Hölder condition with Not of Bounded Variation

Question

Let $\alpha$ $\in ]0, 1]$. $f : [a, b]$ $\to \Bbb R$ is said to satisfy a uniform Holder condition with exponent $\alpha$if there is some positive constant $M$ such that $\mid f(x_1) - f(x_2)\mid \lt M\mid x_1 - x_2\mid$, for all $x_1, x_2 \in [a, b]$

Is there any function that satifies a uniform Holder condition with exponent $\alpha$and which is not of bounded variation?

@Vim if a function is not BV $\Rightarrow$ not differentiable? — delog, May 26 '17 at 08:30
what do you mean by differentiable? At any point or at at least one point? — Vim, May 26 '17 at 08:39
@Vim I had just learned the concept of Bounded Variation, but I 've just been provided with the definition, but not with tis meaning. Why does BV concept is meaningful to mathematicians? — delog, May 26 '17 at 08:52
Where does the $\alpha$ enter? Do you mean $|,f(x_1)-f(x_2),|\lt M|,x_1-x_2,|^\alpha$? — robjohn, Jun 17 '19 at 17:22
If I have proved that $f=x^a\sin(1/x^b)$ on $(0,1]$ ,$f(0)=0$ is of BV if and only if $a>b$. Can we try the function $f=x^\alpha\sin(1/x)$? — mnmn1993, Dec 13 '17 at 12:16

score 3 · Answer 1 · edited May 26 '17 at 08:21

3

The Weierstrass function is Hölder continuous but it is not BV.

edited May 26 '17 at 08:21

Parcly Taxel

103,344

answered May 26 '17 at 08:04

Rigel

14,434

if a function is not BV $\Rightarrow$ not differentiable? – delog May 26 '17 at 08:30
1

If a function is BV then is it differentiable a.e. The Weierstrass' function is not differentiable at any point. – Rigel May 26 '17 at 08:31

Hölder condition with Not of Bounded Variation

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