Consider the functional $F$, which is defined for each Ito process
$$X(t) = \int_0^t \mu(s) \mathrm d s + \int_0^t \sigma(s) \mathrm d W(s)$$
$$F(X) := \mathbb E\bigg(\int_0^T X(s) \mathrm dX(s)\bigg)$$
Now I would like to prove that $F$ is convex, which seems to be intuitive because it is true for absolutely continuous processes.
Due to the Ito formula / product rule,
$$ \int_0^T X(s) \mathrm dX(s) = \frac 1 2 X(T)^2 - \int_0^T \sigma(s)^2 \mathrm ds $$
If we restrict ourselves on Ito processes for which we have $\sigma = 0$, the proof is thus very easy. However, the quadratic variation term for $\sigma \ne 0$ seems to mess everything up. On the other hand, the $\sigma$ is also included in the $X(T)^2$ term, so we don't immediately get a counterexample.
Is there another proof for the claim? Or is the claim wrong and there is a counterexample?