Why is the Holder space with exponent $\alpha>1$ a constant set?

Question

Why is the holder space with $\alpha >1$ a constant set? Is it related to the lipschitz condition? But, this seems to fail, as a continuous function may not be Lipschitz continuous.

See http://www.math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/Holder-spaces.pdf Specifically, the remark near the beginning which ends with "That is why we do not talk about Hoelder spaces with Hoelder exponents larger than 1." — MPW, Oct 12 '14 at 11:39
but what if u is just cts,not diff.?? the proof in the note seems require u to be diff.? — mnmn1993, Oct 12 '14 at 12:12
You don't have to assume f is differentiable, but you can prove that if it satisfies the holder inequality this forces the function to have derivative = 0. (See my answer) — Matthew Levy, Oct 12 '14 at 12:26

score 1 · Answer 1 · edited Oct 12 '14 at 12:36

Assume $\alpha > 1$, this implies that: $$|f(y)-f(x)| \leq C\cdot|x-y|^\alpha$$ for some $C\in\Bbb R$

Then, we divide by $|x-y|$ on both sides: $$\frac{|f(y)-f(x)|}{|x-y|}\leq C\cdot|x-y|^{\alpha - 1}$$

But, this is just the absolute value of the difference quotient for $f$. Since $\alpha - 1 > 0$ and thus the right side of the inequality can be made as small as desired, we have that: $$\lim_{y\to x}{\frac{|f(y)-f(x)|}{|y-x|}} = 0 \Rightarrow f'(x) = 0 \Rightarrow f(x) = D$$

for some constant $D\in\Bbb R$.

Why is the Holder space with exponent $\alpha>1$ a constant set?

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