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.
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Lecter
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What is the definition of "coercive" being used here? Do you mean "norm-coercive" function? – learner Mar 10 '19 at 11:22
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@learner I understand it as: $f$ is coercive iff $f(x)->\infty$ as $||x||->\infty$ – Lecter Mar 10 '19 at 11:28
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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 $\infty$.
Alex Shtoff
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1Then the statement is correct if $\lambda_{\min}(A^T A) > 0$. Considering the Euclidean norm, e.g. $|u|^2 = u^T u$, you have $f(x) = x^T (A^T A) x + 2 b^T A x + |b|^2 \geq \lambda_{\min}(A^T A) |x|^2 + 2b^T A x + ||b||^2$, and by Cauchy-Schwarts this is $\geq \lambda_{\min}(A^T A) ||x||^2 - 2 |x| |A^T b| + |b|^2$. Now, coerciveness is quite easy to prove. – Alex Shtoff Mar 10 '19 at 14:07