definition of BUC($\alpha,\frac{\alpha}{2})$

Question

i found in the book of P;Souplet and P;Quiitner the definition and norm of space $BUC^{(\alpha,\frac{\alpha}{2})}$ .here is the definition: Let $Q=Q_T=\Omega \times(0, T)$ where $\Omega$ is an arbitrary domain in $\mathbb{R}^n$ and $T>0$. Given $\alpha \in(0,1]$ set

$$[f]_{\alpha,Q}=\sup \left\{\frac{\lvert f(x, t)- f(y, s)\rvert}{\lvert x-y\rvert^{\alpha}+\vert t-s \rvert^{\frac{\alpha}{2}}}: x, y \in \Omega, t, s \in (0, T),(x, t) \ne (y, s)\right\} $$

Let $k$ be a nonnegative integer, $\alpha \in(0,1)$ and $a=k+\alpha$. Then we put $$ |f|_{a ; Q}=\sum_{|\beta|+2 j \leq k} \sup _Q\left|D_x^\beta D_t^j f\right|+\sum_{|\beta|+2 j=k}\left[D_x^\beta D_t^j f\right]_{\alpha ; Q} $$ and $B U C^{a, a / 2}(Q):=\left\{f:|f|_{a ; Q}<\infty\right\}$. SPACE BUC$^{(a,\frac{a}{2}}) but i dont see the norm if we have k=0 and second question what additional condition for holder function to belong to this space? thanks

Answer 1

If $k=0$, $a=\alpha$ and your norm only involves the $[\cdot]$ part. In this case, no differentiability is assumed. Just Holder continuity of the function itself (in some sense, more continuous w.r.t. $x$ than w.r.t. $t$. This space is "unbalanced".

If you set $x=y$ you can see that your functions have $\alpha/2$-Holder (in $t$) derivatives of order $\le k$ which have order up to $k/2$ w.r.t. $t$, while if you set $t=s$, those same derivatives are seen to be $\alpha$-Holder in $x$.

The lower the Holder exponent the stronger the condition. Therefore, if you have a function with continuous derivatives of order $\le k$ which have order up to $k/2$ w.r.t. $t$, and the highest order ones are $\alpha/2$- Holder with respect to $(x,t)$, then they belong to $BUC^{a,a/2}$, where $a=k+\alpha$.