I am interested in the value function of a quadratic program of the form
$$
v(y)=\min_x \frac{1}{2} x^\top Q(y) x,
$$
subject to a linear equality constraint
$$
E(y)x=d(y),
$$
and a linear inequality constraint
$$
A x \preceq b,
$$
where $\preceq$ denotes component-wise inequalities.

Notice that $Q$, $E$ and $d$ all depend on a parameter $y\in{\Bbb R}^m_{\geq 0}$. $Q(y)$ is positive definite for all $y$. Importantly, $A$ and $b$ do not depend on $y$. 

For the particular problem I am interested in, I know $Q$, $E$, $d$, $A$ and $b$ but they are a bit complicated and I'm hoping that their specific structure is not important here.

**I would like to show that $v$ is convex.** Given my specific problem, I know that $v$ is convex if we remove the inequality constraint $A x \preceq b$. In that case the problem is simple and I can solve for $v$.

My question is: if $v$ is convex without the inequality constraint, does $v$ remain convex when we add the inequality constraint? Recall that this inequality constraint does not depend on $y$.

Notes:
1. If that helps, in my specific problem $Q$ and $E$ are homogenous in the sense that $Q(\lambda y)=\lambda Q(y)$ and $E(\lambda y)=\lambda E(y)$ for any $\lambda\in{\Bbb R}$, and $d(y)=y-c$ where $c\in{\Bbb R}^m_{\geq 0}$. $E$ is also linear in $y$.
2. I tried to compute $v$ using the dual approach but this seems intractable.
3. I have looked at a few special cases and cannot find a counterexample.

**Value function without inequality constraints**

Without the inequality constraint, the [solution to this problem][1] is given by

$$
\left[\begin{array}{cc}
Q & E'\\
E & 0
\end{array}\right]\left[\begin{array}{c}
x\\
\lambda
\end{array}\right]=\left[\begin{array}{c}
0\\
d
\end{array}\right]
$$
which can be [inverted][2] as
$$
\left[\begin{array}{cc}
Q & E'\\
E & 0
\end{array}\right]^{-1}=\left[\begin{array}{cc}
Q^{-1}-Q^{-1}E'\left(EQ^{-1}E'\right)^{-1}EQ^{-1} & Q^{-1}E'\left(EQ^{-1}E'\right)^{-1}\\
\left(EQ^{-1}E'\right)^{-1}EQ^{-1} & -\left(EQ^{-1}E'\right)^{-1}
\end{array}\right]
$$
Since $Q$ is positive definite it is invertible. Suppose that $EQ^{-1}E'$ is also invertible. Then
$$
x=Q^{-1}E'\left(EQ^{-1}E'\right)^{-1}d
$$
and the objective function at the optimum is
$$
v(y)=d'\left(\left(EQ^{-1}E'\right)^{-1}\right)d
$$

**My particular problem**

In my particular problem $x\in{\Bbb R}_{\geq 0}^{n^2}$ and $y\in{\Bbb R}_{\geq 0}^{n}$. The function $E$ and $d$ are
$$
E(y)=y'\otimes I_n,
$$
and,
$$
d(y)=y-c,
$$
where $c$ is a $n\times 1$ column vector such that $0<c_i<1$. The matrix $Q(y)$ is given by
$$
Q=\left[\begin{array}{ccc}
y_{1}F_{1} &  & 0\\
 & \ddots\\
0 &  & y_{n}F_{n}
\end{array}\right]
$$
where $F_i$ is an $n\times n$ positive definite matrix.

Doing the matrix algebra, and using the expression for $v$ above, we find 
$$
v(y)=\left(y-c\right)'\left(\sum_{i}y_{i}F_{i}^{-1}\right)^{-1}\left(y-c\right)
$$
A proof of convexity for this function can be found [here][3].

**EDIT:**

The inequality constraints I'm interested in are $0\leq x$ and
$$
I_n\otimes 1_n' \leq \bar{x},
$$
where $\bar{x}$ is a $n\times 1$ column vector with elements $0<\bar{x_i}<1$ and $1_n$ is the $n\times 1$ column vector of ones. 

If we think of $x$ as a $n^2\times 1$ column vector made of smaller $n\times 1$ vectors $z_j$ such that $x'=[z_1',\dots,z_n']$ then the last inequality constraint becomes
$$
\sum_j z_{ij}\leq\bar{x}_i 
$$
for all $i$.

  [1]: https://en.wikipedia.org/wiki/Quadratic_programming
  [2]: https://en.wikipedia.org/wiki/Block_matrix
  [3]: https://math.stackexchange.com/questions/4739698/convexity-of-bf-x-mapsto-bf-x-bf-a-top-left-sum-i-x-i-bf-a