question on translation of operator proof Has anyone studied the book 'Nonlinear Partial differential equations with applications' by Tomas Roubicek? I am interested in discussing a point of interest in this book. 
Specifically, on page 52, in the proof of Theorem 2.36 (Leray-Lions) it states that $A_{0}$ inherits coercivity from $A$. How does one prove this?
Thanks for any assistance. 
 A: The setup
For completeness I want to state the parts from the book that are needed:
We are given a (in general nonlinear) operator
$$A: W^{1,p}(\Omega) \rightarrow W^{1,p}(\Omega)^*,$$
which is coercive (see Lemma 2.35 in the given literature). At that, coercivity means the existence of a mapping $\xi: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $\xi$ is unbounded (i.$\,$e. $\lim_{r\to+\infty} \xi(r) = +\infty$) and
$$
 \forall u \in W^{1,p}(\Omega): A(u) \geq \xi(\Vert u\Vert) \Vert u \Vert.
$$
Furthermore, we have the shifted operator
$$
 A_0: \phantom{u} W^{1,p}(\Omega)  \rightarrow W^{1,p}(\Omega)^* \\
 \phantom{A_0: W^{1,p}(\Omega)} u \mapsto A(u+w)
$$
where $w \in W^{1,p}(\Omega)$ is a constant, which is given in the book. (Actually $A_0$ is defined on a linear subspace $V\subseteq W^{1,p}(\Omega)$ but since $A$ is defined on the whole space we can extend $A_0$ to the whole space as well.)

Proof of coercivity
We will show coercivity of $A_0$ on $W^{1,p}(\Omega)$, which then directly implies coercivity of $A_0$ on the subspace $V \subseteq W^{1,p}(\Omega)$ by simple restriction of the operator.
First of all, we state there exist $R \in \mathbb{R}^*$ and $c \in \mathbb{R}^+$ such that for all $u\in W^{1,p}(\Omega)$ with $\Vert u\Vert \geq R$ holds
$$
  \Vert u \Vert \leq c\Vert u+w\Vert. \ \ \ (1)
$$
(Is this plausible enough without a proof? It's rather general and isn't a direct part of the main proof.)
Now define  $\eta: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ by
$$
\begin{align}
&\eta(r) := \begin{cases}  & c\inf\limits_{v \in W^{1,p}(\Omega), \|v\| = r}\xi(\| v+w\|)  ~~~~~~~~~~~~~~~~~~~~\text{ for } r \geq R,\\ &0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{for } r < R  
\end{cases}
\end{align}
$$
(Thanks to John Doe for improving the formatting!) Note that $\eta$ is well-defined because for all $r\in \mathbb{R}^+$ we have $\inf_{v \in W^{1,p}(\Omega), \Vert v\Vert = r} \xi(\Vert v + w\Vert) \geq 0$ as of the definition of $\xi$. Furthermore, we have $\eta(r) \to +\infty$ for $r \to \infty$. Else there would exist a sequence $(u_k)_{k\in \mathbb{N}}$ with $\Vert u_k\Vert = k$ for all $k\in \mathbb{N}$ and $C \geq \limsup_{k\to\infty} \eta(\Vert u_k\Vert) = \limsup_{k\to\infty} \xi(\Vert u_k+ w\Vert)$. In particular we would have $\Vert u_k + w\Vert \to \infty$ for $k\to\infty$ but $\xi(\Vert u_k+ w\Vert) \leq C$ for $k\to \infty$ in contradiction to $\xi(r) \to \infty$ for $r\to\infty$. Consequently this contradiction yields $\eta(r) \to \infty$ for $r\to\infty$ as well.
Just by the definition of $A_0$, then the definition of coercivity applied to the operator $A$, then usage of (1) and finally using the definition of $\eta$ we have for all $u \in W^{1,p}(\Omega)$ with $\Vert u \Vert \geq R$
$$
 A_0(u) = A(u+w) \geq \xi(\Vert u+w\Vert) \Vert u+w\Vert \geq \frac{1}{c}\xi(\Vert u+w\Vert)\Vert u\Vert \geq \eta(\Vert u\Vert) \Vert u\Vert.
$$
and for all $u \in W^{1,p}(\Omega)$ with $\Vert u \Vert < R$ we have (again by definition of $A_0$, coercivity itself and $\eta$)
$$
 A_0(u) = A(u+w) \geq \xi(\Vert u+w\Vert) \Vert u+w\Vert \geq 0 = \eta(\Vert u\Vert) \Vert u\Vert.
$$
Thus $A_0$ is coercive as $\eta$ is a eligible coercivity-function as shown above.

Remarks on the proof
Because of the comments I have a few remarks to add.


*

*The well-definedness part above for $\eta$ is not about showing that $\inf_{v \in W^{1,p}(\Omega), \Vert v\Vert = r} \xi(\Vert v + w\Vert)$ exists (for all $r \geq R$). The existence of the infimum is given by the boundedness from below. But we furthermore ned $\inf_{v \in W^{1,p}(\Omega), \Vert v\Vert = r} \xi(\Vert v + w\Vert) \geq 0$ so that $\eta\: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ is well-defined (because if the infimum was less than zero we wouldn't have a mapping into $\mathbb{R}^+$ anymore). This property is given because $\xi$ is bounded from below by zero (as its range is defined on $\mathbb{R}^+$ as well; this boundary also gives us the mentioned existence of the infimum).

*To the part with "$\limsup_{k\to\infty} \eta(\Vert u_k\Vert) = \limsup_{k\to\infty} \xi(\Vert u_k+ w\Vert)$" (note that the indizes $n$ were corrected to $k$). I'll repeat this part in a more explanatory way. So, before we make any assumptions, we have just by the properties of the infimum (and the definition of $\eta$) that for all $k\in \mathbb{N}$ with $k\geq R$ exists an $u_k \in W^{1,p}(\Omega)$ such that $\Vert u_k\Vert = k$ and
$$
 \xi(\Vert u_k + w\Vert ) - 1 \leq \inf_{u\in W^{1,p}(\Omega), \Vert u\Vert = k} \xi(\Vert u+w\Vert) = \eta(k)= \eta(\Vert u_k\Vert).
$$
(Because we can always choose a sequence that converges to the infimum and thus choose an element that makes the distance to that infimum arbitrarily small. In particular, we find an element that realizes at most the distance 1 to the infimum.)
So, if $\eta$ was bounded by some constant $c_1$ we would have for all $k\in\mathbb{N}$ with $k \geq R$
$$
 \xi(\Vert u_k + w\Vert ) \leq \eta(\Vert u_k\Vert) + 1 \leq c_1+1 =: C.
$$
As $\Vert u_k + w\Vert$ goes to infinity for $k\to \infty$ we get a contradiction because of the definition of $\xi$.
You could now take the $\limsup$ over all the expressions in that inequality and the inequalities would still hold. However, you see that this is actually not necessary at all as we already got our contradiction. I maybe was thinking a little too complicated there (and also was not writing very clearly, I'm sorry).

*For "$\frac{1}{c}\xi(\Vert u+w\Vert)\Vert u\Vert \geq \eta(\Vert u\Vert) \Vert u\Vert$" note that, in the definition of $\eta$, we multiply the whole expression by $c$ from (1). That's why it cancels out. (Also I wrote $C$ instead of $c$ in the beginning of the proof; this has been corrected now.)
