This question came up when I was doing some reading into convolution squares of singular measures. Recall a function $f$ on the torus $T = [-1/2,1/2]$ is said to be $\alpha$-Hölder (for $0 < \alpha < 1$) if $\sup\_{t \in \mathbb{T}} \sup\_{h \neq 0} |h|^{-\alpha}|f(t+h)-f(t)| < \infty$. In this case, define this value, $\omega\_\alpha(f) = \sup\_{t \in \mathbb{T}} \sup\_{h \neq 0} |h|^{-\alpha}|f(t+h)-f(t)|$. This behaves much like a metric, except functions differing by a constant will not differ in $\omega\_\alpha$. My primary question is this: 


1) Is it true that the smooth functions are "dense" in the space of continuous $\alpha$-Hölder functions, i.e., for a given continuous $\alpha$-Hölder $f$ and $\varepsilon > 0$, does there exists a smooth function $g$ with $\omega\_\alpha(f-g) < \varepsilon$?


To be precise, where this came up was worded somewhat differently. Suppose $K\_n$ are positive, smooth functions supported on $[-1/n,1/n]$ with $\int K\_n = 1$. 


2) Given a fixed continuous function $f$ which is $\alpha$-Hölder and $\varepsilon > 0$, does there exist $N$ such that $n \geq N$ ensures $\omega\_\alpha(f-f\*K\_n) < \varepsilon$? 


This second formulation is stronger than the first, but is not needed for the final result, I believe.


To generalize, fix $0 < \alpha < 1$ and suppose $\psi$ is a function defined on $[0,1/2]$ that is strictly increasing, $\psi(0) = 0$, and $\psi(t) \geq t^{\alpha}$. Say that a function $f$ is $\psi$-Hölder if $\sup\_{t \in \mathbb{T}} \sup\_{h \neq 0} \psi(|h|)^{-1}|f(t+h)-f(t)| < \infty$. In this case, define this value, $\omega\_\psi(f) = \sup\_{t \in \mathbb{T}} \sup\_{h \neq 0} \psi(|h|)^{-1}|f(t+h)-f(t)|$. Then we can ask 1) and 2) again with $\alpha$ replaced by $\psi$.


I suppose the motivation would be that the smooth functions are dense in the space of continuous functions under the usual metrics on function spaces, and this "Hölder metric" seems to be a natural way of defining a metric of the equivalence classes of functions (where $f$ and $g$ are equivalent if $f = g+c$ for a constant $c$). Any insight would be appreciated.



Density of smooth functions under "Hölder metric"
In our PDE seminar, we met the same kinds of questions, and 
we think the answer is "WRONG". The smooth functions is NOT 
dense in Hölder spaces.


An example is,
$$f(x) = |x|^{1/2} \quad x \in (-1,1)$$
it is easy to check that $f$ is $1/2$-Hölder continuous.


For details,
for any $g \in C^{1}((-1,1))$, then the derivative of $g$ is continuous
at $0$, so we have 
$$
\lim\_{x \to 0} \frac{|g(x)-g(0)|}{|x|^{1/2}} = \lim\_{x \to 0}
|x|^{1/2}\frac{|g(x)-g(0)|}{|x|} = 0
$$  

and
$$
\omega\_{1/2}(g-f) \ge \frac{|(g(x)-f(x))-(g(0)-f(0))|}{|x|^{1/2}} \ge
|\frac{|(g(x)-g(0)|}{|x|^{1/2}}-\frac{|f(x)-f(0)|}{|x|^{1/2}}|
$$
but  

$$
\frac{|f(x)-f(0)|}{|x|^{1/2}}=1 \quad x \in (-1,1) \quad x \neq 0
$$
let $x \to 0$, we obtain $\omega\_{1/2}(g-f) \ge 1$. 


Thus, for any $g \in C^{1}((-1,1))$, we have $\omega\_{1/2}(g-f)\ge 1$.


For $0< \alpha <1$ we can make similar examples,
but when $\alpha = 1$, the proof of the counter-example 
may be different. 


