Relation between Minimum Holder Coefficient and Holder Semi-norm

continuityfunctional-analysisholder-spacesnormed-spacesreal-analysis

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

Best Answer

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.

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