# What is the suitable tensor product for Holder spaces

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?

• Taking the completion of a weak topology is never a good idea: If $X$ is any locally convex space the completion of $(X,\sigma(X,X'))$ is ${X'}^*$, the algebraic dual of the continuous dual $X'$ endowed with the weak topology $\sigma({X}'^*,X')$. – Jochen Wengenroth Apr 17 '19 at 7:11
• @JochenWengenroth Just for Holder space, in order to resolve the density problem. – yaoliding Apr 17 '19 at 14:24