The game model with multi-task for image denoising and edge extraction

Proof of Theorem 3.1.

When n = 0 , then

Let min 0 = inf ⁡ { u i , j 0 } and max 0 = sup ⁡ { u i , j 0 } . Then

Thus min 0 ≤ u i , j 1 ≤ max 0 , i.e.

This is the same as n = k , so

Thus this scheme satisfies the maximum-minimum principle.

On the other hand, if

then

It is inconsistent with stability. The maximum-minimum principle is not satisfied. ∎

Proof of Theorem 3.2. The denoiser D ~ is bounded.

We have

By the proof of Theorem 3.1, u n satisfies the maximum-minimum principle. Therefore, u n is bounded, A is a bounded nonlinear diffusion operator, and γ n is a scaling factor so that β n is bounded. Therefore, there exists a constant M n such that A ⁢ u n + β n ⁢ ( f - u n ) ≤ M n . Thus,

This concludes the proof. ∎

Proof of Theorem 3.3. The GSG model has bounded gradient.

We define

Then

From the equivalence property theory of the norm, we have ∥ p * ∥ 2 ≤ M . Therefore,

and thus ∥ p ∥ 2 ≤ M .

In Section 3.3, the GSG model, we have defined F ⁢ ( p ) and G ⁢ ( p ) . Next, we will prove that both F ⁢ ( p ) and G ⁢ ( p ) are gradient bounded. According to the Gateaux derivative,

where υ is a small constant in order to prevent

approaching zero.

Then

Since Ω is a bounded closed set and ∥ p ∥ 2 ≤ M , there exists L 21 < ∞ such that ∥ ∇ ⁡ F ⁢ ( p ) ∥ 2 ≤ L 21 . Similarly,

Since Ω is a bounded closed set and

there exists L 22 < ∞ such that ∥ ∇ ⁡ G ⁢ ( p ) ∥ 2 ≤ L 22 .

Thus,

Then

Proof of Theorem 3.4.

Let

Then

Further, ρ n + 1 varies according to two different cases of ρ n + 1 :

1. Case 1: If Δ n + 1 > η ⁢ Δ n , then ρ n + 1 = r ⁢ ρ n .

2. Case 2: If Δ n + 1 ≤ η ⁢ Δ n , then Δ n + 1 ≤ η ⁢ Δ n , where Δ n + 1 ≤ η ⁢ Δ n .

At iteration k, if case 1 holds, by Lemma 3.5, then

On the other hand, if case 2 holds, then

As k → ∞ , one of the following situations will happen:

1. Case 1 occurs infinitely many times but case 2 occurs finitely many times.

2. Case 1 occurs finitely many times but case 2 occurs infinitely many times.

3. Both case 1 and case 2 occur infinitely many times.

When (i) happens, there must exist a K 1 such that for k ≥ K 1 only case 1 happens. Therefore,

When (ii) happens, there must exist a K 2 such that for k ≥ K 2 only case 2 happens. Thus,

(iii) is a union of (i) and (ii). Therefore, as long as we can see that under (i) and (ii) the sequence { θ k } k = 1 ∞ converges, the sequence will also converge under (iii). By Lemma C.1, for any k, we have D ⁢ ( θ k + 1 , θ k ) ≤ C ′′ ⁢ δ k , where k → ∞ is constant and 0 < δ < 1 . Therefore, D ⁢ ( θ k + 1 , θ k ) → 0 as k → ∞ .

In order to verify that { θ k } k = 1 ∞ is a Cauchy sequence, we need to prove D ⁢ ( θ m , θ k ) → 0 , where m > k and k → ∞ :

Thus, D ⁢ ( θ m , θ k ) → 0 as k → ∞ . Then { θ k } k = 1 ∞ is a Cauchy sequence. Therefore, there exists θ * = ( u * , p * ) such that D ⁢ ( θ k , θ * ) → 0 .

Consequently, we have

Lemma C.1.

The sequence { θ k } k = 1 ∞ always satisfies { θ k } k = 1 ∞ for constant C ′′ and 0 < δ < 1 .

Proof.

At iteration k, it holds that

Let

Then