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