The Problem of Image Recovery by the Metric Projections in Banach Spaces

and Applied Analysis 3 Lemma 3. Let p > 1 and E be a p-uniformly convex and smooth Banach space. Then, for each x, y ∈ E, φ p (x, y) ≥ c 0 󵄩󵄩󵄩󵄩x − y 󵄩󵄩󵄩󵄩 p (8) holds, where c 0 is maximum in Remark 2. Proof. Let x, y ∈ E. By Theorem 1, we have ‖x‖ p ≥ 󵄩󵄩󵄩󵄩y 󵄩󵄩󵄩󵄩 p + p⟨x − y, J p y⟩ + c 0 󵄩󵄩󵄩󵄩x − y 󵄩󵄩󵄩󵄩 p , (9) where c 0 is maximum in Remark 2. Hence, we get φ p (x, y) = ‖x‖ p − 󵄩󵄩󵄩󵄩y 󵄩󵄩󵄩󵄩 p − p⟨x − y, J p y⟩ ≥ c 0 󵄩󵄩󵄩󵄩x − y 󵄩󵄩󵄩󵄩 p , (10) which is the desired result. Let C be a nonempty closed convex subset of a strictly convex and reflexive Banach space E, and let x ∈ E. Then, there exists a unique element x 0 ∈ C such that ‖x 0 − x‖ = inf y∈C ‖y − x‖. Putting x 0 = P C x, we call P C the metric projection ontoC (see [24]).We have the following result [25, p. 196] for the metric projection. Lemma 4. Let C be a nonempty closed convex subset of a strictly convex, reflexive, and smooth Banach space E, and let x ∈ E. Then, y = P C x if and only if ⟨y − z, J 2 (x − y)⟩ ≥ 0 for all z ∈ C, where P C is the metric projection onto C. Remark 5. For p > 1, it holds that ‖x‖J p x = ‖x‖ p−1 J 2 x for every x ∈ E. Therefore, under the same assumption as Lemma 4, we have that y = P C x if and only if ⟨y − z, J p (x − y)⟩ ≥ 0 for all z ∈ C.