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Let $f(x)$ be continuous function in R,note $$ h_{n}(x)=2^{n}\left[f\left(x+\frac{1}{2^{n}}\right)-f(x)\right]$$ with $$ |h_{n}(x)|\leq M \qquad (x\in R,n\in N)$$ and $$ h_{n}(x)\rightarrow 0 \qquad (n\rightarrow\infty)$$ Show that $f(x)$ is a constant funciton.

I tired to prove that for any $x,h \in R$, $$ |f(x+h)-f(x)|=0 $$ but I don't know how to deal with the condition.

pxchg1200
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  • Not sure how to do it yet but the fact that $h_n$ tends to 0 gives you a good hint that if the function is derivable, the derivative at any point would be 0. This would imply that $f$ is constant. So can we show that $f$ is derivable? – elaRosca Apr 14 '13 at 14:32
  • I don't think we can show that $f$ is derivable – pxchg1200 Apr 14 '13 at 14:38
  • @xavierm02 Yes, that what I was hinting at. But is this strong enough to say that for any $h->0$ we have a limit? – elaRosca Apr 14 '13 at 14:42
  • $f$ may not be differentiable (that is, you can't just assume it), but that doesn't mean that a one-sided derivative won't exist. In this case, you'd be looking at a right-derivative (the derivative of $|x|$ isn't defined at $x=0$, but the right-derivative at $x=0$ is $1$). Full differentiability shouldn't be necessary for proving that it's constant. – Glen O Apr 14 '13 at 15:16
  • hint $$\dfrac{f(x+\dfrac{1}{2^n})-f(x)}{\dfrac{1}{2^n}}$$ – math110 Apr 14 '13 at 15:22
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    @math110: That's not good enough. $(f(x+h)-f(x))/h$ has to tend to zero for all $h$, not just rational $h$. – TonyK Apr 14 '13 at 15:25
  • You have to use continuity in an essential way. Otherwise you can construct pathological counter-examples using a basis of the reals over the rationals (which requires the Axiom of Choice). – TonyK Apr 14 '13 at 15:26
  • Can someone give a complete correct proof? – pxchg1200 Apr 15 '13 at 00:55

3 Answers3

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Take any $a, b \in \mathbb{R}$ such that $a < b$.

Since $f$ is continuous over $\mathbb{R}$, each $h_n$ is Lebesgue integrable over $[a,b]$. Since $h_n$ converges pointwise to $0$ and all $|h_n|$ are bounded by a constant $M$ (which is trivially Lebesgue integrable) over $[a,b]$. By Lebesgue's dominated convergence theorem,

$$\lim_{n\to\infty} \int_{a}^{b} h_n(x) dx = \int_{a}^{b} \lim_{n\to\infty} h_n(x) dx = \int_{a}^b 0\; dx = 0$$

Notice $$\int_{a}^{b} h_n(x) dx = 2^{n} \int_{a}^{b}\left( f(x+\frac{1}{2^n}) - f(x) \right) dx = 2^{n} \left( \int_{b}^{b+2^{-n}} f(x) dx - \int_{a}^{a+2^{-n}} f(x) dx \right)$$ and by continuity of $f$ at $a$ and $b$, we have:

$$\lim_{n\to\infty} 2^n \int_{a}^{a+2^{-n}} f(x)dx = f(a) \quad\text{ and }\quad \lim_{n\to\infty} 2^n \int_{b}^{b+2^{-n}} f(x)dx = f(b) $$ This implies $$f(b) - f(a) = \lim_{n\to\infty} \int_{a}^{b}h_n(x)dx = 0$$ and hence $f$ is a constant.

achille hui
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The condition $|h_n(x)|\le M$ is needless.

Given $\epsilon>0$, let $f_\epsilon(x)=f(x)+\epsilon x$ on $\mathbb{R}$. From

$$\lim_{n\to\infty}2^n(f_\epsilon(x+2^{-n})-f_\epsilon(x))=\epsilon>0,\quad \forall x\in\mathbb{R},$$ it is easy to see that $f_\epsilon$ is strictly increasing on $\mathbb{R}$. Letting $\epsilon\to 0$, it follows that $f$ is non-decreasing on $\mathbb{R}$. A similar argument also shows that $f$ is non-increasing on $\mathbb{R}$, which completes the proof.

23rd
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  • In the “similar argument”, I guess you consider $f(x)-\varepsilon x$ instead of $f(x)+\varepsilon x$. – Ewan Delanoy Apr 17 '13 at 06:05
  • @EwanDelanoy: Yes, of course. Or you may replace $f$ with $-f$. – 23rd Apr 17 '13 at 06:10
  • I confess I don't see why $f_\epsilon$ must be strictly increasing. What exactly is the argument that $f(y)>f(x)$ when $y-x > 0$ is not a dyadic rational? – Erick Wong Apr 17 '13 at 12:17
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    @ErickWong: The argument is irrelevant to whether $y-x$ is dyadic rational or not. From the positive limit in the displayed equation of my answer, we can see $f_\epsilon$ cannot attain local extreme value at any point, so for any $x<y$, the maximum(resp. minimum) value of $f_\epsilon$ on $[x,y]$ is only attained at $y$(resp. $x$). In other words, $f$ is an opening mapping, so it satisfies maximum/minimum value principle. – 23rd Apr 17 '13 at 13:06
  • @richard +1 Very nice use of continuity. – Erick Wong Apr 17 '13 at 13:16
  • I feel confuse about why you can Let $\epsilon\rightarrow 0$ to get $f$ is non-decreasing function. – pxchg1200 Apr 17 '13 at 14:28
  • @pxchg1200: Given $x<y$, for every $\epsilon>0$, $f_\epsilon(x)<f_\epsilon(y)$. Then let $\epsilon\to 0$. – 23rd Apr 17 '13 at 14:33
  • @richard (+1) I have to agree with Erick this is a very nice use of continuity. the comment above about $f_{\epsilon}$ achieve maximum/minimum at $y/x$ should be incorporated into your answer. – achille hui Apr 17 '13 at 16:08
  • @achillehui: Thank you! Yes, as you said, the omitted details are not so obvious to see, but I originally thought it was a good exercise to fill in the gaps. – 23rd Apr 17 '13 at 16:15
  • @richard, the omitted detail is not that obvious to me. However, since I have also given an answer, I don't want to sound rude and ask ;-p – achille hui Apr 17 '13 at 16:20
  • @achillehui: Sorry for the confusion to you caused by missing details. You should never mind to ask if anything in the proof looks flawed. – 23rd Apr 17 '13 at 16:39
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Assume to the contrary that there is a function $f$ that satisfies the conditions that is not constant, so there are some $x_0,a,b\in \mathbb{R}$ with $a>0, b\ne0$ such that $$ f(x_0+a)=f(x_0)+b $$ Since $f$ is continuous there is a $\delta>0$ such that $|f(x_0+a+d)-f(x_0+a)|<|b|/2$ for all $d\in(-\delta,\delta)$.

Choosing $N$ such that $2^{-N}<\delta$ we can find $D\in[0,\delta)$ so that $2^N(a+D)=K\in \mathbb{Z}$, i.e. so that we can divide the interval $[x_0,x_0+a+D]$ into $K$ equal parts of length $2^{-N}$.

Then $$ |f(x_0+a+D)-f(x_0)|>|b|/2\\ |f(x_0+K 2^{-N}) - f(x_0)|>|b|/2 $$ so considering the changes in $f$ in each interval $[x_0+j2^{-N},x_0+(j+1)2^{-N}]$ for $j=0,1,\ldots,K-1$ there must be a $j$ that satisfies $$ |f(x_0+(j+1)2^{-N})-f(x_0+j2^{-N})|>\frac{|b|}{2K} $$ Similarly, for any $n>N$ we have $2^n(a+D)=2^{n-N}K$ and there must be a $j\in\{0,1,\ldots,2^{n-N}K-1\}$ that satisfies $$ |f(x_0+(j+1)2^{-n})-f(x_0+j2^{-n})|>\frac{|b|}{2^{n-N+1}K} $$ that is, writing $X=x_0+j2^{-n}$, $$ h_n(X) = 2^n|f(X+2^{-n})-f(X)|>\frac{2^{N-1}|b|}{K} $$ But this is impossible, since the right side is a fixed positive number whereas we must have $h_n(X)\rightarrow 0$ as $n\rightarrow \infty$. Hence our initial assumption must be incorrect, and there cannot be a non-constant $f$ with this condition on $h$.

Zander
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