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Solve for $f$: $$\lim_{y\to x}\frac{(x-y)^2}{f(x)-f(y)}=1$$

My try:

$$\lim_{y\to x}{(x-y)^2}=\lim_{y\to x}{f(x)-f(y)}$$

$$\lim_{y\to x}\frac{(x-y)^2}{x-y}=\lim_{y\to x}\frac{f(x)-f(y)}{x-y}$$

$$0=f'(x)$$

$f(x)=c$ but this is obviously wrong. What I did wrong and what is the correct approach?

Relation with 2-Holder continuity:

By the properties of Holder condition, if there exists $C$ such that $|f(x)-f(y)|\leq C(x-y)^2$ (Condition H),

Then, $f$ must be constant.

To apply this, suppose $f(x)>f(y)$, let $C=1$ and take the equity, we have an alternative Holder's condition:

$$\frac{(x-y)^2}{f(x)-f(y)}=1.$$

This condition seems much stronger than my first equation.

OK I think the solution is found here (credit to Zerox): Function on $[a,b]$ that satisfies a Hölder condition of order $\alpha > 1 $ is constant

High GPA
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  • If correct, the sequence of calculations shows that if there is a solution then it has to be constant. It doesn't guarantee that constant functions actually are solutions. Are you sure there are solutions? – Greg Martin Dec 04 '21 at 07:33
  • @GregMartin Well, I see no problem in cancelation (dividing before taking limits) for example $\lim_{x\to 0}\frac{x^2}{x}$. Is everything different in my setting? (this is a reply to a deleted comment) – High GPA Dec 04 '21 at 07:33
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    There is no function $f$ satisfying the equation. In fact there is no solution even if you change $2$ to any $\alpha > 1$. – Zerox Dec 04 '21 at 07:34
  • @Zerox Is this a proposition or a hypothesis? – High GPA Dec 04 '21 at 07:35
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    @HighGPA See the third proposition of the 'Examples' section in this website. – Zerox Dec 04 '21 at 07:38
  • @Zerox Thanks for pointing that out. I am not sure that Holder condition (in your website is implied by my first equation. I am updating your Holder condition in the main question. – High GPA Dec 04 '21 at 07:56

1 Answers1

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There is a mistake, which lies in jumping from$$\lim_{y\to x}\frac{(y-x)^2}{f(y)-f(x)}=1\quad\text{to}\quad\lim_{y\to x}(y-x)^2=\lim_{y\to x}f(y)-f(x).$$For instance,$$\lim_{x\to 0}\frac{\sin\left(\frac1x\right)}{\sin\left(\frac1x\right)}=1,$$but the limit $\lim_{x\to0}\sin\left(\frac1x\right)$ does not exist.

Note that\begin{align}\lim_{y\to x}\frac{(y-x)^2}{f(y)-f(x)}=1&\iff\lim_{y\to x}\frac{f(y)-f(x)}{(y-x)^2}=1\\&\iff\lim_{y\to x}\frac{f(y)-f(x)}{y-x}\cdot\frac1{y-x}=1.\end{align}But, since $\lim_{y\to x}\left|\frac1{y-x}\right|=\infty$, it follows from this that $\lim_{y\to x}\frac{f(y)-f(x)}{y-x}=0$. In other words, $f'$ is the null function. And so, yes, $f$ must be constant. But note that I only proved that if such a function exits, then it must be constant. Since no constant function is a solution of your problem, the problem has no solution.

José Carlos Santos
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