By Peter M. Neumann

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Note that (∇φ)(z(t)) = 0 for t ≥ 0 since if not then (∇φ)(z(t)) = 0 for all t ≥ 0 and hence z is constant and consequently, (3) is violated. 1, if 0 ≤ a < b, b φ(z(a)) = φ(z(b)) + z a Hence ∞ 0 z 2 H exists and therefore ri z lim i→∞ si 2 H = 0. 2 H. 2 H. 26 4 Continuous Steepest Descent in Hilbert Space: Nonlinear Case Consequently, limi→∞ (ri − si ) = ∞ since 2 ri ≤ (ri − si ) z 2 H, i = 1, 2, . . si Since φ(z) (t) = − (∇φ)(z(t) it follows that 2 H ≤ −c2 φ(z(t)), t ≥ 0, φ(z) (t)/φ(z(t)) ≤ −c2 , t ≥ 0 and so for each positive integer i, φ(z(t)) ≤ φ(z(si )) exp(−c2 (t − si )), t ∈ [si , ri ], and in particular, φ(z(ri )) ≤ φ(z(si )) exp(−c2 (ri − si )).

Si Since φ(z) (t) = − (∇φ)(z(t) it follows that 2 H ≤ −c2 φ(z(t)), t ≥ 0, φ(z) (t)/φ(z(t)) ≤ −c2 , t ≥ 0 and so for each positive integer i, φ(z(t)) ≤ φ(z(si )) exp(−c2 (t − si )), t ∈ [si , ri ], and in particular, φ(z(ri )) ≤ φ(z(si )) exp(−c2 (ri − si )). Therefore, since φ(z(ri )) ≥ φ(z(si+1 )), i = 1, 2, . . , it follows that lim φ(z(si )) = 0. , qk+1 . Then k ≤ z(ri ) − z(si ) H k ≤ z(qj+1 ) − z(qj ) H ≤ j=0 si +j+1 k ≤ ( j=0 2 1/2 H) si +j k 1 φ(z(si + j))1/2 ≤ j=0 (φ(z(si )) exp(−c2 j) 2 j=0 k 1 1 j=0 1 2 k exp(−c2 j)) 2 ≤ φ(z(si )) 2 = (φ(z(si )) H (φ(z(si + j)) − φ(z(si + j + 1)))1/2 = j=0 k ≤ z j=0 k z si +j si +j+1 exp(−jc2 /2) j=0 2 −1 ≤ φ(z(si )) (1 − exp(−c /2)) since φ(z(a + 1)) ≤ φ(z(a)) exp(−c2 ), a = si , si + 1, .

42 5 Orthogonal Projections, Adjoints and Laplacians To show 3), note that for x ∈ H Mx ≤ M x H = sup{ z, M x H : z H = 1} = sup{ z, x H : z H = 1} ≤ x H ≤ x H H The second to last inequality is due to the Cauchy-Schwarz inequality. To show 4) let x, y ∈ H. Then M x, y H = M x, M y H = x, M y H. To show 5), let x, y ∈ H . Then M x, y H = x, y H = x, M y H . To show 6), use from [204] the fact that a positive, symmetric bounded linear transformation from a Hilbert space to itself has a unique positive, symmetric bounded square root (which also commutes with the given transformation).

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