x ) 2 R Discrete image: f 2 RN , N = n2 f[i1, i2] = ˜ f(i1/n, i2/n) rf[i] ⇡ r ˜ f(i/n) ˜ f(x + ") = ˜ f(x) + hrf(x), "iR2 + O(||"||2 R2 ) r ˜ f ( x ) = ( @1 ˜ f ( x ) , @2 ˜ f ( x )) 2 R2 Gradient: Images vs. Functionals
f? a solution of min f J ( f ). If f is convex, C1 , rf is L -Lipschitz, Theorem: f(k+1) = f(k) ⌧k rJ(f(k)) f(0) is given. Optimal step size: ⌧k = argmin ⌧2R+ J(f(k) ⌧rJ(f(k))) Proposition: One has hrJ(f(k+1)), rJ(f(k))i = 0 Gradient Descent
shows iterations of the algorithm 1 to solve the inpainting problem on a smooth image using a manifold prior with 2D linear patches, as defined in 16. This manifold together with the overlapping of the patches allow a smooth interpolation of the missing pixels. Measurements y Iter. #1 Iter. #3 Iter. #50 Fig. 3. Iterations of the inpainting algorithm on an uniformly regular image. 5 Manifold of Step Discontinuities In order to introduce some non-linearity in the manifold M, one needs to go log 10 (||f(k) f( )||/||f0 ||) k k E(f(k)) M ( f )[ i ] = ⇢ 0 if i 2 M, f [ i ] otherwise.
s.t. x x (k) 2 span(rE( x (0)) , . . . , rE( x (k))) Intuition: Conjugate Gradient Ax = b () min x 2Rn E( x ) = 1 2 h Ax, x i h x, b i Proposition: 8 ` < k, hrE( x k) , rE( x `)i = 0
(0) = b Ax (0) , p (0) = r (0) r (k) = hrE( x (k)) , d (k)i h Ad (k) , d (k)i d (k) = rE( x (k)) + || v (k)|| || v (k 1)||d (k 1) v (k) = rE( x (k)) = Ax (k) b x (k+1) = x (k) r (k) d (k) Iterations: x (k+1) = argmin E( x ) s.t. x x (k) 2 span(rE( x (0)) , . . . , rE( x (k))) Intuition: Conjugate Gradient Ax = b () min x 2Rn E( x ) = 1 2 h Ax, x i h x, b i Proposition: 8 ` < k, hrE( x k) , rE( x `)i = 0
H Contraint: H = {f ; f = y}. f(k+1) = ProjH ⇣ f(k) ⌧k rJ"( f(k) ) ⌘ ProjH( f ) = argmin g=y ||g f||2 = f + ⇤( ⇤ ) 1(y f) Inpainting: ProjH( f )[ i ] = ⇢ f [ i ] if i 2 M, y [ i ] otherwise. Projected gradient descent: f(k) k!+1 ! f? a solution of ( ? ). (?) Projected Gradient Descent Proposition: If rJ" is L-Lipschitz and 0 < ⌧k < 2/L,