Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Do there exist in Banach space any nonconstant functions satisfying Hoelder condition with power $>1$ defined on open connected set with values in $\mathbb{R}$?

Thanks.

share|improve this question
4  
When I was a grad student, I heard a story about a student defending a PhD thesis in which he had proved some very strong results about Holder continuous functions with exponent greater than 1. He was asked for an example of such a function, and he replied, "well, the constants, of course." His questioner said, well, yes, but what would be a nonconstant example? After some minutes of discussion by the parties present, it was decided that there were no nonconstant examples, and the thesis went up in smoke. But that's just a story I heard, or maybe mis-heard - it proves nothing. –  Gerry Myerson Mar 27 '12 at 22:08
    
@Gerry, Shing Tung Yao tells the same story. On $\mathbb R^n,$ Holder with exponent above $1$ does imply constant. –  Will Jagy Mar 27 '12 at 22:34
    
Do you mean the space of functions $\Lambda^\alpha$ where $\alpha>1$ is not an integer and their $\lfloor\alpha\rfloor$ derivative is in $\Lambda^{\alpha-\lfloor\alpha\rfloor}$ (Hölder or Lipschitz continuous)? –  robjohn Mar 27 '12 at 23:02
1  
Note 2nd paragraph on page 3 of math.ucdavis.edu/~hunter/pdes/ch1.pdf –  Gerry Myerson Mar 27 '12 at 23:43
1  
I note that the "possible duplicate" question is about ${\bf R}^n$, while the current question is about "linear metric spaces (or mabe normed spaces)". Is it still a duplicate? –  Gerry Myerson Mar 27 '12 at 23:47

1 Answer 1

up vote 4 down vote accepted

On a Banach space, finite or infinite dimensional, there are two main definitions of derivative, Frechet and Gateaux. The Frechet derivative for a mapping to $\mathbb R,$ if it exists, is a bounded linear operator (to $\mathbb R$). Your condition, Hölder with exponent larger than one, implies that the Frechet derivative exists everywhere and is the zero mapping. As a result, the Gateaux derivative in every direction is $0.$ Finally, the ordinary mean value theorem for functions from $\mathbb R$ to $\mathbb R$ says that your function is constant.

EDIT, SOME DETAIL: Take the usual $\alpha,$ which you want larger than $1,$ as $\alpha = 1 + \beta$ with $\beta > 0.$ So the condition, with positive constant $C,$ that $$ | f(x_0 + h) - f(x_0) | \leq C \parallel h \parallel^\alpha $$ for $x_0, \alpha \in V,$ a Banach space, becomes $$ | f(x_0 + h) - f(x_0) | \leq C \parallel h \parallel^{1 + \beta}, $$ so $$ \frac{| f(x_0 + h) - f(x_0) |}{\parallel h \parallel} \leq C \parallel h \parallel^\beta. $$ So $$ \lim_{h \rightarrow 0} \frac{| f(x_0 + h) - f(x_0) |}{\parallel h \parallel} = 0. $$ That is, the Frechet derivative of $f$ at $x_0$ is the zero operator, which is certainly bounded. As a result, the Gateaux derivative in any direction is also $0.$

Take $x_0 \in V,$ also $v \in V$ with $v \neq 0.$ Define $g : \mathbb R \rightarrow \mathbb R$ by $$ g(t) = f(x_0 + t v). $$ Now, for real $\delta,$

$$ \lim_{\delta \rightarrow 0} \frac{| f(x_0 + tv + \delta v) - f(x_0 + tv) |}{\parallel \delta v \parallel} = 0, $$

$$ \lim_{\delta \rightarrow 0} \frac{| f(x_0 + (t + \delta )v) - f(x_0 + tv) |}{ |\delta| \parallel v \parallel} = 0, $$

$$ \left( \frac{1}{\parallel v \parallel} \right) \; \lim_{\delta \rightarrow 0} \frac{|g(t+\delta) - g(t)|}{|\delta|} = 0. $$

$$ \lim_{\delta \rightarrow 0} \frac{|g(t+\delta) - g(t)|}{|\delta|} = 0. $$

$$ \lim_{\delta \rightarrow 0} \frac{g(t+\delta) - g(t)}{\delta} = 0. $$

So, $g$ has a derivative, indeed $g' =0,$ thus (ordinary mean value theorem) $g$ and therefore $f$ are constant, as any other element $w \in V$ can be joined with $v$ by a straight line.

EDIT TOOOO: The same proof, with the same conclusion, goes through when $f : V \rightarrow W,$ where both $V,W$ are Banach spaces. The only change is, in the half dozen instances where we have $|f(\mbox {something}) - f(\mbox {something else} )|,$ we switch to the norm in $W,$ the result being $$ \parallel f(\mbox {something}) - f(\mbox {something else} ) \parallel_W.$$

share|improve this answer
    
This is true when using the condition $$ |f(x+h)-f(x)|\le C|h|^\alpha $$ however, if we use the condition $$ |\Delta_h^{k+1}f(x)|\le C|h|^\alpha $$ where $k<\alpha< k+1$, then $f^{(k)}\in\Lambda^{\alpha-k}$ and $f$ does not need to be constant. –  robjohn Mar 27 '12 at 23:35
    
@robjohn, I wanted to typeset the answer for the Yau story before closure. If I ever studied your condition, I do not recall doing so. Is it in Gilbarg and Trudinger, Elliptic Partial Differential Equations of the Second Order? –  Will Jagy Mar 27 '12 at 23:52
    
Gilbarg and Trudinger in Chapter $4$, define the space $C^{k,\alpha}$. Stein in Singular Integrals and Differentiability Properties of Functions starting in Chapter V Section 4.3, defines and considers $\Lambda_\alpha$ for all $\alpha>0$. Proposition $8$ in that section shows the equivalence of $C^{1,\alpha-1}$ and $\Lambda_\alpha$ for $1<\alpha<2$. The corresponding spaces for higher derivatives are also equivalent. Nagel and Stein Lectures on Pseudo-Differential Operators also considers these spaces in Chapter III. –  robjohn Mar 28 '12 at 0:40
    
@robjohn, if you think that is the question the OP asked, or should have asked, you might as well type it up. I would certainly type it offsite in Latex first, then paste it in if the question has not been closed yet. Really frustrating to be halfway through typing here, with a slow editing interface (because it renders the Latex as you go along), and have the software then reject it due to question closure. –  Will Jagy Mar 28 '12 at 1:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.