I am reading through Krylov's Lectures on Elliptic and Parabolic Equations in Hölder Spaces. In an attempt to prove the interpolation inequalities for parabolic PDEs, I've stepped back in the text to reference the relevant proofs in the Elliptic case. Starting with some definitions, we take
$$[u]_{\delta;\Omega} = \sup_{\substack{x,y \in \Omega\\ x\neq y}}\frac{|u(x)-u(y)|}{|x-y|^{\delta}}$$ and $$[u]_{k+\delta;\Omega} = \max_{|\alpha|=k}[D^{\alpha}u]_{\delta;\Omega}.$$
We take $\Omega$ to be an open convex cone, and want to show that (EDIT: taking $|\alpha| = j$)
$$\tag{1}|D^{\alpha}u(x)-D^{\alpha}u(y)| \leq N|x-y|[u]_{j+1;\Omega}$$.
My question is: how does $N$ (a constant dependent only on the dimension of the space $d$) appear here? By definition, we should have that
$$ \frac{|D^{\alpha}u(x)-D^{\alpha}u(y)|}{|x-y|} \leq [u]_{j+1;\Omega}$$
by the definition of the $j+1$ Hölder seminorm. Consequentially, should we not have that
$$\tag{2} |D^{\alpha}u(x)-D^{\alpha}u(y)| \leq |x-y|[u]_{j+1;\Omega} $$ independent of the dimension of the space? Krylov mentions that this is a consequence of the mean value theorem, but I have never seen a version of the MVT using Hölder seminorms as opposed to the value of the derivative along some line.
TL;DR, how does the MVT get you the existence of (1) as opposed to (2) given the context of the problem?
After a bit of effort, it comes down to the way the $\delta$-Holder seminorm is defined for integral values. We note that
$$ [u]_{k+\delta} = \sup_{|\alpha|=k}[D^{\alpha}u]_{\delta}.$$ but if the value of $\delta$ is $1$, then it follows that $k+1$ is also an integer, and this is really just
$$[u]_{k+1} = \sup_{|\alpha|=k}[D^{\alpha}u]_{1}.$$ That is to say, $$[u]_{k+1} = \sup_{|\alpha|=k}\frac{|D^{\alpha}u(x)-D^{\alpha}u(y)|}{|x-y|} = \sup_{|\alpha|=k+1}{|D^{\alpha}u(x)|}.$$
Note that by the mean value theorem (and $k+1$th differentiability of $u$), there is some point $\xi$ in the line $cx+(1-c)y$ for $c\in(0,1)$ for which $$|D^{\alpha}u(x)-D^{\alpha}u(y)| = |\nabla D^{\alpha}u(\xi)\cdot(x-y)|.$$ However, $$|\nabla D^{\alpha}u(\xi)| = \sqrt{\sum_{i=1}^{n}(\partial_{x_i}D^{\alpha}u(\xi)})^2 \leq \sqrt{n(\max_{i,\alpha}\partial_{x_i}D^{\alpha}u(\xi))^2} \leq \sqrt{n}\sup_{|\alpha|=k+1}|D^{\alpha}u(x)|.$$ Therefore, $$|D^{\alpha}u(x)-D^{\alpha}u(y)| \leq \sqrt{n}|x-y|[u]_{k+1}.$$ The result above is, notably, dependent on $n$ (the dimension)! What truly remains left to be proven is why the thought process used above to get a bound independent of $n$ may be incorrect.