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If $|f(x)-f(y)|<(x-y)^2$ for all $x,y \in \mathbb{R}$. Then $f(x)$ is constant.

$\bf{Attempt}$ Put $x=y+h$ where $h \rightarrow 0$

Then $\displaystyle |f(y+h)-f(y)|<(y+h-y)^2 = h^2$

So $\displaystyle \displaystyle \lim_{h\rightarrow 0}\bigg|\frac{f(y+h)-f(y)}{h}\bigg|<\lim_{h\rightarrow 0}h$

So $\displaystyle |f'(y)|<0$

Now how can I calculate $f(x)$? Could someone help me? Thanks

ForgotALot
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DXT
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    Inequalities become inclusive when passed to the limit. So what you get is actually $|f'(y)| \color{red}{\le} 0,$. – dxiv Jun 23 '17 at 04:01
  • Thanks dxiv but i did not understand why we use $=$ sign, please explain me , Thanks – DXT Jun 23 '17 at 04:03
  • why we use = sign Because the absolute value is always $\ge 0$, so once you showed that $|f'(y)| \le 0$ it follows that necessarily $f'(y)=0,$. P.S. Please finish the question. As it stands, it ends in then f(x) is. – dxiv Jun 23 '17 at 04:07
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    $0=|f(x)-f(x)|<(x-x)^2=0$. Is this a contradiction? – user130405 Jun 23 '17 at 04:08
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    @user130405 As written, it is an impossibility, indeed, i.e. there exists no such function. I can only guess the problem had an additional condition $x \ne y,$ which didn't make it into the posted question. My initial comment was based on that unstated assumption. – dxiv Jun 23 '17 at 04:10
  • @DurgeshTiwari : As the problem stands now, the others are right. There is a contradiction buried for $x=y$ in the way it's stated. Once you remove that contradiction, you don't even need Calculus from what I can see to do the proof. – Johnq Jun 23 '17 at 06:07

1 Answers1

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I will amend the hypothesis. Suppose $n$ is a non-negative integer and $$\text {(I).}\quad |f(x)-f(y)|\leq 2^{-n}(x-y)^2 \quad \text { for all $x,y$.}$$ Then, with $z=(x+y)/2$, we have $$|f(x)-f(y)|=|(f(x)-f(z))+(f(z)-f(y))|\leq |f(x)-f(z)|+|f(z)-f(y)|\leq$$ $$ \leq 2^{-n}(x-z)^2+2^{-n}(z-y)^2=2^{-(n+1)}(x-y)^2.$$ The amended hypothesis is that (I) holds when $n=0.$ So by induction, (I) holds for all $n\in \{0\}\cup \mathbb N. $ Therefore $f(x)=f(y)$ for all $x,y.$

Remark: In the Q , verbatim, we cannot have $|f(x)-f(y)|<(x-y)^2$ for all $x,y$ because when $x=y$ this says $0<0^2.$

DanielWainfleet
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