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(1)

Internat. J. Math. & Math. S.

Vol.

6

No.

3

(1983) 609-611

6O9

A NOTE ON A PAPER BY $. HABER

A. NcD. NERCER

Department of Mathematics and Statistics University of Guelph

Guelph, Ontario, Canada NIG 2WI

(Received September 23, 1982 and in revised form February 26, 1983)

ABSTRACT. A technique used by S. Haber to prove an elementary inequality is applied here to obtain a more general inequality for convex sequences.

KEY WORDS AND PHRASES. Convex sequences, Hadamard’s inequality for convex functions, reaa ng em e

1980 MATHEMATICS SUBJECT CLASSIFICATION

C(E.

Primary 26D15, Secondary 40G05.

I.

INTRODUCTION.

Let a and b be non-negative. Then the following elementary inequality was proved in [i].

an

an-i

bn a+b.n

n$I[

+

b+.

+

e

(--) (n=0,1,2,...)...

(i.I)

Now this inequality can be obtained at once by taking f(t) tn

in the well- known result

b

f(t)dt >

f(--)

a+b. (I 2)

b-a a

which holds whenever f is convex in

[a,b].

However, the method used in [i] to obtain

(I.I)

is interesting and it is the purpose of the present note to show that it can be used to prove a considerably more general result about sequences. Indeed this more general result will have (1.2) as a consequence.

2. MAIN RESULTS.

A lemma which we shall use is the following LEMMA. If

>

2

_> >

0

>

I

m

and

m

E 0

=0 V

(2)

610

A.M.

MERCER

and if the ordering of the is such that each positive precedes all the nega- tive ones, then

m

: vl3v ->

0

v=O

This lemma, which is easily proved, is not the one stated by Haber but, essentially, it is what he used. For with b

i defined as in

(i 0

2, []:

n

even)

we do

not

in fact have

Y. b. 0 i=0

which is what is needed to apply the lemma quoted there.

Our result is the following.

n

THEOREM. Let

{u}=

0 be a convex sequence. Then

n n

1 n

11 u _>--

r. _(v)

u

n-{--"-

v=0 2n v

v=O

(2.1)

To

see that

(1.2)

is a consequence of

(2.1)

let the function

f(x)

be bounded and convex (and hence continuous) on

[a,b]

and take

Then

(2.1)

reads

u f(a

+ v__ (b-a))

n

n n

In+l

n--OE f(a

+

vn

(b-a))

> ^n

z v=O

E

()

n f(a

+

vn (b-a))... (2.2)

On letting n the left-hand side here tends to the left-hand side of (1.2). And by virtue of Bernsteln’s result

n

,im Y.

(3)q ( ’) x(l-x)

n-

+(x) (2.3)

n- =0

.a+b.

whenever E

C[O,I]

we see that the right-hand side of

(2.2)

tends to

f[---)

Merely take

(x)

f(a

+ x(b-a))

and x

I/2

in

(2.3) We

now proceed to prove

(2.1)

PROOF. Following Haber let us put Q

[]

and write

Q

, { 70qY + + 7Q

if n Is odd

E

7v 1

v-O

70+71 + + 7Q_ I + 7Q

if n is even

Then

n n

n

Q*

n+-- v=O r.

uv

- v=O r. (v)uv" v-O r. eEu

v

+

un-v

(3)

NOTE ON A PAPER BY S. HABER 611

where

Cv n+-- ’(X)

n Since {u

}n=0

is convex then

uv+ +

u

n-v-I

< u

+

un- (0<<Q-I)

which is to say that the sequence

u +

u

}Q=0

is non-lncreaslng,

we

see too that n-

the sequence {c

}Q=0

is non-lncreasing and that

*

c 0. Appealing to the Lemma

quoted above we find that --0

Q,

c[u

+u

--0 n-

and this complets the proof of

(2.1).

In conclusion I wish to thank the referee for his helpful advice concerning the lemma used here.

REFERENCES

I. HABER,

S. An Elementary Inequality, Internat. J. Math. and Math. Sci. Vol. 2 No. 3

(1979)

531-535.

2.

HARDY,

G.H., LITTLEWOOD,

J.E., POLYA, G., :Inequalities

Cambridge

(1973).

参照

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