$n_0$ must be independent of $t_1$ and $t_2$, of course. If it's not, the processes might be even discontinuous. For instance, $X_n$ is a Poisson process with parameter $1/n$. Then $$E(|X_n(t_1)-X_n(t_2)|^2)\le |t_1-t_2|^2$$ for all $n>|t_1-t_2|^{-1}$ (for all $n\ge 1$ if $t_1=t_2$).
And the same answer works for the second question: when $\alpha=1$, the processes need not to be continuous. In some special cases, where you have higher moments controlled by a lower one polynomially, it may help (e.g. in the Gaussian case $\gamma=2$ and $\alpha=1$ is enough).
$n_0$ must be independent of $t_1$ and $t_2$, of course. If it's not, the processes might be even discontinuous. For instance, $X_n$ is a Poisson process with parameter $1/n$. Then $E(|X_n(t_1)-X_n(t_2)|^2)\le$E(|X_n(t_1)-X_n(t_2)|^2)\le |t_1-t_2|^2$t_1-t_2|^2$$for all$n>|t_1-t_2|^{-1}$(for all$n\ge 1$if$t_1=t_2$). And the same answer works for the second question: when$\alpha=1$, the processes need not to be continuous. In some special cases, where you have higher moments controlled by a lower polynomially, it may help (e.g. in the Gaussian case$\gamma=2$and$\alpha=1$is enough). 1 It should be independent of$t_1$and$t_2$, of course. If it's not, the processes might be even discontinuous. For instance,$X_n$is a Poisson process with parameter$1/n$. Then$E(|X_n(t_1)-X_n(t_2)|^2)\le |t_1-t_2|^2$for all$n>|t_1-t_2|^{-1}$(for all$n\ge 1$if$t_1=t_2$). And the same answer works for the second question: when$\alpha=1$, the processes need not to be continuous. In some special cases, where you have higher moments controlled by a lower polynomially, it may help (e.g. in the Gaussian case$\gamma=2$and$\alpha=1\$ is enough).