Diploma thesis as pdf file - Johannes Kepler University, Linz

1.2 Existence and uniqueness of the solution 5
From this it follows that {η k } is Cauchy sequence, and since H is complete, it
converges to an element y ∈ H. And, since K is closed, y ∈ K and ‖x − y‖ = d.
Now, let’s suppose that there are two projections y 1 and y 2 of the element x.
Then from the previous discussions it follows that :
‖y 1 − y 2 ‖ 2 ≤ 2‖x − y 1 ‖ 2 + 2‖x − y 2 ‖ 2 − 4d 2 = 0.
Therefore, y 1 = y 2 .
Proof of the theorem (1.2.1) We can consider the bilinear form a(·,·) as the
inner product in Hilbert space H, then the norm ‖v‖ a = √ a(v,v) is equivalent to
the given norm in H. By the Riesz representation lemma we can find an element
v ∗ ∈ H such that
〈 f ,v〉 = a(v ∗ ,v) ∀v ∈ H.
Then
J(v) = 1 2 a(v,v) − a(v∗ ,v) = 1 2 ‖v − v∗ ‖ 2 a − 1 2 ‖v∗ ‖ 2 a.
Thus, the problem of minimization of J(v) reduces to finding of the projection of
v ∗ on the closed convex set K. By lemma (1.2.3) there exist unique projection,
therefore unique solution of the minimization problem.
To apply this theorem for the problem (1.1.1) we need to show respective
∫
properties of the set K and a(u,v) = ∇u∇vdω.
Lemma 1.2.4 K is convex and closed.
Ω
Proof Let u,v ∈ K. Then for 0 < t < 1 : tu + (1 − t)v ∈ V and tu + (1 − t)v ≥
tψ + (1 −t)ψ = ψ. This shows convexity of K.
Now let {v k } ∈ K be a convergent sequence. Since from the convergence
in V follows convergence in L 2 (Ω), it contains an a.e. pointwise convergent
subsequence {v kn }. Let {v kn } −→ v pointwise. Suppose that v is not from K.