Let's say we can prove that for a cadlag process, $X, $ there exists $\alpha > 1, \beta > 0, $ and a non-decreasing continuous function $H:[0, 1] \mapsto \mathbb{R} $ such that $$ E\biggl[\bigl|X_s^{(n)} - X_r^{(n)}\bigl|^\beta\cdot \,\bigl|X_t^{(n)}-X_s^{(n)}\bigl|^\beta\biggl]\, \le \bigl(H(t) - H(r)\bigl)^\alpha \quad \mbox{ for } 0 \le r \le s \le t \le 1. $$
How does one prove that from the given condition above, it is possible to infer that there is a constant $C_{\alpha, \beta} $ such that $$ P\bigl[|X_s-X_r|\ge \epsilon, |X_t-X_S|\ge \epsilon] \le\frac{C_{\alpha, \beta}}{\epsilon^{2\beta}}\bigl(H(t)-H(r)\bigl)^\alpha. $$
The inequality about involving the means of $X $ is part of a refinement of Theorem 13.5 in Billingsley's Convergence of Probability Measures, 2nd edition, which establishes sufficient conditions for weak convergence in $D[0, 1]. $ I thought converting from an inequality in the means to one in probabilities was going to be a simple step, by just using some form of Markov inequality, but upon trying it, it does not look that trivial. Unless I am just failing to see something reasonably basic, which I cannot exclude.
Thank you.
Edited on April 27. You are absolutely right Zhoraster. It is as simple as recognizing that for any two random variables, $X $ and $Y, $ and $a> 0, $: $$ a^2\cdot 1_{|X|\ge a, |Y|\ge a} \le |XY| $$ and then going from here it is straightforward.
For one reason or another, I managed to mess up that basic inequality. Sorry everyone for the "false alarm".