A theorem on Riesz summability <span class="mathjax-formula">$(R,\omega , 2)$</span> on Banach space

C OMPOSITIO M ATHEMATICA

S HEN -C HI T ANG

A theorem on Riesz summability (R, ω, 2) on Banach space

Compositio Mathematica, tome 17 (1965-1966), p. 167-171

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167

on

Banach

_space

by

0160hen-Chi Tang

A series

03A3i=~i=0

^{xi is said to} be summable

(R,

^m,

2),

^{to sum s, if}

tends to a

limit s,

^{as 03C9 ~ oo.}

R203C9

^{is called the Riesz mean of second}

order, (R,

^00,

2), of 03A3xi. Similarily,

^we say

that 03A3xi/ip

^{is sum-}

mable

x*,

^if

On the other

hand,

^Cèsaro

summability, (C, 2), of 03A3xi

^or

03A3 xi/ip

is defined as follows:

or

The

relationship

between these two summabilities in our case can be reduced to the

following

^lemma:

LEMMA.

where w ⁼

n+03BB,

0 ~ 03BB ^1. Since its truth is

trivial,

^we omit the _prove here.

168

A normed linear space ^{becomes a} metric space if ^the distance

d(x, y)

is defined

as ~x-y~,

^and it is called a Banach _space if it is

complete

in this metric. Banach space is

generally

^a

complex

space.

However,

it is assumed to be a real Banach _space

throughout

this paper. But ^our conclusions can

readily

^be

generalized

^to

complex

^case.

The purpose ^of this _paper is to prove Theorem A.

THEOREM A. Assume that 0 q ¹ and 0 p q.

1 f

^there

exists an element x

~ X,

^{where X is a Banach} space, such that

then we can choose another element x*

from

^{X such that}

PROOF. Our

proof depends

^{on the}

following

relations. From

(1)

and

hypothesis,

^{we have}

If,

^{based on}

(1),

^{we can} choose x* e X such that

then,

follows. The crucial

point

^{in the}

proof

^{is how to deduce}

(4)

^from

(3).

^{If we} suppose x ⁼ 0,

(3)

^is

Now,

Set m &#x3E; n.

Then,

Set

Then,

By

the "3rd Abel

Transformation",

^{we have}

From

(5)

^and

4,(1)

⁼

0(ip+1-i),

And

Hence,

170

Finally,

Let

Imn

^{denote the first term of the above}

^equality.

^{And we}

employ

^the first Abel Transformation for it.

Hence,

From the

previous results,

^{we have}

where

Similarily,

^{if we set}

then

It is

ready

^to prove

and

where

Hence

where

Finally,

Since X is a

complete

space, there exists x* E X such that

as m ~ 00. We have

completed

^our

proof.

(Oblatum 13-4-64)