How to prove that the norm of an affine function is coercive?

Question

Let $h : \mathbb{R}^n \to \mathbb{R}$ be defined by $h(x) := \|Ax+b\|$. Prove that $h$ is a coercive function.

Maybe I can do it by some inequality, but I can't find wich.

Answer 1

You cannot find such a proof, since this statement is incorrect. $f$ is coercive if for every sequence $\{x_n\}_{n=1}^\infty$ with $\lim_{n \to \infty} \|x_n\| = \infty$ you have $\lim_{n \to \infty} f(x_n) = \infty$ (also written informally as $\lim_{\|x\|\to \infty} f(x) = \infty$).

Suppose that $A$ does not have full rank, meaning that $\ker(A)$ is a subspace. Take any sequence $\{ x_n \}_{n=1}^\infty \subseteq \ker(A)$ with $\lim_{n\to \infty} \|x_n\|=\infty$. For such a sequence you have $$ f(x_n) = \|A x_n+b\|= \|0+b\| =\|b\|, $$ which clearly means that $\lim_{n \to \infty} f(x_n)$ is $\|b\|$ and not $\infty$.