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$\def\R{\mathbb R}f\colon\R^n\to\R$ is a coercive function. Given that $A$ is a positive definite matrix in $\R^{n\times n}$ and $b$ is a vector in $\R^n$,prove the function $g\colon \R^n\to\R$ defined by $g(x)=f(Ax+b)$ is coercive ($x\in \R^n$ in this question).

You may use the fact that $x^\top Ax≥λ_\min (A) x^\top x$ for any nonzero $x\in\R^n$ where $λ_\min (A)$ is the smallest eigenvalue of $A$. ($x^\top$ means transpose of $x$).

Even though there is a hint, I still could not get it through. I was thinking that if I can prove that \|Ax+b\|^2 > P(λ_\min (A)) \cdot \|x\|^2 = P(λ_\min (A)) \cdot x^\top x > P(λ_\min (A)) \cdot r_1^2$$ where $P$ is a positive polynomial, $r_1$ is found by the fact that $f$ is coercive: for any $M>0$, there exist $r_1>0$ such that for any $\|x\|\ge r_1$, $f(x)>M$. If I can construct such $P$, then let $P(λ_\min (A))r_1^2 = r^2$, then for any $M>0$, there exists $r>0$, such that for any $\|x\|>r$, $\|Ax+b\|>r_1$, thus $g(x)=f(Ax+b)>M$. Thus $g$ is coercive. But I just could not find a working $P$.

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dYou are starting out the right way and you are on a good path, but your desired estimate cannot be correct. To see this, let $c = A^{-1}b$ and set $x = -c$. Then the left hand side of your estimate is $0$.

Modify your estimate along the following lines: $ \|Ax + b\| = \|A(x + c)\| \ge \lambda_{min} \|x + c\| $ where as before $c = A^{-1}b$. The right hand side can be estimate from below using the triangle inequality which should give you everything you need.

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    Second comment: Assume $x^tAx \ge \lambda \|x\|^2$ for all $x$, and set $Ax = z$. Then $\lambda\|x\|^2 \le x^Tz \le \|x\| \|z\|$ by Cauchy-Schwarz and therefore $\|\lambda \|x \| \le \|z \| = \|A x\|$.2012-10-29