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 $\Lambda\subseteq\mathbb R^n$, $n\in\mathbb N$.
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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 $$\left.{\mathcal B(\mathbb R)}^{\otimes[0,\infty)}\right|\_\Omega=\sigma(X\_t,t\in[0,\infty))=\mathcal B(\Omega).$$




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$^1$ 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 $A$.



Borel $\sigma$-algebra on the space of Hölder continuous functions
It's not true.


For brevity, let me take $M=[0,1]$ (or $(0,1)$ if you prefer), $E = \mathbb{R}$, $X = C^{0+\alpha}(M,E)$, $\mathcal{B}$ its $\sigma$-algebra, and $\mathcal{F}$ the $\sigma$-algebra appearing on the right side of your desired equation. I claim that $|\mathcal{F}| = \mathfrak{c}$ while $|\mathcal{B}| \ge 2^{\mathfrak{c}}$.


For $t \in [0,1]$, let $\pi\_t : X \to \mathbb{R}$ be the evaluation map $\pi\_t(\omega) = \omega(t)$ which is continuous. Let $\mathcal{U}$ be a countable base for $\mathbb{R}$. Then $\mathcal{F}$ is generated by $\{\pi\_t^{-1}(U) : t \in [0,1], U \in \mathcal{U}\}$. In fact, I claim it is generated by the countable collection $\{\pi\_q^{-1}(U) : q \in [0,1] \cap \mathbb{Q}, U \in \mathcal{U}\}$. For let $\mathcal{F}\_0 \subseteq \mathcal{F}$ be the $\sigma$-algebra generated by the latter collection; note that $\pi\_q$ is $\mathcal{F}\_0$-measurable for $q \in [0,1] \cap \mathbb{Q}$. If $t \in [0,1] \setminus \mathbb{Q}$, choose a sequence $[0,1] \cap \mathbb{Q} \ni q\_n \to t$; then note that $\pi\_{q\_n} \to \pi\_t$ pointwise on $X$, because $X$ is a space of continuous functions. So $\pi\_t$ is $\mathcal{F}\_0$-measurable, and thus $\mathcal{F}\_0$ contains the sets $\pi\_t^{-1}(U)$. Hence $\mathcal{F} \subset \mathcal{F}\_0$.


We have thus shown that $\mathcal{F}$ is countably generated, so its cardinality is $\mathfrak{c}$. However, as noted [here](https://math.stackexchange.com/questions/2564139/is-the-space-c-alpha-separable), the Hölder space contains continuum many functions at pairwise distance $\ge 1$ (in your version of the norm, the distances are a little greater still). Any set of these functions is closed and thus Borel, so there are at least $2^{\mathfrak{c}}$ Borel sets in $X$.


Moral: $C^{0+\alpha}(M)$ is not really similar to $C^0(M)$ and one shouldn't expect the same properties.


