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 $u \in C^{2,\alpha}$.
Showing $\|uv\|_{C^\alpha} \leq C(v) \|u\|_{C^\alpha}$ for Hölder functions
holder-spaces
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Consider the case $\kappa = 0$, $1 < s < \lambda \leqslant 1$, the general one can be easily deduced from this one. Firstly, because $h$ is bounded in $\overline{\Omega}$ we have trivially: $\| hu \|_p \leqslant \| h \|_{\infty} \| u \|_p$. Secondly, for the Gagliardo seminorm $$ [ hu ]_{s, p}^p := \underset{\Omega \times \Omega}{\int \int} \frac{| h (x) u (x) - h (y) u (y) |^p}{| x - y |^{n + sp}} \mathrm{d}x \mathrm{d}y, $$ we use the convexity of $x \mapsto x^p$ twice (that is, we use $| a + b |^p \leqslant 2^{p - 1} (| a |^p + | b |^p)$ and obtain \begin{eqnarray*} \hspace{1em} {| h (x) u (x) - h (y) u (y) |^p} & \leqslant & C \left( | h (x) u (x) - h (x) u (y) |^p + | h (x) u (y) - h (x) u (x) |^p \\ \hspace{2em} + | h (x) u (x) - h (y) u (x) |^p + | h (y) u (x) - h (y) u (y) |^p \right)\\ & = & C \left( | h (x) |^p | u (x) - u (y) |^p + | h (x) |^p | u (y) - u (x) |^p \\ \hspace{2em} + | h (x) - h (y) |^p | u (x) |^p + | h (y) |^p | u (x) - u (y) |^p \right) . \end{eqnarray*} We plug this into the integral but for the third summand we use first that $$ | h (x) - h (y) |^p \leqslant C | x - y |^{\lambda p}, $$ then: \begin{eqnarray*} [ hu ]_{s, p}^p & \leqslant & C \left( \| h \|_{\infty}^p 3 \underset{\Omega \times \Omega}{\int \int} \frac{| u (x) - u (y) |^p}{| x - y |^{n + sp}} \mathrm{d}x \mathrm{d}y + \underset{\Omega \times \Omega}{\int \int} \frac{C | x - y |^{\lambda p} | u (x) |^p}{| x - y |^{n + sp}} \mathrm{d}x \mathrm{d}y \right)\\ & \leqslant & C \left( [ u ]_{s, p}^p + \underset{\Omega \times \Omega}{\int \int} \frac{ | u (x) |^p}{| x - y |^{n + (s - \lambda) p}} \mathrm{d}x \mathrm{d}y \right) . \end{eqnarray*} Notice that the exponent in the denominator of the integrand is $n - \delta < n$ for some $\delta > 0$ so the function $1 / | z |^{n - \delta}$ is integrable in any bounded domain. We use Fubini-Tonelli and change the variable in the $\mathrm{d}y$ integral with $z = x - y$ to obtain: \begin{eqnarray*} [ hu ]_{s, p}^p & \leqslant & C \left( [ u ]_{s, p}^p + \int_{\Omega} | u (x) |^p \int_{x - \Omega} \frac{1}{| z |^{n - \delta}} \mathrm{d}z \mathrm{d}x \right)\\ & \leqslant & C \left( [ u ]_{s, p}^p + \int_{\Omega} | u (x) |^p \int_{\Omega + \Omega} \frac{1}{| z |^{n - \delta}} \mathrm{d}z \mathrm{d} x \right)\\ & \leqslant & C ([ u ]_{s, p}^p + \| u \|_p^p) . \end{eqnarray*} Therefore $$ \| hu \|_{s, p} \leqslant C \| u \|_{s, p}, $$ and keeping track of the constants above we see that indeed $C = C (h, p, \Omega)$.
No chain of balls is needed in the case $|x-y|\ge R$. Just use the estimate $$ \frac{|u(x)-u(y)|}{|x-y|^\alpha}\le 2 R^{-\alpha} \sup_{\Omega'}|u| $$ The appearance of $R^{-\alpha} $ is not a problem, since it only depends on $\alpha$ and on the distance of $\Omega'$ to $\partial \Omega$. And $\sup_{\Omega'}|u|$ is controlled by $\|u\|_{L^2(\Omega)}$ by virtue of Theorem 8.17. (We had to control $\sup_{\Omega'}|u|$ anyway, because it's a part of $C^\alpha$ norm.)
For your Edit2: no, $C$ in (1) cannot be independent of the domain even for harmonic functions. E.g., take $u\equiv 1$ on $B(0,r)$ where $r$ is small. Then $k=0$ (homogeneous equation), and $L^2$ norm is small, but $C^\alpha$ norm is $1$. Simply put, the two sides of (1) do not scale in the same way when the domain is rescaled. So, $C$ cannot be an absolute constant.
I think you misread what the authors wrote. They actually claim that the following constants are domain-independent when $\nu=0$:
- $C$ in (8.46), (8.47), (8.63)
- $\alpha$ in (8.65) and (8.68)
What they claim is correct. Note how (8.46) and (8.47) are set up with powers of $R$, to make both sides scale in the same way. And (8.63) is Harnack's inequality. For harmonic functions, $\alpha$ in (8.65) and (8.68) can be taken to be $1$, because we can uniformly control gradient by the size of the function.
By the way, $\nu=0$ does not mean we have Laplace's equation. The PDE has to be of the form $\operatorname{div}(A(x)\nabla u)=0$ (no lower order terms), but the coefficient matrix need not be identity. The Hölder exponent $\alpha$ can be estimated purely in terms of the ellipticity parameters, with no references to domains. The constant $C$, though, is affected by scaling as explained above.
One more thing: if you are interested in good oscillation estimates for harmonic functions specifically, it's best to deal with the Laplace equation specifically, instead of using results for general elliptic equations.
Best Answer
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 $$\begin{align*}\|uv\|_{C^\alpha} &= \|uv\|_{C^0} + [uv]_{C^\alpha}\\ &\leq \|u\|_{C^0}\|v\|_{C^0} + \|u\|_{C^0}[v]_{C^\alpha} + \|v\|_{C^0}[u]_{C^\alpha}\\ &\leq \|u\|_{C^0}\|v\|_{C^0} + \|u\|_{C^0}[v]_{C^\alpha} + \|v\|_{C^0}[u]_{C^\alpha} + [u]_{C^\alpha}[v]_{C^\alpha}\\ &= \|u\|_{C^\alpha}\|v\|_{C^\alpha}.\end{align*}$$ $\hspace{17.3cm}\Box$