Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula 2) S N fθ) = 1 fθ ϕ)d N ϕ) dϕ, 3) the last identity by virtue of x N T 1 D N ϕ) = 2N k=0 N k= N e ikϕ = e inϕ 2N = k=0 e ikϕ sinn + 1/2)ϕ, sin ϕ/2 x k N 1 x2n+1 = x, 1 x using e iϕ for x, and multiplying numerator and denominator by e iϕ/2. Using sin N + 1 ) ϕ = cos ϕ 2 2 sin Nϕ + sin ϕ cos Nϕ, 2 we deduce that S N fθ) fθ) = 1 [fθ ϕ) fθ)]d N ϕ) dϕ T 4) = 1 1 g θ ϕ) sin Nϕ dϕ + h θ ϕ) cos Nϕ dϕ, T 1 T 1 5) fθ ϕ) fθ) g θ ϕ) =, h θ ϕ) = fθ ϕ) fθ). Clearly, for N 0, 6) f L 1 T 1 ) = ĥθ±n) = ˆf±N) 0 as N, the convergence to 0 by the Riemann-Lebesgue lemma. Applying the Riemann-Lebesgue lemma to ĝ θ ±N) gives the following. 1
2 Proposition 1. Let f L 1 T 1 ). Let K T 1 be compact. Then 7) S N fθ) fθ), uniformly for θ K, provided that 8) {g θ : θ K} is a relatively compact subset of L 1 T 1 ). Proof. The Riemann-Lebesgue lemma plus the compactness hypothesis 8) implies that ĝ θ N) goes to 0 as N, uniformly in θ K. In more detail, take ε > 0. Pick a finite set {θ j : 1 j Mε)} such that, with g j ϕ) = g εj ϕ), 8A) θ K, g j g θ L 1 ε, for some j Mε). The compactness hypothesis 8) guarantees you can do this. The Riemann-Lebesgue lemma says that, for each j {1,..., Mε)}, there exists N j such that 8B) ĝ j N) < ε, N such that N > N j. Now set Ñε) = max{n j : 1 j Mε)}. By 8A) we have, for all θ K, 8C) ) ĝ θ N) min ĝ j N) + ĝ j N) ĝ θ N) j ε + ε, provided N > Ñε). The desired conclusion 7) follows from this, in concert with 4) 6). The following is an important special case. Corollary 2. Let f C ω T 1 ), i.e., 9) fθ ϕ) fθ) Cω ϕ ), θ, ϕ T 1. Assume the modulus of continuity ωt) satisfies 10) Then 7) holds with K = T 1. 0 ωt) t dt <. Proof. We claim the hypotheses 9) 10) imply that 11) g θ is a continuous function of θ with values in L 1 T 1 ).
3 Given this, the compactness condition 8) holds, with K = T 1. T 1, θ ν θ 0. We see that So let θ ν, θ 0 12) g θν ϕ) g θ0 ϕ) for all ϕ T 1 \ 0, and that 13) g θν ϕ) C ω ϕ ) ϕ = Hϕ). Hence g θν ϕ) g θ0 ϕ) 0 for all ϕ T 1 \ 0, and 14) g θν ϕ) g θ0 ϕ) 2Hϕ). Now 10) implies H L 1 T 1 ), so the convergence 15) T 1 g θν ϕ) g θ0 ϕ) dϕ 0 follows by the Dominated Convergence Theorem. The following is a version of Riemann localization. Proposition 3. Take f L 1 T 1 ). Assume f = 0 on O, an open subset of T 1, and let K O be compact. Then S N f f uniformly on K. Proof. Take an interval I = ε, ε) so small that 16) θ K, ϕ I = θ ϕ O, so 17) θ K, ϕ I = g θ ϕ) = 0. Then take ψ CT 1 ) such that ψϕ) = 1 for ϕ < ε/2, ψϕ) = 0 for ϕ ε. Then 18) θ K = ψg θ 0 = g θ ϕ) 1 ψϕ) [fθ ϕ) fθ)]. Since 1 ψϕ))/ tanϕ/2) is continuous on T 1, it follows that 19) θ g θ is continuous from K to L 1 T 1 ). Thus 8) holds, and Proposition 3 follows from Proposition 1. Putting together Corollary 2 and Proposition 3 gives the following.
4 Corollary 4. Take f L 1 T 1 ). Let O T 1 be open and assume f O C ω O), with ω satisfying 10). Let K O be compact. Then S N f f uniformly on K. We now produce another strengthening of Corollary 2. Proposition 5. Take f L 1 T 1 ), and let K T 1 be compact. Assume f K CK) and 20) fθ ϕ) fθ) Cω ϕ ), θ K, ϕ T 1, ω is measurable and satisfies 10). Then 7) holds. Proof. Again it suffices to show that 21) θ g θ is continuous from K to L 1 T 1 ). So let θ ν, θ 0 K and θ ν θ 0. We continue to have 13) 14), i.e., 22) g θν ϕ) g θ0 ϕ) 2Hϕ), H L 1 T 1 ). We need a replacement for 12). In fact, we claim that 23) g θν g θ0 in measure, as ν, i.e., if m denotes Lebesgue measure on T 1, then, for each ε > 0, 24) me εν ) 0 as ν, 25) E εν = {ϕ T 1 : g θν ϕ) g θ0 ϕ) > ε}. In fact, 23) follows directly from the continuity of f K, together with the fact that, with f θ ϕ) = fθ ϕ), 26) f θν f θ0 in measure, as ν, itself a consequence of the fact that 27) f θν f θ0 in L 1 -norm, i.e., 28) T 1 f θν ϕ) f θ0 ϕ) dϕ 0, together with Chebechev s inequality, ) 29) m {ϕ : F ϕ) > ε} 1 ε F L 1. It is a variant of the Dominated Convergence Theorem cf. [T], pp. 37 38), that 22) 23) imply 15). This completes the proof of Proposition 5. Bringing in an argument used for Riemann localization in Proposition 3, we have the following strengthening of Proposition 5.
5 Proposition 6. Take f L 1 T 1 ) and let K T 1 be compact. Assume f K CK). Assume there exists ε > 0 such that 30) fθ ϕ) fθ) < cω ϕ ), θ K, ϕ < ε, ω is measurable on [0, ] and satisfies 10). Then 7) holds. Proof. Take ψ as in 18), and set 31) g θ ϕ) = u θ ϕ) + v θ ϕ), 32) u θ ϕ) = ψϕ) fθ ϕ) fθ), v θ ϕ) = 1 ψϕ) [fθ ϕ) fθ)]. Clearly f L 1 T 1 ) and f K bounded implies {v θ : θ K} is relatively compact in L 1 T 1 ). Meanwhile an analysis parallel to 21) 29) applies here, with g θ replaced by u θ, under the hypotheses given above. Reference [T] M. Taylor, Measure Theory and Integration, AMS, Providence RI, 2006.