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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 $u \in C^{2,\alpha}$.

TheGeekGreek
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  • @user357151 Fixed notation. The notes are unfortunately not that clear about the dependence of $C$, but as this comes from Schauder theory, it is quite important, on what $C$ depends. You are right. – TheGeekGreek Jul 25 '18 at 19:21
  • So you want to show that the multiplication operator $u\mapsto uv$ is bounded on $C^\alpha$. For this it suffices to bound it on the unit ball by some constant, and it looks like you already have that kind of a bound. –  Jul 25 '18 at 19:30
  • @user357151 Thank you! See my answer. – TheGeekGreek Jul 25 '18 at 20:18

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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 $$\begin{align*}\|uv\|_{C^\alpha} &= \|uv\|_{C^0} + [uv]_{C^\alpha}\\ &\leq \|u\|_{C^0}\|v\|_{C^0} + \|u\|_{C^0}[v]_{C^\alpha} + \|v\|_{C^0}[u]_{C^\alpha}\\ &\leq \|u\|_{C^0}\|v\|_{C^0} + \|u\|_{C^0}[v]_{C^\alpha} + \|v\|_{C^0}[u]_{C^\alpha} + [u]_{C^\alpha}[v]_{C^\alpha}\\ &= \|u\|_{C^\alpha}\|v\|_{C^\alpha}.\end{align*}$$ $\hspace{17.3cm}\Box$

TheGeekGreek
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