I need to prove that there is an upper bound on $x^T p$ (both $N$ dimensional vectors) for any value of $p$, where $x$ is the solution to the quadratic programming problem: \begin{equation} \arg \min_{x} \frac{1}{2} {x}^{T} Q x + {x}^T \left( p - d \right) \end{equation} such that $-1 \leq x_i \leq 2$ for all $i$, $Q$ is a positive definite $N \times N$ matrix, and $d$ and $Q$ are fixed.
Without the constraint in the quadratic programming problem, then \begin{equation} x = Q^{-1} (d - p) \end{equation} and so $x^T p = (d - p)^T Q^{-1} p$. By choosing $p$ to maximize this quantity, we find that the maximum value is $\frac{d^T Q^{-1} d}{4}$. Thus there is an upper bound for $x^T p$. Perhaps I am missing something obvious, but if we add the constraint, is $x^T p$ still bounded above, and if so, how do I prove it?
I figured it out. Since $x = 0$ (the column vector of zeros) is in the feasible set of solutions, it must be the case that the solution $x$ for this minimization problem for a given $p$ must be \begin{equation} \frac{1}{2} x^T Q x + x^T (p - d) \leq 0 \implies x^T p \leq x^T d - \frac{1}{2} x^T Q x \end{equation} Since $Q$ is positive definite, \begin{equation} x^T p \leq x^T d \end{equation} Since $d$ is fixed and the feasible set of solutions is bounded, there is an upper bound on $x^T d$, which I'll write as $b$. Thus \begin{equation} x^T p \leq x^T d \leq b \end{equation}