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.

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. $\Box$