Let $f : \mathbb R \to \mathbb R$ be a continuous function such that for any two real numbers $x$ and $y$

Question

Let $f : \mathbb R \to \mathbb R$ be a continuous function such that for any two real numbers $x$ and $y$, $|f(x) - f(y)| \leq 7{|x-y|}^{201}$ then,

(A) $f(101) =f(202) +8$

(B) $f(101) = f(201) +1$

(C) $f(101) = f(200) + 2$

(D) None of the above. Source ISI UGA 2017

I have no idea how to solve questions like this, please help me. Or give me some hint.

Suppose $f$ is the zero function (or any constant function). — quasi, Jun 15 '17 at 04:09
@BaponDas not the answer? Is "None of the above" not an answer? — edm, Jun 15 '17 at 04:13
$|f(x+h) - f(x)| < 7 h^{201} \implies |f'(x)| = \lim_\limits {h\to 0} \frac {|f(x+h) - f(x)|}{|h|} \le 7h^{200} \implies f'(x) = 0$ — Doug M, Jun 15 '17 at 04:14
Existence of the derivative is what you prove, simultaneously showing that it must be zero everywhere. — quasi, Jun 15 '17 at 04:16
@mathmore but it is differentiable... I have proven that the derivative exists. — Doug M, Jun 15 '17 at 04:16
@DougM how can you say that f'(x) =0,from the previous line? — Bapon Das, Jun 15 '17 at 04:23

score 1 · Accepted Answer · edited Sep 11 '17 at 14:52

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Divide by $\lvert\,x-y\,\rvert$ and let $y$ approach $x$ to see that $f'(x)=0$ for every $x$ and thus $f$ is a constant function so (D) is the answer. As Sherlock said, the easiest way to hide something is to put it in plain sight.

edited Sep 11 '17 at 14:52

JohnColtraneisJC

1,919

answered Jun 15 '17 at 05:08

user439545

1,703
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Let $f : \mathbb R \to \mathbb R$ be a continuous function such that for any two real numbers $x$ and $y$

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