Equivalence of norms for the space of Hölder continuous functions I'm currently reading an introduction to rough paths by G. Zanco, and in it he defines the Banach space of $\alpha$-Hölder continuous functions from $[0,T]$ into $E$, with norm $$\|X\|_{C^\alpha} := |X_0|_E + \|X\|_\alpha\thinspace,$$ where $$\|X\|_\alpha := \sup_{t \neq s} \frac{|X_t-X_s|_E}{|t-s|^\alpha}$$ for a function X in this space. The first term on the RHS of the norm has been added to separate constant functions, which all have $\|X\|_\alpha = 0$, making $\|X\|_\alpha$ itself a semi-norm.
It is then stated that this norm is equivalent to $$\|\cdot\|_\infty + \|\cdot\|_\alpha\thinspace.$$ This is the part I don't understand.
I'm not a mathematician by training, and have been visualising things so far to understand them, but I can't understand the meaning of $\|\cdot\|_\infty$. My image of $\|X\|_\alpha$ is essentially to imagine a simple path (say of $x$ in $t$), then take a line of the form $x=t^\alpha$ and drag it along your path, stretching it vertically as required to fit the whole path within its shape. If this can be done along the whole graph, the path is $\alpha$-Hölder continuous and $\|X\|_\alpha$ is the maximum stretching factor that was required. This is $0$ for constant functions.
Thinking of $\|X\|_\infty$ in the same way, I think that it should be $0$ for constant functions, and is surely not finite in all other cases. So how is this norm equivalent to the first one defined?
 A: Here's an attempt at an intuitive explanation.  $\|X\|_\infty$ has a few names, for example the uniform norm or infinity norm. For a continuous function $X$,  $\|X\|_\infty$ is the maximum distance of a point in the image of $X$ from 0. If $X$ is a real-valued constant function, then $\|X\|_\infty$ is the absolute value of its value.
To see why equivalence might be true, start by taking for granted that $\|X\|_\alpha$ being finite implies that $X$ is continuous, and these are the functions we care about. The easy direction follows because $\|X\|_\infty>|X_0|_E$. For thinking about the other direction, $\|X\|_\infty$ is $|X(r)|$ for some $r\in[0,T]$, so $\|X\|_\infty$ cannot be more than $|X_0|+\|X\|_\alpha r^\alpha$ which is no more than $|X_0|+\|X\|_\alpha T^\alpha$.
A: Clearly we have $$|X_0|_E + \|X\|_\alpha \le \|X\|_\infty + \|X\|_\alpha$$
For the other direction, show that $\alpha$-Hölder continuous functions form a Banach space with respect to both norms $\|\cdot\|_\infty + \|\cdot\|_\alpha$ and $X \mapsto |X_0|_E + \|X\|_\alpha$. For one norm I showed it in this answer with $\alpha = \frac12$, and for the other one see here.
Now the inequality $\|X\|_\infty + \|X\|_\alpha \le C(|X_0|_E + \|X\|_\alpha)$ for some $C > 0$ is a consequence of the Bounded Inverse Theorem applied on the identity map on the space of $\alpha$-Hölder continuous functions, where on the domain we have the norm $\|\cdot\|_\infty + \|\cdot\|_\alpha$ and on the codomain we have $X \mapsto |X_0|_E + \|X\|_\alpha$.
