Inf-sup condition only for existence (Necas theorem)? 
So existence and uniqueness follows given (c1) and (c2). Is it possible that existence holds if (c1) only holds? I saw a similar result in other books (Showalter's Monotone Operators in Banach Spaces..)) but am not sure if the inf-sup condition is the same or slightly different.
If this is true, why is not written clearly that (c2) is only there for uniqueness???? And are there other methods that can replace (c2)?
 A: Mr. Banach apparently knows his stuff. Here I will give a longer version of his.
First let $A$ be a bounded linear operator between $W$ and $V'\simeq V$: for any $w\in W$, $v\in V$
$$
\langle Aw,v \rangle_{V',V} := a(w,v) . 
$$
The inf-sup $(c1)$ condition translates to for any $w\in W$
$$
\|A w\|_{V'} = \sup_{v\in V} \frac{a(w,v)}{\|v\|_V} \geq \alpha \|w\|_W ,
$$
which further tells us that the range of $A$, $R(A)$, is closed in $V'\simeq V$. Also, this pins down that the kernel of $A$, $N(A)$, is $\{0\}$. $(c1)$ basically says $A^* \in \mathcal{L}(V,W')$ is surjective (the injectivity is unknown so far), or $A$ has a bounded inverse.
However, this only serves as the sufficient and necessary condition for $Aw = l$ (holding in $V'$) has at least one solution, yet no uniqueness is guaranteed. Because for $\langle Aw,v \rangle_{V',V}=\langle w,A^*v \rangle_{W,W'} = 0$, it could be either (1) $Aw = 0$, or (2) $A^*v = 0$ aka $v\in N(A^*)$.
$(c2)$ essentially rules out the second case, for now $N(A^*)$ is $\{0\}$. This guarantees that the solution is unique.
Example: The solvability of $\operatorname{div} w =l$ in $\mathbb{R}^3$. Then $a(w,v) = (w,\nabla v)$. We have $(w,\nabla v) = 0 \not\Rightarrow v= 0$ because $w$ can be a curl of some vector field. So if there is a solution to $\operatorname{div} w =l$, then adding a curl to the solution is still a solution. However, $(c2)$ rules out this case.
A: Condition (c2) is needed to ensure that the underlying operator A (defined by the condition (Au,v)=a(u,v)) has Range(A)=W. In particular, since Range(A) is a closed subspace of V (thanks to (c1)), it suffices to show that Range(A)^\perp is trivial. The condition (c2) shows this is indeed the case.
