Let

- $(M,d)$ be a separable metric space
- $E$ be a $\mathbb R$-Banach space
- $\alpha\in(0,1]$

Moreover, let $$\left\|f\right\|_{C^{0+\alpha}(K,\:E)}:=\sup_{x\in K}\left\|f(x)\right\|_E+\sup_{\substack{x,\:y\:\in\:K\\x\:\ne\:y}}\frac{\left\|f(x)-f(y)\right\|_E}{{d(x,y)}^\alpha}\;\;\;\text{for }f:M\to E$$ for $K\subseteq M$ and $$C^{0+\alpha}(M,E):=\left\{f:M\to E\mid\left\|f\right\|_{C^{0+\alpha}(K,\:E)}<\infty\text{ for all compact }K\subseteq M\right\}$$ be equipped with the topology generated by $$\left\{\left\|\;\cdot\;\right\|_{C^{0+\alpha}(K,\:E)}:K\subseteq M\text{ is compact}\right\}.$$

I want to show that $$\mathcal B\left(C^{0+\alpha}(M,E)\right)=\left.{\mathcal B(E)}^{\otimes M}\right|_{C^{0+\alpha}(M,\:E)};$$ at least in the special case $M=\Lambda$ for an open subset $\Lambda\subseteq\mathbb R^n$, $n\in\mathbb N$.

The desired result is motivated by the following fact: If $M=[0,\infty)$ and $E=\mathbb R$, the space $\Omega:=C^0([0,\infty))$ of continuous functions from $[0,\infty)$ to $\mathbb R$, equipped with the topology of uniform convergence on compact subsets, is a Polish space. Let $$X_t:\Omega\to\mathbb R\;,\;\;\;\omega\mapsto\omega(t)$$ for $t\ge0$. Then we are able to show that $$\left.{\mathcal B(\mathbb R)}^{\otimes[0,\infty)}\right|_\Omega=\sigma(X_t,t\in[0,\infty))=\mathcal B(\Omega).$$

$^1$ Let $\mathcal B(X)$ denote the Borel $\sigma$-algebra on a topological space $X$. If $I$ is a nonempty set and $(\Omega_i\mathcal A_i)$ is a measurable space for all $i\in I$, let $\bigotimes_{i\in I}\mathcal A_i$ denote the product $\sigma$-algebra on $\prod_{i\in I}\Omega_i$. If $(\Omega_i\mathcal A_i)=(\Omega,\mathcal A)$ for all $i\in I$, write ${\mathcal A}^{\otimes I}$ instead of $\bigotimes_{i\in I}\mathcal A$. Moreover, for any measurable space $(\Omega,\mathcal A)$ and $A\subseteq\Omega$, let $\left.\mathcal A\right|_A$ denote the trace $\sigma$-algebra on $A$.