Relation between Minimum Holder Coefficient and Holder Semi-norm

Question

I have the following two definitions pertaining to Holder continuity:

For both, let $\alpha \in (0,1]$ and consider the path $X:[0,T] \rightarrow \mathbb{R}^d$

1. $X$ is $\alpha$-Holder continuous if there exists a $C \in \mathbb{R}$ with $C>0$ such that $$ |X_t - X_s| \leq C|t-s|^{\alpha} $$ $C$ is known as a Holder coefficient of $X$ with respect to $\alpha$, and the smallest such constant is known as the minimum Holder coefficient, and is denoted $[X]_{\alpha}$

2. $X$ is $\alpha$-Holder continuous if $||X||_{\alpha} < \infty$ where the semi-norm is defined as

$$ ||X||_{\alpha} = \sup_{0\leq s \leq t \leq T} \frac{|X_t - X_s|}{|t-s|^{\alpha}} $$

I am wondering what is the relation between the minimum Holder coefficient $[X]_{\alpha}$ and the seminorm $||X||_{\alpha}$. To me it seems like they are equal to each other - but is there a proof for this or is it too obvious? Sorry if this is a very trivial question.

Thanks

Answer 1

They are equal. If $$ |X_t - X_s| \le C |t-s|^\alpha \quad \forall s,t $$ then (trivially), $\|X\|_\alpha \le C$. This implies $\|X\|_\alpha \le [X]_\alpha$.

On the other hand $$ |X_t - X_s| \le \frac{ |X_t-X_s|}{|t-s|^\alpha} |t-s|^\alpha \le \|X\|_\alpha |t-s|^\alpha, $$ which implies the reverse inequality.