please I have confusion if someone can help me: We define the Grammian operator as follows: \begin{equation}Q_{T}:=L_{T} L_{T}^{*}=\int_{0}^{T} S(T-s) B B^{*} S^{*}(T-s) d s, \quad T>0\end{equation} when B and S are self-adjoin the expression of $Q_{T}$ be come: \begin{equation}Q_{T}:=L_{T} L_{T}^{*}=\int_{0}^{T} |S(T-s) B|^{2} d s, \quad T>0\end{equation} or \begin{equation}Q_{T}:=L_{T} L_{T}^{*}=\int_{0}^{T} \big(S(T-s) B\big) \omicron \big(B\,\, S (T-s) \big)d s, \quad T>0\end{equation} I really don't know !!
also between $ S (T-s) $ and $ B $ is this product or composition?
Best regards
If $B$ and $S$ are self-adjoint then $B^* = B$ and $S^*=S$. If we substitute this into, \begin{align} Q_{T}:=L_{T} L_{T}^{*}=\int_{0}^{T} S(T-s) B B^{*} S^{*}(T-s) d s, \quad T>0 \end{align} you get \begin{align} Q_{T}:=L_{T} L_{T}^{*}=\int_{0}^{T} S(T-s) B B S(T-s) d s, \quad T>0 \end{align}
If this is the Controllability Gramian then $S(T-s) = e^{A(T-s)}$ and is a matrix. Therefore, $B S(T-s)$ is the product of two matrices.
Just as a side remark, if you have two linear operators $p$ and $q$ then the composition of these two operators is $p\circ q$. So for an input, $x$, $(p\circ q)(x) = p(q(x))$. However, if the linear operators have a matrix representation, $p(x) = Px$ and $q(x) = Qx$ where $P$ and $Q$ are matrices then the composition of linear operators is the same is the product of two matrices, $(p\circ q)(x) = (PQ)(x)$