Equivalent definition of coercivity of a bilinear form on a normed vector space

Question

I am searching for the reason why the coercivity of a bilinear form on a normed vector space $a\colon V\times V\to\mathbb{R}$ is defined like this: "$\exists\, \alpha>0\,$ s. t. $a(\xi,\xi)\geq\alpha\|\xi\|^2$ for every $\xi\in V$". Why the square of the norm? It is easy to prove that the definition is equivalent to require that "$\exists\,\alpha>0$ s. t. $a(\xi,\xi)\geq\alpha$ for every $\xi\in S_V:=\{x\in V\colon\|x\|=1\}$", so the coercivity measures how far from the origin the unit sphere is mapped by the bilinear form, but is it all here? Is there nothing else? I tried to prove that the definition is equivalent to "$\lim\limits_{\|\xi\|\to+\infty}a(\xi,\xi)=+\infty$", in order to have an analogy with the coercitivity of a scalar function of real variables, such as $x\mapsto x^2$. Do you think is it true? Obviously, $(\Rightarrow)$ holds, but is it true that if $\lim\limits_{\|\xi\|\to+\infty}a(\xi,\xi)=+\infty$ then $\exists\, \alpha>0\,$ s. t. $a(\xi,\xi)\geq\alpha\|\xi\|^2$ for every $\xi\in V$?

Answer 1

Assume that for all $\alpha$ there is $x_\alpha$ such that $$ a(x_\alpha,x_\alpha) < \|x_\alpha\|^2. $$ Hence for all $n$ there $x_n$ with $$ a(x_n,x_n) < n^{-3} \|x_n\|^2. $$ This implies $x_n\ne0$, and we can scale the sequence to satisfy $\|x_n\|=1$. Then $a( nx_n,nx_n) < n^{-1}$, and the limit $a(x,x)$ for $\|x\|\to\infty$ cannot be $+\infty$.