# Infinitly differentiable hölder continuous function

I want to find an example of an infinitly differentiable hölder continuous function with parameter $\alpha=1$, i.e. $f\in C^{0,1}$ for $f:A\to B\subset \mathbb{R}$, where $A$ is closed set in $\mathbb{R}^n$. For simplicity let $A=[a,b]^n$. Now if $n=1$, I would take the simple polynomial $f(x)=x$, which is infinetly differentiable and, since it is lipschitz, it is hölder continuous with parameter $\alpha =1$. My problem is that the functions range should be a subset of $\mathbb{R}$. So I do not see how to generalize this in higher dimension. For example the norm function, would be Lipschitz and it also has the right range, but is however not infinitely differentiable.

• @nayrb Sorry for the confusion! It does not have to be whole $\mathbb{R}$. But the evaluation of the function should be a real number. – user20869 Oct 11 '12 at 13:44
• Will you be satisfied with a constant function $f:R^n \rightarrow R$ as an infinitely differentiable lipschitz function? – abnry Oct 11 '12 at 13:53
Take $f$ a smooth bounded function with bounded derivative, and take $g(x):=f(\lVert x\rVert^2)$, where $\lVert \cdot\rVert$ is the Euclidian norm. Then $$|g(x)-g(y)|=\left|\int_0^1\partial_t g(tx+(1-t)y)dt\right|=\left|\sum_{j=1}^n2(x_i-y_i)f'(\lVert tx+(1-t)y\rVert^2)dt\right|\leq 2n\lVert x-y\rVert \sup_{v\in\Bbb R^n}|f'(v)|.$$ Example: $f(t)=e^{-t^2}$.