$g$ is coercive for $g(x)=x^TAx+b^Tx+c$

Question

Suppose $A$ is a symmetric positive definite matrix $A\in \Bbb{R}^{n\times n}$, $b \in \Bbb{R}^n$, and c is a real number. Let $$g(x)=x^TAx + b^Tx + c$$ Show that $g$ is coercive.

Because $A$ is positive definite, then $x^TAx\gt0\ \ , \forall x \ne0$

I tried expanding out $g(x)$ and got $$g(x)=(a_{11}x_1^2+a_{22}x_2^2+...+a_{nn}x_n^2+2x_1(a_{12}x_2+a_{13}x_3+...+a_{1n}x_n)+2x_2(a_{23}x_3+...+a_{2n}x_n)+...+2a_{n-1,n}x_{n-1}x_n)+(b_1x_1+b_2x_2+...+b_nx_n)+c$$

Now I'm stuck as to how to show that $$\lim_{|x|\to\infty} g(x)=\infty$$

Answer 1

As $A$ is positive definite, let $C>0$ so that $x^T Ax\ge C|x|^2$ for all $x$. Then

$$|g(x)| = |x^TA x + bx + c| \ge C|x|^2 - |b| \cdot |x| - |c| \to \infty$$

as $|x| \to \infty$.

Answer 2

Recall that $\,g:\mathbb R^n \to \mathbb R\,$ is coercive if for any $\, x\in \,\mathbb R^n\,$ $%there exists a constant \,c\in \mathbb R\, such that$ $ %\big\langle \,g\left(x\right),\,x\,\big\rangle \ge c\, \left\| \, x \, \right\|^2 %= c \,\big\langle \,x,\,x\,\big\rangle \left\| \, x \, \right\| \to \infty \implies \left\lvert \,g\left(x\right)\,\right\rvert \to +\infty.$

Now, in your case $\,g\left(x\right)=x^TAx + b^Tx + c\,$ so that \begin{align} \big\lvert \,g\left(x\right)\big\rvert & = \left\lvert \,x^TAx + b^Tx + c\,\right\rvert \\ & \geq \left\lvert\,x^TAx\,\right\rvert-\left\lvert\,b^Tx\,\right\rvert - \left\lvert \,c\,\right\rvert \\ & = %\left\lvert\,x^TAx\,\right\rvert-\left\|\,b^Tx\,\right\| - \left\lvert\,c\,\right\rvert \big\langle \,x,\, Ax\,\big\rangle - \big\langle \,b,\, x\,\big\rangle - \left\lvert \,c\,\right\rvert \\ & \ge a\, \left\| \,x\,\right\|^2 - \left\| \,b \,\right\|_\infty \left\| \,x\,\right\| - \left\lvert \,c\,\right\rvert , \end{align} where $\displaystyle\, a:= \min_{i,j=1,\dots, n} \left\lvert\, A_{ij} \,\right\rvert $ is the minimum absolute value of entries of matrix $\,A$.

Therefore we conclude that $\, g\,$ is coercive: $$ \lim_{\left\|x\right\| \to \infty} \big\lvert \,g\left(x\right)\big\rvert =\lim_{\left\|x\right\| \to \infty} \left( a \left\| \,x\,\right\|^2 - \left\| \,b \,\right\|_\infty \left\| \,x\,\right\| - \left\lvert \,c\,\right\rvert \right) = + \infty, $$

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EDIT: Thanks to comment by John MA I correct a mistake and elaborate on the last inequality. Let

$$ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{bmatrix}, \quad x = \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n}\end{bmatrix}, $$

so that $$ Ax = \begin{bmatrix} \displaystyle\sum_{i=1}^n a_{1i} \,x_{i} \\ \displaystyle\sum_{i=1}^n a_{2i} \,x_{i} \\ \vdots \\ \displaystyle\sum_{i=1}^n a_{ni} \,x_{i} \end{bmatrix}, $$

Denote $\displaystyle\, a:= \min_{a_{ij}\in A} a_{ij}\ge 0\,$ (since $\,A\,$ is symmetric positive-definite), then

\begin{align} \big\langle \,x,\, Ax\,\big\rangle &= x_1 \sum_{j=1}^n a_{1j} \,x_{j} + x_2 \sum_{j=1}^n a_{2j} \,x_{j} + \cdots + x_n \sum_{j=1}^n a_{nj} \,x_{j} \\ &= x_1^2 \sum_{i=1}^n a_{i1} + x_2^2 \sum_{i=1}^n a_{i2} + \cdots x_n^2 \sum_{i=1}^n a_{i1} + \sum_{\substack{i,j=1\\i\neq j}}^n a_{ij} \,x_i \, x_j \\ & = \sum_{j=1}^n \, x_j^2 \, \sum_{i=1}^n a_{ij} + \sum_{\substack{i,j=1\\i\neq j}}^n a_{ij} \,x_i \, x_j \\ &\ge \sum_{j=1}^n\,x_j^2 \,\sum_{i=1}^n a_{ij} =a\cdot n\sum_{j=1}^n x_j^2 \\ & = a\cdot n\cdot \left\|\,x\, \right\|^2 \end{align} Therefore we conclude that \begin{align} \big\lvert \,g\left(x\right)\big\rvert & \ge \big\langle \,x,\, Ax\,\big\rangle - \big\langle \,b,\, x\,\big\rangle - \left\lvert \,c\,\right\rvert \\ & \ge a\, \left\| \,x\,\right\|^2 - \left\| \,b \,\right\|_\infty \left\| \,x\,\right\| - \left\lvert \,c\,\right\rvert, \end{align} and therefore $\,g\left(x\right)\,$ is coercive.