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A function $f$ is called Hölder-continuous on the interval $I$, when there exist a constant $C$ in $\mathbb{R}_{\geq0}$ and an exponent $\alpha>0$ such that $$|f(x)-f(y)|\leq C|x-y|^\alpha$$ for every $x,y$ in $I$.

What is to be proven is that if $f$ is Hölder-continuous with $a>1$, $f$ must be constant.

What I have found out so far:

I need to show that $$a>1 \iff f(x)=f(y)$$ for all $x,y$ in $I$.

After dividing the inequation by $|x-y|^\alpha$ I have

$$\frac{|f(x)-f(y)|}{|x-y|} \frac{1}{|x-y|^{\alpha-1}}\leq C$$

Is there a way to deduct that there is no constant $C$ such that this inequation can be satisfied except when the numerator is $0$?

B.Swan
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1 Answers1

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HINT

$$\frac{|f(x+h)-f(x)|}{|h|}\leq C|h|^{\alpha-1}$$

Now use squeeze theorem.

  • I.E show the derivative is $0$ at every point and then use mean value theorem. – Asinomás Dec 13 '16 at 05:44
  • Just for clarity: I let $h \rightarrow 0$ and show that the value of the derivative approaches 0 for all $x$ in $I$? – B.Swan Dec 13 '16 at 05:51
  • @B.Schnebbler The value of the derivative is exactly $0$ for all $x\in I$, yup! –  Dec 13 '16 at 05:51