$$f(x) = y^\top x - \frac{1}{2} x^\top Q x$$
where $Q$ is both symmetric and positive definite is shown here to be bounded above. However, is the following proof correct for $f(x)$ being unbounded below?
Proof that $f(x)$ is unbounded below:
Since $Q$ is symmetric, the spectral theorem tells us that, $x^\top Q x \geq \lambda_{\min}(Q) \ x^\top x$, where $\lambda_{\min}(Q) > 0 $ is the smallest eigenvalue. However $x^\top Q x = \lambda(Q) \ x^\top x$ when $x$ is an eigenvector, since $Q x = \lambda(Q) x$.
In this case, there is a linear term in the quadratic equation; therefore, let $x = \alpha v$, where $v$ is the eigenvector associated with $\lambda_{\min}(Q)$, such that $\|v\|_2 = 1$ and $\alpha > 0$. Substituting $\alpha v$ in the original quadratic equation yields
$f(\alpha v) = \alpha y^\top v - \frac{\alpha^2 \lambda_{\min}(Q)}{2} v^\top v$, and since $\| v\|_2 = 1$, therefore,
$f(\alpha v) = \alpha y^\top v - \frac{\alpha^2 \lambda_{\min}(Q)}{2}$
now as $\alpha \rightarrow \infty$, $f(\alpha v) \rightarrow -\infty$
Thank you.