Density of smooth functions under "Hölder metric" 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.
 A: 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. 
A: Smooth functions are not dense in the space of Hölder continuous functions, but it is possible to characterize those functions that can be approximated. This is done below.

Definition. Let $(X,d)$ be a metric space and $0<\alpha\leq 1$. The space $C^{0,\alpha}(X)$ of  Hölder continuous functions is a space
  of bounded real valued functions such that $$
 [f]_{C^{0,\alpha}}:=\sup\left\{\frac{|f(x)-f(y)|}{d(x,y)^\alpha}:\,
 x\neq y\right\}<\infty. $$

$C^{0,\alpha}(X)$ is a Banach space with respect to the norm
$$
\Vert f\Vert_{C^{0,\alpha}}=\Vert f\Vert_\infty+[f]_{C^{0,\alpha}}.
$$

Definition. Let $(X,d)$ be a metric space. For $0<\alpha\leq 1$ we define the space $C^{0,\alpha+}(X)$ to be a subspace of
  $C^{0,\alpha}(X)$ (equipped with the  same norm) that consists of
  functions $f\in C^{0,\alpha}(X)$ such that for every compact set $K$, $$
 \lim_{t\to 0+}\ \sup\left\{\frac{|f(y)-f(x)|}{d(x,y)^\alpha}:\ x,y\in
 K,\ d(x,y)\leq t,\ x\neq y\right\}= 0. $$

In other words $C^{0,\alpha+}(X)$ is a subspace of $C^{0,\alpha}(X)$ consisting of functions that, on every compact set, 
have a slightly better modulus of continuity than 
that one in the definition of the $C^{0,\alpha}$ norm.

Theorem. Let $0<\alpha<1$. Then a function  $f\in C^{0,\alpha}(\mathbb{R}^n)$  can be approximated by a sequence smooth
  functions $f_k\in C^\infty$  so that for every compact set
  $K\subset\mathbb{R}^n$, $\Vert f_k-f\Vert_{C^{0,\alpha}(K)}\to 0$ as
  $k\to\infty$ if and only if $f\in C^{0,\alpha+}(\mathbb{R}^n)$.

Remark. The result is not true for $\alpha=1$. Lipschitz functions that are not $C^1$ cannot be approximated by smooth functions in the Lipschitz norm, because the Lipschitz norm is the same as $C^1$ norm.

Example. $|x|^{1/2}\in C^{0,1/2}(-1,1)\setminus C^{0,1/2+}(-1,1)$ and hence it cannot be approximated by smooth functions in the
  $C^{0,1/2}$ norm (see also the accepted answer).

Proof of the theorem.
Suppose that $f\in C^{0,\alpha}(\mathbb{R}^n)$ can be approximates by smooth functions. We need to show that $f\in C^{0,\alpha+}(\mathbb{R}^n)$.
Let $B_R$ be a ball of any radius. Let $\varepsilon>0$ be given. Then for a sufficiently large $k$,
$$
|(f_k-f)(y)-(f_k-f)(x)|\leq\frac{\varepsilon}{2}|x-y|^\alpha
\quad
\text{for all $x,y\in B_R$.}
$$
Let $M=\sup_{B_R}|\nabla f_k|$. Hence the mean value theorem yields
$$
|f(y)-f(x)|\leq\frac{\varepsilon}{2}|x-y|^\alpha +|f_k(y)-f_k(x)|\leq
\left(\frac{\varepsilon}{2}+M|x-y|^{1-\alpha}\right)|x-y|^\alpha
$$
so
$$
|f(y)-f(x)|\leq\varepsilon |x-y|^\alpha
\quad
\text{for all $x,y\in B_r$ satisfying $|x-y|<(\varepsilon/2M)^{1/(1-\alpha)}$.}
$$
This proves that $f\in C^{0,\alpha+}$.
Suppose now that
$f\in C^{0,\alpha+}(\mathbb{R}^n)$. We will show that the approximation by mollification $f_t$ has the desired property i.e., 
for every ball $B_R$, $\Vert f_t-f\Vert_{C^{0,\alpha}(B_R)}\to 0$ as $t\to 0$. 
Since $f_t\to f$ uniformly, it remains to estimate the constant in the H\"older estimate of the difference $f_t-f$.
Let $\varepsilon>0$ be given. It follows from the definition of $C^{0,\alpha+}$
that there is $R>\tau>0$ such that if $x,y\in B_{2R}$, $|x-y|<\tau$, then $|f(x)-f(y)|\leq \frac{1}{2}\varepsilon|x-y|^\alpha$. Hence for $0<t<R$, 
$|f_t(x)-f_t(y)|\leq\frac{1}{2}\varepsilon|x-y|^\alpha$ for $x,y\in B_R$ satisfying $|x-y|<\tau$. This easily follows from the definition of $f_t$, because $f_t(x)$ 
is a weighted average of $f$ on the ball $B(x,t)\subset B_{2R}$.
Therefore
$$
|(f_t-f)(x)-(f_t-f)(y)|\leq \varepsilon |x-y|^\alpha 
\quad
\text{for all $x,y\in B_R$ satisfying $|x-y|<\tau$.}
$$
Let $0<\delta<R$ be such that 
$\Vert f_t-f\Vert_\infty<\varepsilon\tau^\alpha/2$ for $0<t<\delta$. 
If $x,y\in B_R$, $|x-y|\geq\tau$, then 
$$
|(f_t-f)(x)-(f_t-f)(y)|\leq 2\Vert f_t-f\Vert_\infty<\varepsilon\tau^\alpha\leq\varepsilon |x-y|^\alpha.
$$
We proved that if $0<t<\delta$, then
$$
|(f_t-f)(x)-(f_t-f)(y)|\leq \varepsilon |x-y|^\alpha 
\quad
\text{for all $x,y\in B_R$}
$$
as desired. This completes the proof. 
$\Box$
A: The answer is yes (Edit: my "proof" below seems incomplete; read the comment below).
 I will prove it in $\mathbb{R}^N$; the proof is easily adapted to the torus. Let $\phi\colon\mathbb{R}^N\to\mathbb{R}^N$ be a smooth function with compact support, non-negative, and such that $\int_{\mathbb{R}^N}\phi(x)dx=1$. Define $\phi_\varepsilon(x)=\varepsilon^{-N}\phi(\varepsilon^{-1}x)$. Given an $\alpha$-Hölder function $f\colon\mathbb{R}^N\to\mathbb{R}^N$, $\phi_\varepsilon*f$ is smooth and $\lim_{\varepsilon\to0}w_\alpha(f-\phi_\varepsilon*f)=0$. To prove it, let $g_\varepsilon=f-\phi_\varepsilon*f$. Then, for any $x,y\in\mathbb{R}^N$,
$$
|g_\varepsilon(x)-g_\varepsilon(y)|\le\int_{\mathbb{R}^N}\phi(t)|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|dt.
$$
Let
$$
G_\varepsilon(t)=\sup_{x\ne y}\frac{|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|}{|x-y|^\alpha}
$$
$$
\qquad\qquad\qquad\le\sup_{x\ne y}\frac{|f(x-\varepsilon t)-f(y-\varepsilon t)|+|f(x)-f(y)|}{|x-y|^\alpha}\le2w_\alpha(f).
$$
Then, for any $x,y\in\mathbb{R}^N$, $x\ne y$
$$
\frac{|g_\varepsilon(x)-g_\varepsilon(y)|}{|x-y|^\alpha}
\le\int_{\mathbb{R}^N}\phi(t)\frac{|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|}{|x-y|^\alpha}dt\le\int_{\mathbb{R}^N}\phi(t)G_\varepsilon(t)dt.
$$
Taking the supremum over $x,y\in\mathbb{R}^N$ we get
$$
w_\alpha(g_\varepsilon)\le\int_{\mathbb{R}^N}\phi(t)G_\varepsilon(t)dt.
$$
As we have seen, $G_\varepsilon(t)$ is bounded and, since $f$ is continuous, $\lim_{\varepsilon\to0}G_\varepsilon(t)=0$ for all $t\in\mathbb{R}^N$. The Lebesgue dominated convergence theorem implies the result.
