Hi I was given the following definition of coercivity:
Let $V$ be a Banach space.
The first definition: $A:V \rightarrow V^{*}$ is coercive iff $\exists \zeta: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}: \lim\limits_{s \rightarrow + \infty}\zeta(s) = + \infty$ and $\langle A(u),u \rangle \geq \zeta(\Vert u \Vert)\Vert u \Vert$.
Alternatively we could state, $A:V \rightarrow V^{*}$ is coercive means $$\lim\limits_{\Vert u \Vert \rightarrow \infty}\frac{\langle A(u),u \rangle}{\Vert u \Vert} \ = +\infty$$
Can anyone see how to show that the alternative definition implies the first definition? I can see how the first implies the alternative.
Thanks.
This is actually not difficult to show: Assume that $A$ is coercive in the second sense, so it holds $\langle A(u_k), u_k\rangle/\Vert u_k \Vert \to \infty$ for all unbounded sequences $(u_k)$. Define $\xi: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ by $$ \xi(r) := \inf_{\Vert u\Vert = r} \frac{\langle A(u),u\rangle}{\Vert u\Vert} $$ Then just by definition of $\xi$ we have for all $u\in V$ $$ \langle A(u), u\rangle\geq \xi(\Vert{u}\Vert)\Vert{u}\Vert. $$ For the first definition of coercivity to be fulfilled we still have to show $\xi(r) \to \infty$ for $r \to \infty$. For this assume that $\xi$ is bounded by some constant $C$. Then it implies, that for all $k \in \mathbb{N}$ there exists a $u_k \in V$ with $\Vert u \Vert = k$ and $$ \frac{\langle A(u_k),u_k\rangle}{\Vert u_k\Vert} \leq \inf_{\Vert u\Vert = k} \frac{\langle A(u),u\rangle}{\Vert u\Vert} + 1 = \xi(k) + 1 \leq C+1. $$ But then we'd have an unbounded sequence $(u_k)_{k\in\mathbb{N}}$ for which $\langle A(u_k), u_k\rangle/\Vert u_k \Vert \not\to \infty$ for $k\to\infty$. This is a contradiction to the coercivity of $A$ in the second sense. Thus $\xi$ is unbounded and the first definition holds as well.
I hope this proof works for you. I don't know if $\xi$ has to fulfill further properties. One should maybe be able to show continuity.
For coercivity there is often used the definition where $\xi$ is just a multiple of the identity. Then the alternative definition does not imply the first definition, at least not for general operators. Simply think of a nonlinear operator that vanishes for small $\Vert u\Vert$ but is just the identity for bigger $\Vert u \Vert$. Then the second condition is given but not the first one.
As far as I can tell, the first definition with $\xi$ constant is often used when one thinks about quadratic functions that could be induced by a bilinear function, so $A$ is usually linear then. The "alternative" is for example important for the theory on monotone operators where you could also consider highly nonlinear operators $A$. So it's kind of an attempt to generalize coercivity there.
EDIT: One might argue if that's really a generalization. One just tries to reduce the definition of coercivity to what's seen to be important to the further theory.