Proving quadratic function is bounded (vector input) I am reading Boyd's Convex Optimization textbook and I am looking at example 3.22, line 2.
It says $y^T x - \frac12 x^T Q x$ is bounded from above for all possible values of $y$. Also, it is important to note $Q$ is a symmetric positive definite. So for all values of $x$, $x^T Q x$ is a positive number. Why is $y^T x - \frac12 x^T Q x$ bounded above? I am not sure how to compare the quantity $y^T x$ with the quantity $\frac12 x^T Q x$. For $y=0$, I can see why.. then there are two cases, $y>0$ and $y<0$.
Thanks

Hint:  Presumably $x$ is a vector and $y$ is a vector of the same length as $x$.  As $Q$ is symmetric positive definite, it has a complete set of eigenvectors, so you can express $y$ as a linear sum of them.  Then note the order of quantifiers-$y$ is chosen, then the function is bounded above.
