2
$\begingroup$

$f:\mathbb{R}\to\mathbb{R}$ is a Lipschitz function satisfying $$\forall x\in\mathbb{R}, \lim_{n\to\infty} n(f(x+\frac1n)-f(x))=0$$ I want to show that $f$ is constant everywhere. I proved $$\lim_{y\to x^+} \frac{|f(y)-f(x)|}{|y-x|}=0$$ by choosing $y\in[x+\frac1{n+1},x+\frac1n]$ and using Lipschitz condition. How can I finish the rest?

$\endgroup$
  • $\begingroup$ How did you arrive at the first part? $\endgroup$ – Anthony Peter Mar 9 '16 at 6:54
  • 1
    $\begingroup$ Fix $\epsilon>0$. First choose $n_1$ such that $$\forall n>n_1,\ (n+1)|f(x+\frac1{n+1}-f(x)|<\frac{\epsilon}2$$. Then, choose $n_2$ such that $C/n<\epsilon/2$, where $C$ is the Lipschitz constant of $f$. Set $n_0=\max\{n_1,n_2\}$. For $n>n_0$, $$\forall y\in[x+\frac1{n+1},x+\frac1n],\qquad |f(y)-f(x)|\le|f(y)-f(x+\frac1{n+1})|+|f(x+\frac1{n+1})-f(x)|$$$$<C|y-(x+\frac1{n+1})|+\frac1{n+1}\cdot\frac\epsilon{2}\le C\frac1{n(n+1)}+\frac1{n+1}\cdot\frac\epsilon{2}=\epsilon\frac1{n+1}=\epsilon|y-x|$$ $\endgroup$ – Emre Mar 9 '16 at 7:09
  • 1
    $\begingroup$ The least interesting proof is to use the full assumption that $f$ is Lipschitz (see below if you have greater ambitions). Then at every point $x$ at which $f'(x)$ exists this condition easily implies that $f'(x)=0$. A Lipschitz function has a derivative a.e. and with an a.e. derivative equal to zero is constant. $\endgroup$ – B. S. Thomson Mar 9 '16 at 21:32
5
$\begingroup$

We can prove a more general result. Suppose $f:\mathbb R\to \mathbb R$ and $x\in \mathbb R.$ Let's say $Df(x) =0$ if there is a sequence $x_n\to x^+$ such that

$$\frac{f(x_n)-f(x)}{x_n-x} \to 0.$$

Thm: If $f:\mathbb R\to \mathbb R$ is continuous and $Df(x) = 0$ for all $x\in \mathbb R,$ then $f$ is constant.

Proof: Suppose $a<b.$ We want to show $f(b) = f(a).$ To prove this, let $\epsilon>0.$ It suffices to show $|f(b)-f(a))|\le \epsilon(b-a).$

Set $E = \{x\in [a,b]: |f(x)-f(a)|\le \epsilon(x-a)\}.$ Clearly $a\in E,$ so $E$ is nonempty. The continuity of $f$ shows $E$ is closed. Suppose, to reach a contradiction, that $b\not \in E.$ Then $E$ contains a largest element $x_0 < b.$ Because $x_0\in E,$ we have $|f(x_0) - f(a)|\le \epsilon(x_0-a).$ But since $Df(x_0) = 0,$ there is $x_1, x_0<x_1 < b,$ such that

$$|f(x_1)-f(x_0)| \le \epsilon(x_1-x_0).$$

We then have

$$|f(x_1)-f(a)| \le |f(x_1)-f(x_0)| +|f(x_0)-f(a)| \le \epsilon(x_1-x_0)+\epsilon(x_0-a) = \epsilon(x_1-a).$$

Thus $x_1\in E,$ and $x_1>x_0,$ contradicting the definition of $x_0.$ This shows $b\in E,$ proving the theorem.

$\endgroup$
1
$\begingroup$

This kind of question has been asked and answered here before and a search would doubtless find the necessary ideas. This asks whether a certain function with a very weak type derivative equal to zero at every point would be constant.

Unfortunately the assumption that $f$ is Lipschitz is misleading and would foul up the search. As @zhw points out, continuity is enough and, moreover, all one needs is that at every point there is a right-hand derived number equal to zero. (It doesn't have to be translation invariant as here using the same sequence $\frac1n$ at each point.)

The correct historical reference for this problem is a monotonicity theorem that is implicit here. It is due to Dini and is quite ancient. His proof is essentially just the same "last point" argument that @zhw has just presented.

Theorem (Dini 1878). If $f:[a,b]\to\mathbb R$ is continuous and $D^+ f(x) \geq 0$ for all $a\leq x < b$ then $f$ is nondecreasing. [Fondamenti per la teorica delle funzioni di variabili reali (Pisa, T. Nistri, 1878)]

Note the use of the upper Dini derivative here, which seems a very weak condition. The assumption in the given problem ensures that $D^+ f(x) \geq 0$ everywhere. The assumption that $f$ is Lipschitz is overkill: we need only continuity. Thus $f$ is nondecreasing. But $-f$ satisfies all the same conditions so $-f$ is also nondecreasing. And one concludes $f$ is constant. But we realize here that this is about monotonicity and not about constancy after all. Generally when someone asks you to prove a function is constant there is a pair of monotonicity problems lurking.

The standard modern reference is Saks, Theory of the Integral, pp. 203-204 where he gives, in addition to the original, a useful refinement due to Zygmund.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.