$L_p$ space inclusion for Riemann-Stieltjes integral The integral here is defined in the Riemann-Stieltjes sense, the interval $[a,b]$ contains $\psi$'s support, and $\alpha<1$. $X$ is $\alpha$-Hölder continuous.
\begin{align*}  \left|\int_a^b \psi(v)\mathrm{d}X(v)\right|&\leq \|\psi\|_{\infty}\lim\limits_{\left|\mathcal{P}\subseteq{[a,b]}\right|\to 0}\sum\limits_{i=0}^{n-1}\left|X(v_{i+1})-X(v_i)\right|\\
&\leq \|\psi\|_{\infty}\mu([a,b])^{1-\frac{1}{\alpha}}\lim\limits_{\left|\mathcal{P}\subseteq [a,b]\right|\to 0}\left(\sum\limits_{i=0}^{n-1}\left|X(v_{i+1})-X(v_i)\right|^{\frac{1}{\alpha}}\right)^{\alpha},
\end{align*}
where $\mu$ is the standard Lebesgue measure. Have I correctly used the result that, for $1\leq p<q\leq\infty$, where $A$ is a finite measure space and $f$ is measurable (with measure $\mu$),we have
$$\|f\|_p\leq \mu(A)^{\frac{1}{p}-\frac{1}{q}}\|f\|_q$$
?
 A: As far as I understand you are claiming that
$$\sum_{i=0}^n |a_i| \leq c\left( \sum_{i=0}^n |a_i|^{1/\alpha} \right)^{\alpha} \tag{1}$$
for some absolute constant $c=c(\alpha)$ and $\alpha \in (0,1)$. If we choose $a_i := 1$ for all $i=0,\ldots,n$, this would imply
$$n = \sum_{i=0}^n |a_i| \leq c \left( \sum_{i=0}^n |a_i|^{1/\alpha} \right)^{\alpha} = c n^{\alpha};$$
hence,
$$c \geq n^{1-\alpha} \xrightarrow[]{n \to \infty} \infty.$$
This shows that there cannot exist a constant $c=c(\alpha)$ such that $(1)$ holds for any sequence $(a_i)_{i=0,\ldots,n} \subseteq \mathbb{R}$, $n \in \mathbb{N}$.
This, in turn, means that your reasoning doesn't work since the third "$\leq$" doesn't hold true.
Remark: The reason why $(1)$ does not hold true is that $n$ can become arbitrary large. In fact, Jensen's inequality shows that there exists a constant $c=c(\alpha,n)$ such that
$$\sum_{i=0}^n |a_i| \leq c \left( \sum_{i=0}^n |a_i|^{1/\alpha} \right)^{\alpha}$$
for any $(a_i)_{i=0,\ldots,n}$. As the example in the first part of my answer shows, the constant $c=c(n,\alpha)$ explodes if we let $n \to \infty$.
Example Consider $[a,b] := [0,1]$ and the deterministic process $X_t := t$. If we set $t_i := i/n$ for fixed $n \in \mathbb{N}$, then
$$X_b-X_a = 1 = \sum_{i=0}^n (X_{t_{i+1}}-X_{t_i})$$ and
$$\left( \sum_{i=0}^n |X_{t_{i+1}}-X_{t_i}|^{1/\alpha} \right)^{\alpha} = n^{\alpha-1} \xrightarrow[]{n \to \infty} 0$$
and therefore there cannot exist a constant $c>0$ such that
$$\left| \int_a^b dX_t \right| = |X_b-X_a| \leq c \left( \sum_{i=0}^n |X_{t_{i+1}}-X_{t_i}|^{1/\alpha} \right)^{\alpha}$$
for all $n \in \mathbb{N}$. So even in this very simple setting we cannot expect that we can bound $\left\|\int_a^b \psi(t) \, dX_t \right\|_{\infty}$ by the $1/\alpha$-variation of $(X_t)_t$.
A: If i well understand your last comment, i will give an answer using basic propriety of Riemann-Stieltjes integral :
\begin{eqnarray}
\left|\int_{a}^{b} \varphi(t) dX_t(t)  \right|&=& \left|\int_{a}^{b} X_t(t) d\varphi(t)  \right| \rm{ because }\,  \varphi \,\rm{ vanish\,  in } \, a \, \rm{ and }\, b.\\
&\leq & \int_{a}^{b}\left| X_t(t)\right| |d\varphi(t)|\\
&\leq& c \left( \int_{a}^{b}\left| X_t(t)\right|^{1/\alpha} |d\varphi(t)| \right)^\alpha
\end{eqnarray}
Where $c=\left(\int_a^b |d\varphi(t)|\right)^{1-\alpha}$
