By Fowles G.R., Cassiday G.L.

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37), which concludes the ﬁrst part of the lemma. 8. We proceed through our analysis with the following weak convergence result. 17 The whole sequence {uρ } converges weakly in X to the element u. 37). 15 that the sequence {uρ } is bounded in X. 20. The last step is provided by the following strong convergence result. 54) uρ → u in X as ρ → 0. Proof Let ρ > 0. 50) to obtain that lim (Auρ , uρ − u)X = 0. 17, it follows that lim (Au, uρ − u)X = 0. 46), it follows that m uρ − u 2 X ≤ (Auρ − Au, uρ − u)X = (Auρ , uρ − u)X − (Au, uρ − u)X .

30 that, if (X, (·, ·)X ) is an inner product space then the function v → v 2X = (v, v)X is strictly convex and lower semicontinuous. We now recall the deﬁnition of Gˆ ateaux diﬀerentiable functions. 31 Let (X, (·, ·)X ) be an inner product space, ϕ : X → R and u ∈ X. Then ϕ is Gˆ ateaux diﬀerentiable at u if there exists an element ∇ϕ(u) ∈ X such that lim t→0 ϕ(u + t v) − ϕ(u) = (∇ϕ(u), v)X t ∀ v ∈ X. 66) is unique and is called the gradient of ϕ at u. The function ϕ : X → R is said to be Gˆ ateaux diﬀerentiable if it is Gˆ ateaux diﬀerentiable at every point of X.

35) is understood in the following sense: for every sequence {ρn } ⊂ R+ converging to 0 as n → ∞ one has uρn → u in X as n → ∞. Proof Let ρ > 0. 33) and add the resulting inequalities to obtain (Auρ − Au, uρ − u)X ≤ jρ (u) − jρ (uρ ) + j(uρ ) − j(u). 34)(a) to ﬁnd that m uρ − u X ≤ G(ρ). 34)(b). 22) in the case when K = X and we investigate its unique solvability by using a regularization method. Thus, we assume in what follows that A : X → X, j : X → R and 44 Elliptic variational inequalities f ∈ X are given and we consider the problem of ﬁnding an element u such that u ∈ X, (Au, v − u)X + j(v) − j(u) ≥ (f, v − u)X ∀ v ∈ X.