On a Result of Levin and Stečkin

We note here the constant (p/(1−r))p is best possible, as shown in [9] by setting an = n −1−(1−r)/p−ǫ and letting ǫ → 0+. This implies inequality (1.2) for 0 < p ≤ 1/3 with the best possible constant cp = (p/(1 − p)) p. On the other hand, it is also easy to see that inequality (1.2) fails to hold with cp = (p/(1 − p)) p for p ≥ 1/2. The point is that in these cases p/(1 − p) ≥ 1 so one can easily construct counter examples. A simpler proof of Levin and Stečkin’s result (for 0 < r = p ≤ 1/3) is given in [3]. It is also pointed out there that using a different approach, one may be able to extend their result to p slightly larger than 1/3, an example is given for p = 0.34. The calculation however is more involved and therefore it is desirable to have a simpler approach. For this, we let q be the number defined by 1/p+1/q = 1 and note that by the duality principle (see [10, Lemma 2] but note that our situation is slightly different since we have 0 < p < 1 with an reversed inequality), the case 0 < r < 1, 0 < p < 1 of inequality (1.3) is equivalent to the following one for an > 0: (1.4) ∞