is a weak solution of Neumann's problem

By Holder inequality for any u, v ∈ ˆ H 1 (U)
|B[u, v]| ≤ �Du� L 2 (U)�Dv� L 2 (U) ≤ �u� ˆ H 1 (U) �v� ˆ H 1 (U) . (5)
From definition of B[u, v], for any u ∈ ˆ H 1 (U)
�Du� 2
L2 (U) = B[u, u]. (6)
Also, since (4) holds, then denoting (u)U := 1
�
|U| U u(x) dx, we get (u)U
using Poincare inequality, we estimate
= 0. Thus,
�u� 2
L2 (U) =
�
u 2 �
dx = (u − (u)U) 2 �
dx ≤ C |Du| 2 dx = CB[u, u],
U
where C depends only on U. Combining this and (6), get
1
C + 1 �u�2 Hˆ 1 = (U) 1
�
C + 1
Also, if f ∈ ˆ L2 , then the functional
�
F (v) =
U
U
u
U
2 + |Du| 2 dx ≤ B[u, u]. (7)
U
fv dx
is linear bounded on ˆ H 1 (U). Thus (5) and (7) allow to apply Lax-Milgram theorem
to obtain a unique u ∈ ˆ H 1 (U) such that
B[u, v] = F (v) for all v ∈ ˆ H 1 (U).
This implies that (2) holds for all v ∈ H1 (U) satisfying �
v dx = 0.
U
Recall that u satisfies (4) and f satisfies (3). Thus if (2) holds for some v ∈ H1 (U),
then (2) also holds for v replaced by v + c, where c
�
is any constant. Then (2) holds
for all v ∈ H 1 (U): indeed, if v ∈ H 1 (U) and c = 1
|U|
then we know that (2) holds for v − c, thus for v.
Thus u is a weak solution of (1).
U v dx, then v − c ∈ ˆ H 1 (U), and