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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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    @Gerry, finished typesetting answer for Banach spaces and what I know as the definition of Holder continuous.2012-03-27

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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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    @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.2012-03-28