Theorem : Suppose a stochastic process $X=\{X(t) : 0\le t\le T\}$ takes values in a Banach space $B$ so that for some $p>1$, $C>0$, $r>0$,
$$\mathbb E\left[\left\lVert X(t)-X(s)\right\rVert_B^p\right]^{\frac 1p} \le C\lvert t-s\rvert^{\frac 1p +r}$$
for all $s,t\in [0,T]$. Then there exists a family $\left\{\tilde{X}(t) : t\in [0,T]\right\}$ of random variables such that $X(t)=\tilde{X}(t)$ almost surely for all $t\in [0,T]$ and $t\in [0,T] \mapsto \tilde{X}(t,\omega)\in B$ is continuous for all $\omega\in \Omega$. In fact, for each $\alpha\in (0,r)$, we have
$$\mathbb E\left[\sup_{0\le s<t\le T} \left(\frac{\left\lVert \tilde{X}(t) - \tilde{X}(s)\right\rVert_B}{(t-s)^\alpha}\right)^p\right]^{\frac 1p} \le KT^{\frac 1p +r-\alpha}$$
for a suitable constant $K=K(\alpha, r, C)$.
Proof : By rescaling time, we can assume without loss of generality, that $T=1$.
Now, for $n\ge 0$, we define
$$M_n := \max_{1\le m\le 2^n} \left\lVert X\left (m2^{-n} \right) - X\left((m-1)2^{-n}\right)\right\rVert_B$$
and observe that
$$\mathbb E\left[M_n^p\right]^{\frac 1p} \le \mathbb E \left[\left(\sum_{m=1}^{2^n}\left\lVert X\left (m2^{-n} \right) - X\left((m-1)2^{-n}\right)\right\rVert_B^p\right)^{\frac 1p}\right]$$
which is bounded by $C2^{-rn}$.
Next, consider the path $t\rightsquigarrow X(t)$ on each interval $\left[(m-1)2^{-n}, m2^{-n}\right]$ and linearize it to $t\rightsquigarrow X_n(t)$. It is easy to see that
$$\max_{t\in [0,1]} \left\lVert X_{n+1}(t) - X_n(t)\right\rVert_B = \max_{1\le m\le 2^n} \left\lVert X\left((2m-1)2^{-n-1}\right) - \frac{X\left((m-1)2^{-n}\right) - X(m2^{-n})}{2}\right\rVert$$
which is bounded by $M_{n+1}$.
So,
$$\mathbb E \left[\sup_{t\in [0,1]} \left\lVert X_{n+1}(t) - X_n(t)\right\rVert_B^p\right]^{\frac 1p} \le C 2^{-rn}$$
and so there exists a measurable $\tilde{X} : [0,1]\times \Omega \to B$ satisfying
$$\mathbb E \left[\sup_{t\in [0,1]} \left\lVert \tilde{X}(t) - X_n(t)\right\rVert_B^p\right]^{\frac 1p} \le \frac{C 2^{-rn}}{1-2^{-r}}$$
and $t\rightsquigarrow \tilde{X}(t,\omega)$ is continuous for all $\omega\in \Omega$.
Next, notice that for all $t\in [0,1]$,
$$\left\lVert \tilde{X}(\tau) - X(t) \right\rVert_B \xrightarrow[\tau\to t]{p} 0$$
and hence $\tilde{X}(\tau) = X(t)$ almost surely.
Finally, note that for $2^{-n-1}\le t-s\le 2^{-n}$, we have
\begin{align*}\left\lVert \tilde{X}(t)- \tilde{X}(s)\right\rVert_B &\le \left\lVert \tilde{X}(t)-X_n(t)\right\rVert_B + \left\lVert X_n(t)-X_n(s)\right\rVert_B + \left\lVert X_n(s) - \tilde{X}(s)\right\rVert_B\\&\le 2 \sup_{\tau\in [0,1]} \left\lVert \tilde{X}(\tau) - X_n(\tau)\right\rVert_B + 2^n(t-s)M_n\end{align*}
and hence
$$\frac{\left\lVert\tilde{X}(t) - \tilde{X}(s) \right\rVert_B}{(t-s)^\alpha}\le 2^{\alpha n +\alpha+1} \sup_{\tau\in [0,1]} \left\lVert \tilde{X}(\tau)-X_n(\tau)\right\rVert_B + 2^{n+\alpha n}M_n$$
which implies
\begin{align*}\mathbb E \left[\sup_{0\le s<t\le 1} \left(\frac{\left\lVert\tilde{X}(t) - \tilde{X}(s) \right\rVert_B}{(t-s)^\alpha}\right)^p\right]^{\frac 1p} &\le C\sum_{n=0}^\infty \left(\frac{2^{\alpha n +\alpha +1-rn}}{1-2^{-r}}+2^{\alpha n-rn}\right)\\ &\le \frac{5C}{(1-2^{-r})(1-2^{\alpha-r})} \end{align*}
hence completing the proof.
