Let $H \in (0, 1)$, $D \in \mathbb{R}$ and assume that the following function
$$
r ( t, s ) = \frac{1}{2} \, \Big[ t^{2H} + s^{2H} - | t - s |^{2H} \Big] + D \, t^H s^H,
\quad t, \, s \geq 0
$$
is positive definite. Then, clearly $D \geq -1$, since otherwise $r(t, t)$ is negative. It is also clear that if $D \geq 0$, then this is a covariance of $B_H (t) + \sqrt{D} \, t^H \, \xi$, where $\xi$ is independent of $B_H$ standard normal rv. Hence, $r$ is again positive definite.

**Question:** can $D \in (-1,0)$?

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May be related to https://mathoverflow.net/q/44528/496689, but I didn't find the connection.

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**Update.** If $r$ is to be positive definite, it has to satisfy in particular
$$
g(t)=
\det
\begin{pmatrix}
r ( t, t ) & r ( t, t + 1 ) \\
r ( t, t + 1 ) & r ( t+1, t+1)
\end{pmatrix}
\geq 0.
$$
Expanding this function near $t = 0$, we obtain
$$
g ( t ) \sim ( 1 + 2 D ) \, t^{2H} - D \, t^{3H} -2D \, t^{H+1}.
$$
Hence, $D$ has to satisfy $D \geq -1/2$.