Prove that $g: \mathbb{R}^n \to \mathbb{R}$ is bounded below

Question

Suppose that A is a regular symmetric real matrix of order n. Prove that function $g: \mathbb{R}^n \to \mathbb{R}$

$g(x)=\frac{1}{2}x^TAx-b^Tx$

is bounded below $\forall b\in\mathbb{R}^n$ iff A is a positive definitive matrix.

Answer 1

Suppose $A$ is positive definite.

Note that $g(x)= \frac 12 \sum_{i=1}^n x_i^2\lambda_i -b^Tx \geq \frac 12 \sum_{i=1}^n x_i^2\lambda_i-\|b\|\|x\|\geq \frac{\min \lambda_i}2\|x\|^2-\|b\|\|x\|$

where $\lambda_i>0$ are the eigenvalues of $A$. Hence $\lim_{\|x\|\to \infty} g(x) = \infty$. Thus $g$ is bounded below by $1$ for $\|x\|$ larger than some $R$. Since $g$ is continuous it's also bounded below over the closed ball with radius $R$. Combine these two lower bounds and you're done.

Something stronger can be proven: if $g$ is coercive and continuous over a closed set, it reaches a global minimum.

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If $A$ is not positive definite, either $A=0$ and a suitable $b$ is easily found or WLOG $\lambda_1< 0$. Consider $x_k = (k,0,\ldots,0$). Then $g(x_k)=\frac{\lambda_1}{2}k^2-b_1k\to -\infty$ regardless of $b$, thus $g$ is not bounded below.