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One can apply a deterministic result, called Garsia--Rodemich--Rumsey inequality, to estimate $\mathrm{E}[||X||^\alpha_{\gamma;[0,T]}]$. Here is a particular form of this result, which is most convenient for us.

For any $\alpha >1$, $\delta> 1/\alpha$, there is a constant $C(\delta,\alpha)$ such that for any $f\in C([0,T],S)$ and $t,s\in[0,T]$ $$ |f(t) - f(s)|^\alpha\le C(\delta,\alpha)|t-s|^{\delta\alpha - 1} \int_s^t \int_s^t \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv. $$ In particular, for $\gamma = \delta - 1/\alpha$ $$ ||f||_{\gamma;[0,T]}:= \sup_{0\le t<s\le T}\frac{d(f(t),f(s))}{|t-s|^\gamma}\le C(\delta,\alpha)^{1/\alpha}\bigg(\int_0^T\int_0^T \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv\bigg)^{1/\alpha}. $$

For any $\gamma<\beta/\alpha$ take $\delta\in (1/\alpha,\gamma+1/\alpha)$ to get $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\delta,\alpha) \int_0^T\int_0^T \frac{\mathrm{E} \left[d(X_u,X_v)^\alpha\right]}{|u-v|^{\delta\alpha + 1}}du\,dv \\ \le C(\delta,\alpha) K\int_0^T\int_0^T |u-v|^{\beta-\delta\alpha}du\,dv = C(\delta,\alpha,\beta,T)K. $$ As a result, $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\alpha,\beta,\gamma,T) K, $$ and the constant may be written explicitly.

This is slightly better than @Kostya_I's estimate $\mathrm{P}(||X||_{\gamma;[0,T]}^\alpha > x)\le CKx^{-1}$, which does not imply integrability.

One can apply a deterministic result, called Garsia--Rodemich--Rumsey inequality, to estimate $\mathrm{E}[||X||^\alpha_{\gamma;[0,T]}]$. Here is a particular form of this result, which is most convenient for us.

For any $\alpha >1$, $\delta> 1/\alpha$, there is a constant $C(\delta,\alpha)$ such that for any $f\in C([0,T],S)$ and $t,s\in[0,T]$ $$ d(f(t), f(s))^\alpha\le C(\delta,\alpha)|t-s|^{\delta\alpha - 1} \int_s^t \int_s^t \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv. $$ In particular, for $\gamma = \delta - 1/\alpha$ $$ ||f||_{\gamma;[0,T]}:= \sup_{0\le t<s\le T}\frac{d(f(t),f(s))}{|t-s|^\gamma}\le C(\delta,\alpha)^{1/\alpha}\bigg(\int_0^T\int_0^T \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv\bigg)^{1/\alpha}. $$

For any $\gamma<\beta/\alpha$ take $\delta\in (1/\alpha,\gamma+1/\alpha)$ to get $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\delta,\alpha) \int_0^T\int_0^T \frac{\mathrm{E} \left[d(X_u,X_v)^\alpha\right]}{|u-v|^{\delta\alpha + 1}}du\,dv \\ \le C(\delta,\alpha) K\int_0^T\int_0^T |u-v|^{\beta-\delta\alpha}du\,dv = C(\delta,\alpha,\beta,T)K. $$ As a result, $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\alpha,\beta,\gamma,T) K, $$ and the constant may be written explicitly.

This is slightly better than @Kostya_I's estimate $\mathrm{P}(||X||_{\gamma;[0,T]}^\alpha > x)\le CKx^{-1}$, which does not imply integrability.

One can apply a deterministic result, called Garsia--Rodemich--Rumsey inequality, to estimate $\mathrm{E}[||X||^\alpha_{\gamma;[0,T]}]$. Here is a particular form of this result, which is most convenient for us.

For any $\alpha >1$, $\delta> 1/\alpha$, there is a constant $C(\delta,\alpha)$ such that for any $f\in C([0,T],S)$ and $t,s\in[0,T]$ $$ |f(t) - f(s)|^\alpha\le C(\delta,\alpha)|t-s|^{\delta\alpha - 1} \int_s^t \int_s^t \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv. $$ In particular, for $\gamma = \delta - 1/\alpha$ $$ ||f||_{\gamma;[0,T]}:= \sup_{0\le t<s\le T}\frac{d(f(t),f(s))}{|t-s|^\gamma}\le C(\delta,\alpha)^{1/\alpha}\bigg(\int_0^T\int_0^T \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv\bigg)^{1/\alpha}. $$

For any $\gamma<\beta/\alpha$ take $\delta\in (1/\alpha,\gamma+1/\alpha)$ to get $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\delta,\alpha) \int_0^T\int_0^T \frac{\mathrm{E} \left[d(X_u,X_v)^\alpha\right]}{|u-v|^{\delta\alpha + 1}}du\,dv \\ \le C(\delta,\alpha) K\int_0^T\int_0^T |u-v|^{\beta-\delta\alpha}du\,dv = C(T,\delta,\alpha)K. $$ In particular, $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\alpha,\beta,\gamma,T) K; $$ the constant may be written explicitly.

This is slightly better than @Kostya_I's estimate $\mathrm{P}(||X||_{\gamma;[0,T]}^\alpha > x)\le CKx^{-1}$, which does not imply integrability.

One can apply a deterministic result, called Garsia--Rodemich--Rumsey inequality, to estimate $\mathrm{E}[||X||^\alpha_{\gamma;[0,T]}]$. Here is a particular form of this result, which is most convenient for us.

For any $\alpha >1$, $\delta> 1/\alpha$, there is a constant $C(\delta,\alpha)$ such that for any $f\in C([0,T],S)$ and $t,s\in[0,T]$ $$ |f(t) - f(s)|^\alpha\le C(\delta,\alpha)|t-s|^{\delta\alpha - 1} \int_s^t \int_s^t \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv. $$ In particular, for $\gamma = \delta - 1/\alpha$ $$ ||f||_{\gamma;[0,T]}:= \sup_{0\le t<s\le T}\frac{d(f(t),f(s))}{|t-s|^\gamma}\le C(\delta,\alpha)^{1/\alpha}\bigg(\int_0^T\int_0^T \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv\bigg)^{1/\alpha}. $$

For any $\gamma<\beta/\alpha$ take $\delta\in (1/\alpha,\gamma+1/\alpha)$ to get $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\delta,\alpha) \int_0^T\int_0^T \frac{\mathrm{E} \left[d(X_u,X_v)^\alpha\right]}{|u-v|^{\delta\alpha + 1}}du\,dv \\ \le C(\delta,\alpha) K\int_0^T\int_0^T |u-v|^{\beta-\delta\alpha}du\,dv = C(\delta,\alpha,\beta,T)K. $$ As a result, $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\alpha,\beta,\gamma,T) K, $$ and the constant may be written explicitly.

This is slightly better than @Kostya_I's estimate $\mathrm{P}(||X||_{\gamma;[0,T]}^\alpha > x)\le CKx^{-1}$, which does not imply integrability.

One can apply a deterministic result, called Garsia--Rodemich--Rumsey inequality, to estimate $\mathrm{E}[||X||^\alpha_{\gamma;[0,T]}]$. Here is a particular form of this result, which is most convenient for us.

For any $\alpha >1$, $\delta> 1/\alpha$, there is a constant $C(\delta,\alpha)$ such that for any $f\in C([0,T],S)$ and $t,s\in[0,T]$ $$ |f(t) - f(s)|^\alpha\le C(\delta,\alpha)|t-s|^{\delta\alpha - 1} \int_s^t \int_s^t \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv. $$ In particular, for $\gamma = \delta - 1/\alpha$ $$ ||f||_{\gamma;[0,T]}:= \sup_{0\le t<s\le T}\frac{d(f(t),f(s))}{|t-s|^\gamma}\le C(\delta,\alpha)^{1/\alpha}\bigg(\int_0^T\int_0^T \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv\bigg)^{1/\alpha}. $$

For any $\gamma<\beta/\alpha$ take $\delta\in (1/\alpha,\gamma+1/\alpha)$ to get $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\delta,\alpha) \int_0^T\int_0^T \frac{\mathrm{E} \left[d(X_u,X_v)^\alpha\right]}{|u-v|^{\delta\alpha + 1}}du\,dv \\ \le C(\delta,\alpha) K\int_0^T\int_0^T |u-v|^{\beta-\delta\alpha}du\,dv = C(T,\delta,\alpha)K. $$ In particular, $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\alpha,\beta,\gamma,T) K; $$ the constant may be written explicitly.

This is slightly better than @Kostya_I's estimate, which gives $\mathrm{P}(||X||_{\gamma;[0,T]}^\alpha > x)\le CKx^{-1}$, and integrability cannot be deduced from here.

One can apply a deterministic result, called Garsia--Rodemich--Rumsey inequality, to estimate $\mathrm{E}[||X||^\alpha_{\gamma;[0,T]}]$. Here is a particular form of this result, which is most convenient for us.

For any $\alpha >1$, $\delta> 1/\alpha$, there is a constant $C(\delta,\alpha)$ such that for any $f\in C([0,T],S)$ and $t,s\in[0,T]$ $$ |f(t) - f(s)|^\alpha\le C(\delta,\alpha)|t-s|^{\delta\alpha - 1} \int_s^t \int_s^t \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv. $$ In particular, for $\gamma = \delta - 1/\alpha$ $$ ||f||_{\gamma;[0,T]}:= \sup_{0\le t<s\le T}\frac{d(f(t),f(s))}{|t-s|^\gamma}\le C(\delta,\alpha)^{1/\alpha}\bigg(\int_0^T\int_0^T \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv\bigg)^{1/\alpha}. $$

For any $\gamma<\beta/\alpha$ take $\delta\in (1/\alpha,\gamma+1/\alpha)$ to get $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\delta,\alpha) \int_0^T\int_0^T \frac{\mathrm{E} \left[d(X_u,X_v)^\alpha\right]}{|u-v|^{\delta\alpha + 1}}du\,dv \\ \le C(\delta,\alpha) K\int_0^T\int_0^T |u-v|^{\beta-\delta\alpha}du\,dv = C(T,\delta,\alpha)K. $$ In particular, $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\alpha,\beta,\gamma,T) K; $$ the constant may be written explicitly.

This is slightly better than @Kostya_I's estimate $\mathrm{P}(||X||_{\gamma;[0,T]}^\alpha > x)\le CKx^{-1}$, which does not imply integrability.