By Martin Schechter

The strategies used to resolve nonlinear difficulties vary vastly from these facing linear good points. Deriving all of the worthy theorems and rules from first ideas, this textbook offers higher undergraduates and graduate scholars a radical figuring out utilizing as little history fabric as attainable.

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This means that u ∈ H, that u is continuous, that u = u − f in the usual sense, and that (u ) = u − f . 9. 17. If f (x) is continuous in I and (f, ϕ¯k ) = 0, k = 0, ±1, ±2, . . 51) then f (x) = 0 in I. 18. 52) then f (x) is constant in I. Proof. Let 2π 1 α0 = √ f (x) dx, 2π 0 √ and take g(x) = f (x) − (α0 / 2π). Then √ βk = (g, ϕ¯k ) = (f, ϕ¯k ) − (α0 2π) 2π ϕ¯k dx = 0 0 for k = 0. Moreover, √ β0 = (g, 1)/ 2π = α0 − α0 = 0. 17. Hence, f (x) ≡ α0 / 2π. 19. e. in I. Proof. Deﬁne x F (x) = f (t) dt.

50) holds in the limit. 50) holds for all v ∈ C 1 (I). This means that u ∈ H, that u is continuous, that u = u − f in the usual sense, and that (u ) = u − f . 9. 17. If f (x) is continuous in I and (f, ϕ¯k ) = 0, k = 0, ±1, ±2, . . 51) then f (x) = 0 in I. 18. 52) then f (x) is constant in I. Proof. Let 2π 1 α0 = √ f (x) dx, 2π 0 √ and take g(x) = f (x) − (α0 / 2π). Then √ βk = (g, ϕ¯k ) = (f, ϕ¯k ) − (α0 2π) 2π ϕ¯k dx = 0 0 for k = 0. Moreover, √ β0 = (g, 1)/ 2π = α0 − α0 = 0. 17. Hence, f (x) ≡ α0 / 2π.

17. If f (x) is continuous in I and (f, ϕ¯k ) = 0, k = 0, ±1, ±2, . . 51) then f (x) = 0 in I. 18. 52) then f (x) is constant in I. Proof. Let 2π 1 α0 = √ f (x) dx, 2π 0 √ and take g(x) = f (x) − (α0 / 2π). Then √ βk = (g, ϕ¯k ) = (f, ϕ¯k ) − (α0 2π) 2π ϕ¯k dx = 0 0 for k = 0. Moreover, √ β0 = (g, 1)/ 2π = α0 − α0 = 0. 17. Hence, f (x) ≡ α0 / 2π. 19. e. in I. Proof. Deﬁne x F (x) = f (t) dt. 36), and F (2π) = 0 = F (0). Hence F is periodic in I. Let √ 2πα0 = k = 0, ±1, ±2, . . γk = (F, ϕ¯k ), Then √ √ γk = (F, e−ikx / 2π) = (F, (e−ikx / 2π) /(−ik)) = − (F , ϕ¯k )/(−ik) = (f, ϕ¯k )/ik = 0 for k = 0.