Duality in quadratically constrained quadratic program I have been given the primal quadratic program with a single quadratic constraint as given below:
$$ \text{min} ~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{1}{2}x^{T}Qx $$
\begin{align*}
\text{subject to}~~~~~~~~~\frac{1}{2}x^{T}x &\leq \frac{1}{2}\\
x &\in \mathbb{R}^n
\end{align*}
where Q is an indefinite, symmetric, rational matrix of size n. The task is to show that the optimal values of the primal problem as given and its dual agree. That is, there is no duality gap.
I have (perhaps incorrectly) determined the Lagrangian of this program to be
$$L(x,u) = \frac{1}{2}x^{T}Qx+\frac{1}{2}u(x^{T}x-1) = \frac{1}{2}x^{T}(Q+uI)x-\frac{1}{2}u. $$
At this point, after chasing through some circular algebra attempting to solve the dual problem to maximize (over $u$) the infimum (over $x \in \mathbb{R}^n$) of $L(x,u)$ and doing some KKT analysis on the primal problem as given, I can't seem to arrive at the desired result. Further, I understand there is a theorem which says that if there exists a saddle point of the Lagrangian function, then there is no duality gap (Bazaraa, Sherali, and Shetty), yet I can't seem to reconcile any sort of "mathematically tight" guarantee that there exists a saddle point for $L$ (although I have a hunch that there ought to be since $Q$ is indefinite, that is, $Q$ is not PD, not PSD, not ND, not NSD).
If anyone happens to see something particularly insightful in all of this, I would really appreciate the input.
 A: Let $\lambda_{\min}$ be the minimum eigenvalue of $Q$.
The dual function is
\begin{align}
g(u) &= \inf_x L(x,u) \\
&= \begin{cases}
-\infty \quad \text{if } u < -\lambda_{\min} , \\
-\frac12 u \quad \text{otherwise}.
\end{cases}
\end{align}
The dual problem is
\begin{align}
\operatorname*{maximize}_{u} & \quad -\frac12 u \\
\text{subject to} & \quad u \geq -\lambda_{\min}.
\end{align}
The optimal value for the dual problem is
\begin{equation*}
d^\star = \frac12 \lambda_{\min}.
\end{equation*}
And this is equal to the primal optimal value, according to the standard variational characterization of the eigenvalues of $Q$.
Note: To derive the dual function, I used the following facts.


*

*The minimum eigenvalue of $Q + u I$ is $\lambda_\min + u$.

*Let $M \in \mathbb R^{n \times n}$ be symmetric.  The quadratic function $h(x) = \frac12 x^T M x$ is unbounded below if the minimum eigenvalue of $M$ is negative.  Otherwise, the minimum value of $M$ is $0$.

A: Hint: 
Your problem is partially addressed in this question
Note that this is one the rare non-convex problems that has a closed form solution and has zero duality gap.
