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Let $\lambda \in (0,1]$.

Let $f:[a,b]\rightarrow \mathbb{R}$.

Define the space $H^\lambda[a,b]$ as those functions such that there exists a constant $A$ such that $|f(x_1)-f(x_2)|\le A|x_1-x_2|^\lambda, x_1,x_2 \in [a,b]$.

Assume a function has the property that

$\frac{f(x_1)-f(x_2)}{|x_1-x_2|}\rightarrow 0, x_1\rightarrow x_2$.

I want to show that $f \in H^\lambda$. From what I have read, $f$ is in the little Hölder space and it should also be contained in the big Hölder-space.

From what I see $H^1\subset H^\lambda$. So I need to show that $f\in H^1$?

I am struggling to show that $f\in H^1$.

What I did was to assume it didn't then there must be sequences $x_1^n,x_2^n$, such that $\frac{|f(x_1^n)-f(x_2^n)|}{|x_1^n-x_2^n|}\rightarrow \infty$. By compactness I can assume that the sequences converge. And since $f$ must be continuous(and then bounded) they must converge to the same point, or else $\frac{|f(x_1^n)-f(x_2^n)|}{|x_1^n-x_2^n|}$ will be bounded. So I can assume that both sequences converge to the point $x$.

The idea is then to bound $\frac{|f(x_1^n)-f(x_2^n)|}{|x_1^n-x_2^n|}$ in some way and get a contradiction. But I am not able to do that.

I get $\frac{|f(x_1^n)-f(x_2^n)|}{|x_1^n-x_2^n|}\le \frac{|f(x_1^n)-f(x)|}{|x_1^n-x_2^n|}\frac{|x_1^n-x|}{|x_1^n-x|}+\frac{|f(x)-f(x_2^n)|}{|x_1^n-x_2^n|}\frac{|x_2^n-x|}{|x_2^n-x|}$. Where I have assumed that both $x_1^n\ne x$ and $x_2^n \ne x$, if one of them are equal for an infinite amound of points, the result I want follows easily.

However, I do not have a bound on $\frac{|x_1^n-x|}{|x_1^n-x_2^n|}$ or $\frac{|x-x_2^n|}{|x_1^n-x_2^n|}$, and if one of them blows up I fail.

Do you have any hints on how to show this?

Under is the scan from the book, where it comes up that this property must hold, I have a red line under the statement:

enter image description here

1 Answers 1

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There is a typo in $(1.3)$: the denominator should be $|x_1-x_2|^\lambda$.

Assuming (1.3) holds, define $F:[a,b]\times[a,b]\to \mathbb R$ by $$ F(x_1,x_2) = \begin{cases} \frac{|f(x_1)-f(x_2)|}{|x_1-x_2|^\lambda},\quad & x_1 \ne x_2 \\ 0,\quad & x_1=x_2 \end{cases} $$ By construction, $F$ is continuous. Since $[a,b]\times [a,b]$ is compact, $F$ is bounded. Hence, $f\in H^\lambda$.

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    Thanks, but how do you get continuity on the plane? – 2017-01-29
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    On an unbounded interval, (1.3) does not imply (1.1). Notice that the definition of $H^\lambda$ is made only for a "finite interval", and referring back to the beginning of the section, we see that it means a "finite closed interval". – 2017-01-29
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    Yes, but I meant how do you get continuity on the bounded plane. It tried to prove it: We must prove if $(x_n,y_n) \rightarrow (x,y)$ then $F(x_n,y_n)\rightarrow F(x,y)$. If $x\ne y$ then when n is big enough, we are away from the identity-line, and then I see that we get the convergence. But if $x=y$, I do not see how we get convergence, in order to get convergence we must then show that if $(x_n,y_n)\rightarrow(x,x)$, then $F(x_n,y_n)\rightarrow 0$, but I am not able to do that. – 2017-01-29