# Prove that $f$, such that $\forall\delta\gt0$ $|f(y)-f(x)|\lt\delta^2$ $\forall x,y\in\mathbb{R}$ and $|y-x|\lt\delta$, is a constant function.

The function $f$ is defined on $\mathbb{R}$ such that for every $\delta\gt0$, $|f(y)-f(x)|\lt\delta^2$ for all $x,y\in\mathbb{R}$ and $|y-x|\lt\delta$. Prove that $f$ is a constant function.

So, what I know, is that I need to show that $f(a)=f(b)$ for all points $a,b\in\mathbb{R}$. Or for every $\epsilon\gt0$, $|f(a)-f(b)|\lt\epsilon$.

I'm at a loss here. I got a hint to divide the interval $[a, b]$ into $n$ smaller intervals. But I don't understand why, and how this would help me.

Hints and preferably a proof are very much appreciated. Thanks in advance.

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Maybe you can think about the derivative of $f$, since the condition implies $|(f(y)-f(x))/(y-x)|<\delta$. So it must be zero for every $x$. –  fmoura2005 Oct 24 '12 at 0:39

I have what I think is an equivalent situation:

Suppose $f$ is such that $f(x)-f(y)\leq (x-y)^2$ for all $x,y$. Then $f$ is constant.

PROOF Choose any $x,y$. The hypothesis means that

$f(y)-f(x)=-(f(x)-f(y))\leq (y-x)^2$ whence $|f(x)-f(y)|\leq (x-y)^2$ for each $x,y$. Divide $[x,y]$ into $n$ intervals $[x_{k-1},x_k]$ such that $x_k-x_{k-1}=\dfrac{y-x}n$, $x_0=x$; $x_n=y$. That is, $$x_k=x+\frac k n (y-x)$$

Then

$$|f(x)-f(y)|=\left|\sum_{k=1}^n f(x_{k-1})-f(x_k)\right|\\ \leq \sum_{k=1}^n\left|f(x_{k-1})-f(x_k)\right| \\ \leq \sum\limits_{k = 1}^n {{{\left( {{x_{k - 1}} - {x_k}} \right)}^2}} \\ = \frac{1}{{{n^2}}}\sum\limits_{k = 1}^n {{{\left( {y - x} \right)}^2}} = \frac{{{{\left( {y - x} \right)}^2}}}{n}$$

Thus

$$\tag 1 \left| {f\left( x \right) - f\left( y \right)} \right| \leq \frac{{{{\left( {y - x} \right)}^2}}}{n}$$

for every $n\in \Bbb N$. Now suppose $f(x)\neq f(y)$. $(1)$ means $$\tag 2 \frac{{\left| {f\left( x \right) - f\left( y \right)} \right|}}{{{{\left( {y - x} \right)}^2}}} \leq \frac{1}{n}$$

But since $|f(x)-f(y)|>0$ we have $$\frac{{\left| {f\left( x \right) - f\left( y \right)} \right|}}{{{{\left( {y - x} \right)}^2}}} > 0$$whence there must be an $n$ such that $$\frac{1}{n} < \frac{{\left| {f\left( x \right) - f\left( y \right)} \right|}}{{{{\left( {y - x} \right)}^2}}}$$

But this contradicts $(2)$. Then it must be $f(x)=f(y)$ for each choice of $x,y$.

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Hint: Show that $f$ is differentiable in each point and the same time that the derivative is $0$.

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It is natural to think that the given condition implies the inequality $$\left| f(y) - f(x) \right| \leq C \left| y - x \right|^2. \tag{1}$$ But this implication requires some justification. So here is a proof.

Let $\epsilon > 0$ be any positive real number. To prove $(1)$, we decompose the set $\{ 0 < \left| y - x \right| < \epsilon \}$ in a dyadic manner as follow:

$$\{ 0 < \left| y - x \right| < \epsilon \} = \bigcup_{n=1}^{\infty} \big\{ 2^{-n} \epsilon \leq \left|y - x\right| < 2^{-(n-1)}\epsilon \big\}.$$

Then whenever $n \geq 1$ and $2^{-n} \epsilon \leq \left|y - x\right| < 2^{-(n-1)}\epsilon$, we have

$$\left|f(y) - f(x)\right| < 2^{-2(n-1)} \epsilon^2 \leq 4 \left| y - x \right|^2.$$

Thus whenever $0 < \left| y - x \right| < \epsilon$, we have $2^{-n} \epsilon \leq \left|y - x\right| < 2^{-(n-1)}\epsilon$ for some $n \geq 1$ and hence

$$\left|f(y) - f(x)\right| \leq 4 \left| y - x \right|^2.$$

Then it follows that

$$\left|\frac{f(y) - f(x)}{y - x}\right| \leq 4 \left| y - x \right|,$$

proving $(1)$ with $C = 4$.

Now the rest follows in various ways as many people pointed out. For example, taking the limit as $y \to x$, we find that $f$ is differentiable at any point $x$ with vanishing derivative. Therefore $f$ is constant.

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What does $\{ 0 < \left| y - x \right| < \epsilon \}$ symbolize? –  Pedro Tamaroff Oct 24 '12 at 1:32
@PeterTamaroff, Excuse me for that notation. It is a widely used convention in measure theoretic context to omit the explicit range of the variables whenever the ambient set is unambiguous. Thus in our case, it is the set of points $(x, y) \in \Bbb{R}^2$ satisfying the condition $0 < \left| y - x \right| < \epsilon$. –  Sangchul Lee Oct 24 '12 at 1:36
I imagined, but just to be sure. =) Do you think you can take a look at this? –  Pedro Tamaroff Oct 24 '12 at 1:39

Step 1. Let $x,y\in\mathbb{R}$ with $x \ne y$. For all $r>1$ we have $|x-y|<r|x-y|$, so $|f(x)-f(y)|<r^2|x-y|^2$. This is true for all $r>1$, so we must have $|f(x)-f(y)| \le |x-y|^2$.

Step 2. This is just Exercise 5.1 in Rudin; rewrite the above as $$\left\vert\frac{f(x)-f(y)}{x-y}\right\vert \le |x-y|$$ and it's easy to see that $f'(x)=0$ for all $x\in\mathbb{R}$ directly from the definition of the derivative as a limit. Apply the mean value theorem, and you're done.

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