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52 2. Finite Element Method for Poisson Equation

$ % 1/2
1 1
n 2 2 √
= n dx + n dx = 2n →∞ for n →∞ .
v n a
0 1− 1
n
Therefore, there exists no constant C> 0 such that v a ≤ C v 0 for all
v ∈ V .
However, as we will show in Theorem 2.18, there exists a constant C> 0
such that the estimate
for all v ∈ V
v 0 ≤ C v a
holds; i.e., · a is the stronger norm.
It is possible to enlarge the basic space V without violating the previous
statements. The enlargement is also necessary because, for instance, the
proof of the existence of a solution of the variational equation (2.13) or
the minimization problem (2.14) requires in general the completeness of V.
However, the actual definition of V does not imply the completeness, as
the following example shows:
Example 2.8 Let Ω = (0, 1) again and therefore

1
a(u, v):= u v dx .

0

α α 1
For u(x):= x (1−x) with α ∈ , 1 we consider the sequence of functions
2
 & 1 1 '
u(x) for x ∈ , 1 − ,
 n n
 & '
1
u n(x):= nu( ) x for x ∈ 0, 1 ,
n n
 & '
 1 1
nu(1 − )(1 − x) for x ∈ 1 − , 1 .
n n
Then
u n − u m a → 0 for n, m →∞ ,
u n − u a → 0 for n →∞ ,
but u/∈ V ,where V is defined analogously to (2.7) with d =1.
˜
In Section 3.1 we will see that a vector space V normed with · a exists
˜
˜
such that u ∈ V and V ⊂ V . Therefore, V is not complete with respect
to · a ;otherwise, u ∈ V must be valid. In fact, there exists a (unique)
completion of V with respect to · a (see Appendix A.4, especially (A4.26)),
but we have to describe the new “functions” added by this process. Besides,
integration by parts must be valid such that a classical solution continues to
be also a weak solution (compare with Lemma 2.1). Therefore, the following
idea is unsuitable.
Attempt of a correct definition of V :
Let V be the set of all u with the property that ∂ i u exists for all x ∈ Ω
without any requirements on ∂ i u in the sense of a function.