Full Length Article
On the optimal relationships between LP-norms for the Hardy operator and its dual for decreasing functions

https://doi.org/10.1016/j.jat.2019.105362Get rights and content

Abstract

We prove sharp inequalities between Lpnorms (1<p<) of functions Hf and Hf, where H is the Hardy operator, H is its dual, and f is a nonnegative nonincreasing function on (0,). In particular, we extend one result obtained for integer p2 by Boza and Soria (2019), to the whole range of values p2.

Section snippets

Introduction and main results

Denote by M+(R+) the class of all nonnegative measurable functions on R+(0,+). Let fM+(R+). Set Hf(x)=1x0xf(t)dtand Hf(x)=xf(t)tdt.These equalities define the classical Hardy operator H and its dual operator H. By Hardy’s inequalities [4, Ch. 9], these operators are bounded in Lp(R+) for any 1<p<. Furthermore, it is easy to show that for any fM+(R+) and any 1<p< 1pHfpHfppHfpfor1<p<(as usual, p=p(p1)).

However, the constants in (1.1) are not optimal. Sharp constants are

Proof of Theorem 1.3

Let 1<p<. Taking into account (1.1), we may assume that Hf and Hf belong to Lp(R+). Denote Ip=01x0xf(t)dtpdx.Since HfLp(R+), we have Hf(x)=o(x1p)asx0+orx+.Thus, integrating by parts, we obtain Ip=p0x1pf(x)0xf(t)dtp1dx.

Further, set Ip=0tf(x)xdxpdt, Φ(t,x)=txf(u)udu,0<tx,and G(t,x)=Φ(t,x)p. Since G(t,t)=0, we have tf(x)xdxp=tGx(t,x)dx=ptf(x)xΦ(t,x)p1dx.Thus, by Fubini’s theorem, Ip=p0tf(x)xΦ(t,x)p1dxdt =p0f(x)x0xΦ(t,x)p1dtdx.Set g(x)=0xΦ(t,x)p1dtandΛ(t

References (8)

There are more references available in the full text version of this article.

Cited by (0)

This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

View full text