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I'm having a bit of trouble with the title problem out of Davidson and Donsig's Real Analysis. I'll state it again:

Suppose $f$ is continuous on $\mathbb{R}$ such that $$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{h} = 0\ \forall x\in\mathbb{R}.$$ Prove that $f$ is constant.

They provide the following hint which I have been trying to apply.

HINT: Fix $\epsilon > 0$. For each $x$, find a $\delta > 0$ so that $|f(x+h)-f(x-h)| \leq \epsilon h$ for $0\leq h \leq \delta$. Let $\Delta$ be the supremum of all such $\delta$. Show that $\Delta = \infty$.

Here's how I've started. Fix $\epsilon > 0,\ x\in\mathbb{R}$. By the definition of the limit, $$(\forall x\in\mathbb{R})(\forall\epsilon>0)(\exists\delta>0)(0<|h|\leq\delta\implies\Bigg|\frac{f(x+h)-f(x-h)}{h}\Bigg|<\epsilon).$$

Therefore, we immediately get a $\delta>0$ for our $x,\epsilon$ such that $$|f(x+h)-f(x-h)|\leq\epsilon h,\ 0\leq h \leq\delta.$$

I don't know how to proceed from here. I'm not even sure conceptually how showing that $\Delta=\infty$ would give us that $f$ is constant. Any help would be appreciated!

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marked as duplicate by Jack, Arnaud D., Nosrati, B. Goddard, Mark Bennet Nov 23 '17 at 17:06

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  • $\begingroup$ If u keenly see the given condition, it itself says it is constant. at any point $x$ we have same value for f in its neighbouthood. It means f(x) = same value at all x. So f(x) = constant $\endgroup$ – Magneto Nov 23 '17 at 1:29
  • $\begingroup$ @anirudhb: "at any point x we have same value for f in its neighbouthood." What??? $\endgroup$ – Jack Nov 23 '17 at 1:55
  • $\begingroup$ @Jack at any point x, we have in h-neighbouthood of x, $f(x+h)=f(x-h) = c(say)$ as $h \to 0$ $\endgroup$ – Magneto Nov 23 '17 at 3:23
  • $\begingroup$ @anirudhb: your argument does not make sense. It is true that $\lim_{h\to 0}f(x+h)=\lim_{h\to 0}f(x-h)=f(x)$. This does not imply $f(x)$ is constant in a neigbourhood of $x$. Moreover, it does not make sense to say "$f(x+h)=f(x-h)=c$ as $h\to 0$". $\endgroup$ – Jack Nov 23 '17 at 3:42
  • $\begingroup$ @Jack yes, u r right. I got it. It should make f continuous. Thanks for correcting me. $\endgroup$ – Magneto Nov 23 '17 at 3:47
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I don't understand the hint in your book. But here is an alternative way to do it.

The limit $$SDf(x):=\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h},$$ if exists, is called the symmetric derivative of $f$ at $x$. In Brian Thomson's Symmetric Properties of Real Functions, it is shown that

Theorem. A continuous function is necessarily increasing in any interval in which its symmetric derivative exists and is positive.

As a corollary one can deduce that continuous functions with zero symmetric derivatives are constant:

For any $\epsilon>0$, $SD(f(x))=0$ means that $f(x)+\epsilon x$ has a positive symmetric derivative and so, by the theorem, must be increasing; similarly $-f(x)+\epsilon x$ has a positive symmetric derivative and must also be increasing. As $\epsilon>0$ is arbitrary $f$ must be constant as required.


Here is the proof of the theorem in Thomson's book:

enter image description here enter image description here

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  • $\begingroup$ Very nice proof. +1 $\endgroup$ – Paramanand Singh Nov 23 '17 at 3:53
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We are going to show that the set of points where $f$ is differentiable is dense. Let $(a,b)$ an open interval, since $f$ is continuous on $[a,b]$, it attains its maximal at $x_0$, suppose that $x_0\in (a,b)$

$\frac{f(x_0+h)-f(x)}{h}\geq 0$ and $\frac{f(x_0-h)-f(x_0)}{-h}\geq 0$, this implies that $\frac{f(x_0+h)-f(x_0-h)}{h}\geq \frac{f(x_0+h)-f(x_0)}{h}\geq 0$, simce $lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}=0$ and $f$ is derivable at $x_0$, if the maximum is $f(a)$ or $f(b)$, we consider all the intervals $(c,d)\subset (a,b)$ and if for every $(c,d)\subset (a,b)$ the maximum is $f(c)$ or $f(d)$, $f$ is monotone, we deduce that the set of points where $f$ is differentiable on $(a,b)$ is dense.

Monotone+continuous but not differentiable

Suppose $f$ is differentiable at $x$

$$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{h} = 0=2f'(x)\ \forall x\in\mathbb{R}.$$

since $\frac{f(x+h)-f(x-h)}{h}=$ $\frac{f(x+h)-f(x)}{h}$+ $\frac{f(x-h)-f(x)}{-h}$

so $f'(x)=0$ implies that $f$ is constant.

Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$

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    $\begingroup$ How do you know that $f$ is differentiable? Couldn't the symmetric limit exist without the derivative existing? $\endgroup$ – carmichael561 Nov 23 '17 at 0:58
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    $\begingroup$ For instance, $f(x)=|x|$ fails to be differentiable at zero, but $\lim_{h\to0}\frac{f(h)-f(-h)}{h}=0$. $\endgroup$ – carmichael561 Nov 23 '17 at 1:06
  • $\begingroup$ Yeah, I thought about arguing this way, but I agree with carmichael in that I don't think you can assume the derivative exists. I looked at the link you posted about the uses of this limit, and it seems that the authors on that post assume the existence of the derivative at x to give the equality you gave in the first line. $\endgroup$ – Jarkrandel Nov 23 '17 at 1:34
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Consider a compact interval $[a,b]$

Fix $\epsilon>0$ For each $x \in [a,b]$, there exist $\delta(x)$, such that $|f(x+h)-f(x-h)|\le h\epsilon$ whenever $|h|<\delta(x)$

Consider the open intervals $\left\{I_x=(x-\delta(x),x+\delta(x)) \right\}_{x\in[a,b]}$, they clearly cover $[a,b]$ which is compact, hence there exists a finite sub-cover$\left\{I_{x_n}\right\}_{n=1}^{N}$ for $[a,b]$

Assume $x_1<x_2<...<x_N$ and choose a sequence of points $y_n$ such that $y_n$ lies in the intersection of $I_{x_n}$ and $I_{x_{n+1}}$ so that

$|f(y_{n+1})-f(y_n)| <\delta(x_n)\epsilon$,

and $|f(y_{n+2})-f(y_{n+1})| <\delta(x_{n+1})\epsilon$

Let $y_0= a$ and $y_{N+1}=b$

Now

$|F(b)-F(a)|= \left|\sum_{i=0}^{N}f(y_{i+1})-f(y_i)\right|\le\sum_{i=0}^{N}\left|f(y_{i+1})-f(y_i)\right|\le\sum_{i=0}^{N}\delta(x_i)\epsilon\le(b-a)\epsilon$

The last inequality holds since we can choose any $\eta(x)<\delta(x)$ and the first inequality will still hold. Finally since $\epsilon$ is arbitrary $|f(b)-f(a)|=0$, hence $f(b)=f(a)$, letting $b$ vary, we conclude that $f(x)$ is constant.

Note: Anybody is welcome to edit the answer to get a better format and possiby a better way of writing the proof

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  • $\begingroup$ The potatohead's argument is ignored :( Sorry, potato :( $\endgroup$ – MereMortal47 Nov 23 '17 at 3:55

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