Showing $\|uv\|_{C^\alpha} \leq C(v) \|u\|_{C^\alpha}$ for Hölder functions Suppose $\Omega \subseteq \subseteq \mathbb{R}^n$ and $0 < \alpha < 1$. As my notes suggest, there is an estimate of the form $$\|uv\|_{C^\alpha} \leq C(v) \|u\|_{C^\alpha}$$ for $u,v \in C^\alpha$ where we consider the Hölder space equipped with the usual norm $$\|\cdot\|_{C^\alpha} = \|\cdot\|_{C^0} + [\cdot]_{C^{\alpha}}$$ with the seminorm $[\cdot]_{C^\alpha}$. However, I was only able to show that $$\|uv\|_{C^\alpha} \leq C(u,v)(\|v\|_{C^\alpha} + \|u\|_{C^\alpha})$$ If it helps, we can also assume $u \in C^{2,\alpha}$.
 A: 
Lemma. If $u,v \in C^\alpha$ then $uv \in C^\alpha$ with $\|uv\|_{C^\alpha} \leq \|u\|_{C^\alpha}\|v\|_{C^\alpha}$.

Proof. Note that $\|uv\|_{C^0} \leq \|u\|_{C^0}\|v\|_{C^0}$. Moreover, for any $x,y \in \Omega$ we estimate $$\begin{align*} |u(x)v(x) - u(y)v(y)| &= |u(x)v(x) - u(y)v(y) + u(x)v(y) - u(x)v(y)|\\
&\leq |u(x)||v(x) - v(y)| + |v(y)||u(x) - u(y)|\\
&\leq \|u\|_{C^0}|v(x) - v(y)| + \|v\|_{C^0}|u(x) - u(y)|\end{align*}$$ Thus assuming $x \neq y$ and dividing above by $|x - y|^\alpha$ yields $$\frac{|u(x)v(x) - u(y)v(y)|}{|x- y|^\alpha} \leq \|u\|_{C^0}[v]_{C^\alpha} + \|v\|_{C^0}[u]_{C^\alpha}$$ Thus taking the supremum yields $$[uv]_{C^\alpha} \leq \|u\|_{C^0}[v]_{C^\alpha} + \|v\|_{C^0}[u]_{C^\alpha}.$$ Hence we compute $$\begin{align*}\|uv\|_{C^\alpha} &= \|uv\|_{C^0} + [uv]_{C^\alpha}\\
&\leq \|u\|_{C^0}\|v\|_{C^0} + \|u\|_{C^0}[v]_{C^\alpha} + \|v\|_{C^0}[u]_{C^\alpha}\\
&\leq \|u\|_{C^0}\|v\|_{C^0} + \|u\|_{C^0}[v]_{C^\alpha} + \|v\|_{C^0}[u]_{C^\alpha} + [u]_{C^\alpha}[v]_{C^\alpha}\\
&= \|u\|_{C^\alpha}\|v\|_{C^\alpha}.\end{align*}$$
$\hspace{17.3cm}\Box$
