Let $(X\_t, t \in \mathbb{N})$ be a martingale, and let $a \leq b \leq T \in \mathbb{N}$ be constants. Is there something like Doob's inequality for $\mathbb{E} \sup\_{a \leq t \leq b} X\_t(X\_T-X\_t)$, i.e. is it possible to bound this supremum by something involving just a deterministic variance? I am hoping for one of the following inequalities to be true:
$$
\mathbb{E} \sup\_{a \leq t \leq b} \left(X\_t(X\_T-X\_t)\right)^2 \leq C\mathbb{E} \left(X\_b(X\_T-X\_a)\right)^2
$$
or maybe even
$$
\mathbb{E} \sup\_{a \leq t \leq b} \left(X\_t)(X\_T-X\_t)\right)^2 \leq C\mathbb{E} X\_b^2 \mathbb{E}(X\_T-X\_a)^2
$$
for some deterministic constant $C$. In my application, $X\_t$ is the sum process of iid mean-0 random variables; perhaps that makes life easier?



Doob's inequality for martingale "convolution"
Assume that $X\_t$ have independent and centered increments, but not necessarily identically distributed. 
Let $D\_i=X\_i-X\_{i-1}$ for $i\geqslant a+1$ and $D\_a=X\_a$. Let 
$
S\_t=X\_t\left(X\_T-X\_t\right).
$
Then $$S\_t=\left(\sum\_{i=a}^tD\_i\right)\left(\sum\_{j=t+1}^TD\_j\right)$$
and 
\begin{align}
S\_{t+1}-S\_t= \left(\sum\_{i=a}^{t+1}D\_i\right)\left(\sum\_{j=t+2}^TD\_j\right)-\left(\sum\_{i=a}^tD\_i\right)\left(\sum\_{j=t+1}^TD\_j\right)\\
= \left(\sum\_{i=a}^{t}D\_i\right)\left(\sum\_{j=t+2}^TD\_j\right)+ D\_{t+1}\left(\sum\_{j=t+2}^TD\_j\right)-\left(\sum\_{i=a}^{t}D\_i\right)\left(\sum\_{j=t+2}^TD\_j\right)-\left(\sum\_{i=a}^{t}D\_i\right) D\_{t+1} \\
= D\_{t+1}\left(\sum\_{j=t+2}^TD\_j\right) -\left(\sum\_{i=a}^{t}D\_i\right) D\_{t+1}
\end{align} 
hence letting 
$$
D'\_t= -\left(\sum\_{i=a}^{t}D\_i\right) D\_{t+1};\quad D''\_t:= D\_{t+1}\left(\sum\_{j=t+2}^TD\_j\right),
$$
the following equality holds 
$$
S\_u= S\_a+\sum\_{t=a}^{u-1}D'\_t+\sum\_{t=a}^{u-1}D''\_t.
$$
Then 
$$
\mathbb E\left[\max\_{a\leqslant t\leqslant b}S\_t^2\right]
\leqslant 4\mathbb E\left[D\_a^2\left(\sum\_{j=a+1}^TD\_j\right)^2\right]
+4\mathbb E\left[\max\_{a\leqslant u\leqslant b}\left(\sum\_{t=a}^{u-1}D'\_t\right)^2\right]
+4\mathbb E\left[\max\_{a\leqslant u\leqslant b}\left(\sum\_{t=a}^{u-1}D''\_t\right)^2\right].
$$
Using independence, the first term can be simplified. For the second one, we have maximum of a martingale, and for the third of a reversed martingale. At the end, we get 
$$
\mathbb{E}\left[\max\_{a \leqslant t \leqslant b} \left(X\_t(X\_T-X\_t)\right)^2 \right]\leq C\mathbb{E} X\_a^2 \mathbb{E}(X\_T-X\_b)^2.
$$


