# Hoelder functions with power $>1$

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.

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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
Note 2nd paragraph on page 3 of math.ucdavis.edu/~hunter/pdes/ch1.pdf – Gerry Myerson Mar 27 '12 at 23:43
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

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 $$\parallel f(\mbox {something}) - f(\mbox {something else} ) \parallel_W.$$

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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