Is this parametrized semidefinite program convex? I am considering an optimization problem of the form:
\begin{equation}
\begin{split}
f(s)   &=  \min_{X} \mathrm{tr}(C(s)X) \\
&\;\;\;\;\;\;\;\;\;\;\; X \ge 0, \\
&\;\;\;\;\;\;\;\;\;\;\; \mathrm{tr}(A_iX) = a_i, \;\; 1 \le i \le M,
\end{split}
\end{equation}
where the minimization is over $n\times n$ Hermitian matrices $X$. Further, $A_i$ for $1 \le i \le M$ denote some $n\times n$ Hermitian matrices which together with $a_i \in \mathbb{R}$ determine linear constraints on $X$. Finally, the matrix-valued function $C(s)$ is of the block form:
\begin{equation}
C(s) = \left( \begin{array}{cc} C_{1}(s) & 0 \\ 0 & 0\end{array} \right),
\end{equation}
where the upper left block $C_1(s)$ is of size $(n_1 + 1) \times (n_1 + 1)$ for some $n_1 < n$, and is given by:
\begin{equation}
C_1(s) = \left( \begin{array}{ccccc}  I_{n_1\times n_1} & -ic \mathbb{I}_{n_1\times n_1} & \cdot & \cdot \\ i c \mathbb{I}_{n_1\times n_1} & \cdot & -i \frac{s}{2} \mathbb{I}_{n_1\times n_1} & \cdot \\ \cdot & i \frac{s}{2} \mathbb{I}_{n_1\times n_1} & \cdot & \cdot \\ \cdot & \cdot & \cdot & s^2\end{array} \right).
\end{equation}
Here, $c \in \mathbb{R}$, $I_{n_1\times n_1}$ is the $n_1 \times n_1$ matrix of ones and $\mathbb{I}_{n_1\times n_1}$ denotes the $n_1 \times n_1$ identity matrix (whereas all entries indicated by $\cdot$ vanish).
Can it be shown that $f(s)$ is convex?
If not, which further requirements has the optimization to fulfill in order to guarantee convexity of $f(s)$?
 A: Yes, this is convex because the objective function and all constraints are convex.
The objective function is affine (linear), which is convex. The semidefinjite constraint on X is convex. The trace equality constraint on X is affine (linear), and therefore is convex.
A: Consider the following argument for a slightly changed problem (with $s^2$ in $C(s)$ replaced by $-s^2$) . Not sure if this would be of any help, but writing it anyway. Note that due to concavity of $-s^2$ in C(s) and the rest of the terms being either constant or linear in $s$ (in $C(s)$), we have:
$$
C(\lambda s_1 + (1-\lambda)s_2) \succeq  \lambda C(s_1) + (1-\lambda)C(s_2).
$$
Therefore,
$$
f(\lambda s_1 + (1-\lambda)s_2) \geq \min_{X\in \Gamma} \left\{ \lambda \mbox{Tr}(C(s_1)X) + (1-\lambda)\mbox{Tr}(C(s_2)X) \right\} \geq  \lambda \min_{Y\in \Gamma} \left\{  \mbox{Tr}(C(s_1)Y)\right\} + (1-\lambda) \min_{Z\in \Gamma} \left\{  \mbox{Tr}(C(s_2)Z)\right\} = \lambda f(s_1) + (1-\lambda)f(s_2). 
$$
And hence $f(.)$ is concave in $s$.
