In Theorem 4, that ϕ(t) = ½{f(x+t)+f(x−t)−2s}→0 (C,β) as t→0 whenever the Denjoy-Fourier series of f(x) is summable (C, α) to s, where η>α+1 and a ⩾−1, my proof for the case −1⩽α<0,β<1, was incomplete*. The proof may be completed by the following lemma, valid also for 0⩽a⩽1. LEMMA. If Σan cos nt is the Denjoy-Fourier series of ϕ(t) and an =0(na), then ϕβ(t)=βΣanγβ(nt), for t≠0, where β > a+1 and − 1 ⩽ a < 0. Proof. Suppose that 0 < β < 2. Then, writing γ β , ɛ ( x ) = ∫ 0 1 - ɛ ( 1 - u ) β - 1 cosxy du, we have†, since (1−u)β−1 is of bounded variation in (0, 1 − ɛ), ∫ 0 1 - ɛ ( 1 - u ) β - 1 φ ( t u ) d u = Σanγβ, ɛ(nt. Moreover, for a given t≠0, it is easily verified that γβ, ɛ(nt), and hence ∑ a n { γ β ( n t ) - γ β , ɛ ( n t ) } = ∑ n ⩽ 1 / ɛ o ( n α ) O ( ɛ β ) + ∑ n > 1 / ɛ o ( n α - β ) = o ( ɛ β - α - 1 The result follows by letting ɛ → 0.