PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 98(112) (2015), 91–96 DOI: 10.2298/PIM140722005E

INTEGRAL PROPERTIES OF RAPIDLY AND REGULARLY VARYING FUNCTIONS

Nebojša Elez and Vladimir Vladičić

Abstract. Regularly and rapidly varying functions are studied as well as the asymptotic properties related to several classical inequalities and integral sums.

1. Introduction

Regular and rapid variation of functions was initiated by Karamata [3]. It is sometimes called Karamata theory. Nowadays, it is a well developed theory used in asymptotic analysis of functions, Tauberian theorems, probability and analytic number theory.

Recall that a measurable function f : [a, ∞) → (0, ∞), a > 0 is called regularly varying in the sense of Karamata if for some α ∈ R it satisfies

lim_{x→∞} \frac{f(λx)}{f(x)} = λ^{α}

for every λ > 0, and we denote f ∈ R_{α}. The classes R_{α}, α ∈ R were introduced in [3], where it is proved (see also [1]) that if a function f ∈ R_{α}, α > 0, is locally bounded, then

∫_{a}^{x} f(t) dt ∼ \frac{x}{α + 1} f(x), x → ∞.

We will use Potter’s theorem (see e.g., [1]): If f ∈ R_{α}, α > 0, then for every μ > 1 and ε > 0 there exists x_{0} > 0 such that

\frac{f(y)}{f(x)} < μ( \frac{y}{x})^{α+ε} , x_{0} ⩽ x < y

Recall [1], a measurable function f : [a, ∞) → (0, ∞), a > 0, is called rapidly varying in the sense of de Haan, with the index of variability ∞, if it satisfies

lim_{x→∞} \frac{f(λx)}{f(x)} = ∞

2010 Mathematics Subject Classification: 26A12; 26D15.

Partially supported by Ministry of Science and Technology of the Republic of Srpska.

Communicated by Stevan Pilipović.

for every λ > 1. This functional class is denoted by R_{∞} (see e.g., [1]). Example: If f(x) = x^{r(x)} , r(x) → ∞, and r is a nondecreasing function, then f ∈ R_{∞} .

Here we consider functions f ∈ R_{∞} defined on the interval [0, ∞). Analogous results can be obtained if the domain of a function f is [a, ∞), a > 0. Let f ∈ R_{∞} if f(x) = O(\underline{f} (x)), x → ∞, where \underline{f} (x) = inf\{f(t) | x ⩽ t\}, then we say f ∈ MR_{∞} . In this case we also know that \underline{f} is a nondecreasing function and \underline{f} ⩽ f. We will use the following properties of rapidly varying functions:

(1) If f ∈ R_{∞} , then lim_{x→∞} \frac{\underline{f} (λx)}{f(x)} = ∞ and \underline{f} ∈ R_{∞} .

(2) Using (1) we have: for f ∈ R_{∞} lim_{x→∞} \frac{f(φ(x))}{f(ψ(x))} = ∞, if ψ(x) → ∞ and

lim inf_{x→∞} \frac{φ(x)}{ψ(x)} > 1.

(3) If λ > 1 and f ∈ R_{∞} is locally bounded on [0, ∞), then

∫_{0}^{x} f(t)dt ∼ ∫ x_{\frac{x}{λ}} f(t)dt, x → ∞.

(4) If f ∈ R_{∞} , φ ∈ R_{α} and f, φ are locally bounded on [0, ∞), then

∫_{0}^{x} f(t)φ(t)dt ∼ φ(x) ∫_{0}^{x} f(t) dt, x → ∞.

(5) If f ∈ MR_{∞} is locally bounded on [0, ∞), then

\frac{1}{x} ∫_{0}^{x} f(t) dt = o(f(x)), x → ∞.

We use notation (f(x) ≫ g(x), x → a for g(x) = o(f(x)), x → a.

2. Results

Our first theorem is connected with Chebyshev’s inequality: if f, g : [a, b] → R are monotonic functions of the same monotonicity, then

∫_{a}^{b}f(t) g(t) dt ⩾ \frac{1}{b − a} ∫_{a}^{b}f(t) dt ∫_{a}^{b}g(t) dt.

If f, g are of different monotonicity, then the above inequality holds in the opposite direction.

Theorem 2.1. Let f, g ∈ MR_{∞} be locally bounded on [0, ∞). Then

∫_{0}^{x} f(t) g(t) dt ≫ \frac{1}{x} ∫_{0}^{x} f(t) dt ∫_{0}^{x} g(t) dt ≫ ∫_{0}^{x} f(t) g(x − t) dt, x → ∞.

The following theorem is connected with Jensen’s inequality: let f : R → R be a convex function and φ : [a, b] → (0, +∞) a nondecreasing function. Then

f( \frac{1}{b − a} ∫_{a}^{b}φ(x) dx) ⩽ \frac{1}{b − a} ∫_{a}^{b}f(φ(x)) dx.

Theorem 2.2. Let f ∈ R_{∞} , φ ∈ R_{α}, α > 0 be locally bounded on [0, ∞). Then

f ( \frac{1}{x} ∫_{0}^{x} φ(t) dt) ≪ \frac{1}{x} ∫_{0}^{x} f(φ(t)) dt, x → ∞.

Theorem 2.3. Let f ∈ R_{∞} and ε > 0. Then

f( \frac{n}{1 + ε} ) ≪ \frac{f(1) + f(2) + ⋯+ f(n)}{n} , n → ∞.

Theorem 2.4. Let f ∈ R_{∞} be a locally bounded on [0, ∞). Then for every p > 1,

\frac{1}{x} ∫_{0}^{x} f(t) dt ≪ ( \frac{1}{x} ∫_{0}^{x} f(t)^{p}dt) ^{\frac{1}{p}} , x → ∞.

Theorem 2.5. If f ∈ R_{∞} , φ ∈ R_{α} are locally bounded functions on [0, ∞), then

∫_{0}^{x} f(t)φ(t) dt ≪ ( ∫_{0}^{x} f(t)^{p}dt) ^{\frac{1}{p}} ( ∫_{0}^{x} φ(t)^{q}dt) ^{\frac{1}{q}} , x → ∞

where \frac{1}{p} + \frac{1}{q} = 1, p > 0, q > 0.

Using Theorem 2.4 for the function f([x] + 1) ∈ R_{∞} on (0, n) we get:

Corollary 2.1. If f ∈ R_{∞} and p > 1, then

\frac{f(1) + f(2) + ⋯+ f(n)}{n} ≪ ( \frac{f(1)^{p} + f(2)^{p} + ⋯+ f(n)^{p}}{n} ) ^{\frac{1}{p}} , n → ∞.

Theorem 2.6. Let f ∈ R_{∞} be a locally bounded function on [0, ∞) and n ∈ N. Then

\frac{f(\frac{x}{n}) + f(\frac{2x}{n} ) + ⋯+ f(\frac{(n−1)x}{n} )}{n} ≪ \frac{1}{x} ∫_{0}^{x} f(t) dt, x → ∞.

Theorem 2.7. Let f ∈ MR_{∞} be a locally bounded function on [0, ∞) and let n ∈ N. Then

\frac{f(\frac{x}{n}) + f(\frac{2x}{n} ) + ⋯+ f(x)}{n} ≫ \frac{1}{x} ∫_{0}^{x} f(t) dt, x → ∞.

3. Proofs

Proof of Theorem 2.1. It is enough to prove that for every M > 1

\frac{1}{M} ∫_{0}^{x} f(t) g(t) dt > \frac{1}{x} ∫_{0}^{x} f(t) dt ∫_{0}^{x} g(t) dt > M ∫_{0}^{x} f(t) g(x − t) dt

for sufficiently large x. There exists m ∈ (0, 1) and x_{1} > 0 such that

mf(x) ⩽ \underline{f} (x), mg(x) ⩽ \underline{g}(x), x > x_{1}.

Let 1 < λ < \frac{4M}{4M−m^{2}} ; then \frac{m^{2}}{4M(1− \frac{1}{λ} )} > 1. Further on, there exists x_{2} > 0, such that if x > x_{2}, then from (1) and (3) we obtain

\frac{1}{2} ∫_{0}^{x} \underline{f} (t) dt ⩽ ∫ x_{\frac{x}{λ}} \underline{f} (t) dt, \frac{1}{2} ∫_{0}^{x} \underline{g}(t) dt ⩽ ∫ x_{\frac{x}{λ}} \underline{g}(t) dt.

Now for x > x_{0} = max{x_{1}, x_{2}}, by Chebyshev’s inequality for the nondecreasing functions \underline{f} , \underline{g} on the interval ( \frac{x}{λ} , x), we have

\frac{1}{M} ∫_{0}^{x} f(t) g(t) dt ⩾ \frac{1}{M} ∫_{0}^{x} \underline{f} (t) \underline{g}(t) dt ⩾ \frac{1}{M} ∫ x_{\frac{x}{λ}} \underline{f} (t) \underline{g}(t) dt

⩾ \frac{1}{M} \frac{1}{x − \frac{x}{λ}} ∫ x_{\frac{x}{λ}} \underline{f} (t) dt ∫ x_{\frac{x}{λ}} \underline{g}(t) dt ⩾ \frac{1}{x} \frac{1}{M(1 − \frac{1}{λ})} \frac{1}{4} ∫_{0}^{x} \underline{f} (t) dt ∫_{0}^{x} \underline{g}(t) dt.

⩾ \frac{1}{4xM(1 − \frac{1}{λ})} m^{2} ∫_{0}^{x} f(t) dt ∫_{0}^{x} g(t) dt > \frac{1}{x} ∫_{0}^{x} f(t) dt ∫_{0}^{x} g(t) dt

In a similar way we can show \frac{1}{x} ∫_{0}^{x} f(t) dt ∫_{0}^{x} g(t) dt > M ∫_{a}^{x} f(t)g(x − t) dt for sufficiently large x. □

Proof of Theorem 2.2. If φ ∈ R_{α}, then for every λ > 0

φ(λx) ∼ λ^{α} φ(x), x → ∞; ∫_{0}^{x} φ(t) dt ∼ \frac{x}{α + 1} φ(x), x → ∞.

Let i : [0, +∞) → (0, ∞) be a locally integrable function such that lim_{x→∞} i(x) = 1 and \frac{1}{x} ∫_{0}^{x} φ(t) dt = \frac{1}{1+α} φ(x)i(x), for every x > 0. Let 1 < λ < (1 + α)^{\frac{1}{α}} ; then \frac{α+1}{λ^{α}} > 1. For sufficiently large x and t ⩾ \frac{x}{\sqrt{λ}}, using f ∘ φ ∈ R_{∞} , we have

\frac{1}{x} ∫_{0}^{x} f(φ(t)) dt > \frac{1}{x} ∫ x_{\frac{x}{\sqrt{λ}}} f(φ(t)) dt ⩾ \frac{1}{x} ( x − \frac{x}{\sqrt{λ}} ) \underline{f ∘ φ} ( \frac{x}{\sqrt{λ}} ) .

Now by (1), we obtain

\frac{\underline{f ∘ φ} (\frac{x}{\sqrt{λ}}))}{f(φ(\frac{x}{λ}))} → ∞, x → ∞.

There is a function j : (0, +∞) → R, lim_{x→∞} j(x) = 1 such that, for every x > 0

φ ( \frac{x}{λ}) = \frac{1}{λ^{α}} φ(x)j(x).

Using

lim inf_{x→∞} \frac{\frac{1}{λ^{α}} φ(x)j(x)}{\frac{1}{1+α} φ(x)i(x)} = \frac{1 + α}{λ^{α}} > 1,

and (2), we have

lim_{x→∞} \frac{f(φ ( \frac{x}{λ} ))}{f ( \frac{1}{x} ∫_{0}^{x} φ(t) dt)} = lim_{x→∞} \frac{f ( \frac{1}{λ^{α}} φ(x)j(x))}{f ( \frac{1}{1+α} φ(x)i(x))} = ∞.

Finally

f ( \frac{1}{x} ∫_{0}^{x} φ(t) dt) ≪ \frac{1}{x} ∫_{0}^{x} f(φ(t)) dt, x → ∞.  □

Proof of Theorem 2.3. Let k ∈ N so that (1+2k)^{\frac{1}{k}} < 1+ϵ, g(x) = f(\sqrt[k]{x}) and φ(x) = [x + 1]^{k} (where [·] denotes the integer part). Obviously φ ∈ R_{k} and g ∈ R_{∞} . Using Theorem 2.2 we get

\frac{1}{n} ∫_{0}^{n} g(φ(x)) dx ≫ g ( \frac{1}{n} ∫_{0}^{n} φ(x) dx) , n → ∞.

RAPIDLY AND REGULARLY VARYING FUNCTIONS 95

This leads to

\frac{g(1) + g(2^{k}) + ⋯+ g(n^{k})}{n} ≫ g ( \frac{1 + 2^{k} + ⋯+ n^{k}}{n} ) , n → ∞.

Since lim_{n→∞} \frac{(1+2^{k} +···+n^{k})(1+2k)}{n^{k+1}} = \frac{2k+1}{k+1} > 1, by (2) we have

lim_{n→∞} \frac{g ( \frac{1+2^{k} +···+n^{k}}{n} )}{g ( \frac{n^{k}}{1+2k} )} = ∞.

Finally

\frac{f(1) + ⋯+ f(n)}{n} = \frac{g(1) + g(2^{k}) + ⋯+ g(n^{k})}{n} ≫ g ( \frac{n^{k}}{1 + 2k} ) , n → ∞,

and now by (2) and (1 + 2k)^{\frac{1}{k}} < 1 + ε we obtain

g ( \frac{n^{k}}{1 + 2k} ) = f ( \frac{n}{(1 + 2k)^{\frac{1}{k}}} ) ≫ f ( \frac{n}{1 + ε} ) , n → ∞  □

Proof of Theorem 2.4. We will use the inequality

(*)  \frac{1}{b − a} ∫_{a}^{b}f(t) dt ⩽ ( \frac{1}{b − a} ∫_{a}^{b}f(t)^{p}dt) ^{\frac{1}{p}}

where p > 1 and f is a nonnegative function. Let M > 1 be an arbitrary number and p > 1. Let 1 < λ ⩽ _{\frac{1}{1−(\frac{1}{2M} )^{\frac{p}{p−1}}}} ; then 2M (1 − \frac{1}{λ} ) ^{1− \frac{1}{p}} ⩽ 1. For sufficiently large x, using (3) and (*), we get

\frac{M}{x} ∫_{0}^{x} f(t) dt ⩽ \frac{2M}{x} ∫ x_{\frac{x}{λ}} f(t) dt = 2M \frac{λ − 1}{λ} \frac{1}{x − \frac{x}{λ}} ∫ x_{\frac{x}{λ}} f(t) dt

⩽ 2M \frac{λ − 1}{λ} ( \frac{1}{x − \frac{x}{λ}} ∫ x_{\frac{x}{λ}} f(t)^{p}dt) ^{\frac{1}{p}}

= 2M ( \frac{λ − 1}{λ} ) ^{1− \frac{1}{p}} ( \frac{1}{x} ∫ x_{\frac{x}{λ}} f(t)^{p}dt) ^{\frac{1}{p}} ⩽ ( \frac{1}{x} ∫_{0}^{x} f(t)^{p}dt) ^{\frac{1}{p}} . □

Proof of Theorem 2.5. Note that φ(x)^{q} ∈ R_{αq}, by (4) we have

∫_{0}^{x} φ(t)^{q}dt ∼ \frac{xφ(t)^{q}}{αq + 1} , x → ∞

∫_{0}^{x} f(t)φ(t) dt ∼ φ(x) ∫_{0}^{x} f(t) dt, x → ∞.

Now we apply Theorem 2.4 and obtain

φ(x) ∫_{0}^{x} f(t) dt = xφ(x)\frac{1}{x} ∫_{0}^{x} f(t) dt ≪ xφ(x) ( \frac{1}{x} ∫_{0}^{x} f(t)^{p}dt) ^{\frac{1}{p}} , x → ∞.

Finally

xφ(x) ( \frac{1}{x} ∫_{0}^{x} f(t)^{p}dt) ^{\frac{1}{p}} = x^{\frac{1}{q}} φ(x) ( ∫_{0}^{x} f(t)^{p}dt) ^{\frac{1}{p}}

∼ ( ∫_{0}^{x} f(t)^{p}) ^{\frac{1}{p}} x^{\frac{1}{q}} ( \frac{αq + 1}{x} ) ^{\frac{1}{q}} ( ∫_{0}^{x} φ(t)^{q}) ^{\frac{1}{q}} , x → ∞

∫_{0}^{x} f(t)φ(t) dt ≪ (αq + 1)^{\frac{1}{q}} ( ∫_{0}^{x} f(t)^{p}) ^{\frac{1}{p}} ( ∫_{0}^{x} φ(t)^{q}dt) ^{\frac{1}{q}} , x → ∞.  □

Proof of Theorem 2.6. Let M > 0, n ∈ N and λ = \frac{n− \frac{1}{2}}{n−1} > 1. For a sufficiently large x we have

\underline{f} ( \frac{k − \frac{1}{2}}{n} x) ⩾ \underline{f} ( λ\frac{k − 1}{n} x) > 2Mf ( \frac{k − 1}{n} x)

for every k ∈ {2, 3, …, n}. Now it follows

\frac{1}{x} ∫_{0}^{x} f(t) dt = \frac{1}{x} ∑_{k=1}^{n} ∫ _{\frac{k−1}{n} x}^{\frac{k}{n} x} f(t) dt > \frac{1}{x} ∑_{k=2}^{n} ∫ _{\frac{k− \frac{1}{2}}{n} x}^{\frac{k}{n} x} f(t) dt

⩾ \frac{1}{x} ∑_{k=2}^{n} \frac{x}{2n} \underline{f} ( \frac{k − \frac{1}{2}}{n} x) > \frac{1}{2n} ∑_{k=2}^{n} 2Mf( \frac{k−1}{n} x) = M \frac{f(\frac{x}{n}) + ⋯+ f(\frac{n−1}{n} x)}{n} . □

Proof of Theorem 2.7. Let n ∈ N and M > 1. Then for a sufficiently large x and every k ∈ {1, 2, …, n} we have by (5),

f( \frac{k}{n}x) ⩾ Mn^{2} \frac{k}{nx} ∫_{0}^{\frac{k}{n} x} f(t) dt ⩾ \frac{Mn}{x} ∫ _{\frac{k−1}{n} x}^{\frac{k}{n} x} f(t) dt.

Summing by k ∈ 1, 2, …, n we have

\frac{f ( \frac{x}{n} ) + f ( \frac{2x}{n} ) + ⋯+ f(x)}{n} ⩾ M \frac{1}{x} ∫_{0}^{x} f(t) dt.  □

References

1. N. H.Bingham, C. M. Goldie, J. L. Teugels, Regular Variation, Cambridge University Press, London, 1987.

2. N. K. Elez, D. Ž Djurčić, Some properties of rapidly varying functions, J. Math. Anal. Appl 401(2) (2013), 888–895.

3. J. Karamata, Sur un mode de croissance reguliere des functions, Mathematica, Cluj 4 (1930), 38–53.

4. D. S. Mitrinovic, J. E. Pecaric, A. N. Fink, Classical and New Inequalities in Analysis, Kluwer, 1993.

5. S. K. Simic, Integral kernels with regular variation property, Publ. Inst. Math., Nouv. Sér. 72(86), (2002), 55–61.

Department for Mathematics, Informatics and Physics (Received 09 12 2013) University East Sarajevo (Revised 15 12 2014) East Sarajevo, Bosnia and Herzegovina

    nelez@ffuis.edu.ba
    vvladicic@ffuis.edu.ba