We know that for $X\subset\mathbb R^m,Y\subset\mathbb R^n$ open, then $C^0(\bar X\times\bar Y)=C^0(\bar X)\hat\otimes_\varepsilon C^0(\bar Y)$ where $V\hat\otimes_\varepsilon W$ is the injective tensor product given by the completion of $\|\sum_{i=1}^nv_i\otimes w_i\|:=\sup\limits_{\|\alpha\|_{V^*},\|\beta\|_{W^*}\le1}\|\sum_{i=1}^n\alpha(v_i)\beta(w_i)\|$.

Let $\alpha\in\mathbb R_+\backslash\mathbb Z$. For Holder spaces I believe we have $c^\alpha(\bar X\times\bar Y)=c^\alpha(\bar X)\hat\otimes_\varepsilon c^\alpha(\bar Y)$. But $C^\alpha(\bar X\times\bar Y)\supsetneq C^\alpha(\bar X)\hat\otimes_\varepsilon C^\alpha(\bar Y)$ just like $L^\infty(\bar X\times\bar Y)\supsetneq L^\infty(\bar X)\hat\otimes_\varepsilon L^\infty(\bar Y)$.

But smooth function is weak dense in $C^\alpha$ and $C^\alpha$ is weak topology complete, so we define a "tensor product" $C^\alpha(\bar X)\boxtimes C^\alpha(\bar Y)$ by the weak topology completion of $C^\alpha(\bar X)\hat\otimes_\varepsilon C^\alpha(\bar Y)$. In such case $C^\alpha(\bar X\times\bar Y)=C^\alpha(\bar X)\hat\otimes_\varepsilon C^\alpha(\bar Y)$ holds.

**My question** (may be soft) is, let $\alpha,\beta\in\mathbb R_+\backslash\mathbb Z$, what is the direct characterization of such tensor product for $C^\alpha(\bar X)\boxtimes C^\beta(\bar Y)$? Since taking weak completion is abstract.

And is there any general frame of characterize such tensors?

algebraicdual of the continuous dual $X'$ endowed with the weak topology $\sigma({X}'^*,X')$. $\endgroup$ – Jochen Wengenroth Apr 17 '19 at 7:11