ at $x$ when $x\to0^+$ is the opposite of the sense of variation of $g(t)$ at $t=x^{u-1}$, that is, when $t\to+\infty$. Since the derivative of $g$ is $g'(t)=1-z\sin(t)$ and the sine oscillates between $-1$ and $+1$, the sense of variation of $f(x)/x$ at $x$ when $x\to0^+$ changes infinitely many times if $z>1$.

Finally, any $u$ and $z$ do as soon as $$ z/(z+1)<u<1<z. $$

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Another fine example. I'm very grateful for all your help. –  John Nov 5 '11 at 18:49

For any positive integer $n$, let $$f(x)=x\sqrt{2n}\qquad\text{when}\qquad \frac{1}{2n}\le x\le \frac{1}{2n-1}.$$ For $\dfrac{1}{2n+1} \le x \le \dfrac{1}{2n}$, define $f(x)$ by interpolating linearly between the values of $f$ at $\dfrac{1}{2n+1}$ and at $\dfrac{1}{2n}$. Finally, let $f(0)=0$.

It is clear that $f$ is continuous on the interval $[0,1]$. We verify that $f$ is strictly increasing. The only thing that needs to be checked is that $f\left(\dfrac{1}{2n+1}\right)< f\left(\dfrac{1}{2n}\right)$. Note that $$f\left(\dfrac{1}{2n+1}\right)=\frac{\sqrt{2n+2}}{2n+1}\qquad\text{and}\qquad f\left(\dfrac{1}{2n}\right)=\frac{\sqrt{2n}}{2n}.$$ Square both sides. A little manipulation yields the desired inequality.

Now we examine $\dfrac{x}{f(x)}$. Between $\dfrac{1}{2n}$ and $\dfrac{1}{2n-1}$, this has constant value $\dfrac{1}{\sqrt{2n}}$. Thus any interval $(0,b)$ contains infinitely many intervals on which $\dfrac{x}{f(x)}$ is constant.

For the final condition, we need to verify that $\displaystyle\lim_{x\to 0^+}\dfrac{x}{f(x)}=0$. This is easy: $\dfrac{x}{f(x)}$ approaches $0$ roughly like $\sqrt{x}$.

Comment: We arranged for $\dfrac{x}{f(x)}$ to be constant on the intervals $\dfrac{1}{2n} \le x \le \dfrac{1}{2n-1}$. By a small modification of the definition of $f(x)$, we can even arrange for $\dfrac{x}{f(x)}$ to go down a little on these intervals.

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Thanks a lot, André! Really nice example. I noticed only one typo: in the fourth line it should be $f(0)=0$. Could you, please, explain in details what do you mean by a "small modification" mentioned in Comment? –  John Nov 5 '11 at 14:10
    
@John: Thanks for the typo fix. Will try to expand comment, probably not today (full). –  André Nicolas Nov 5 '11 at 15:24
    
@John: I think if you make the first line $f(x)=(x+x^2/n^2)$ you get there. You want $f$ to grow just a bit faster than linearly. –  Ross Millikan Nov 5 '11 at 15:29

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