Volume 2011, Article ID 179695,9pages doi:10.1155/2011/179695
Research Article
A New Proof of Inequality for
Continuous Linear Functionals
Feng Cui and Shijun Yang
College of Statistics and Mathematics, Zhejiang Gongshang University, Hangzhou 310018, China
Correspondence should be addressed to Feng Cui,[email protected]
Received 23 January 2011; Accepted 2 March 2011
Academic Editor: Andrei Volodin
Copyrightq2011 F. Cui and S. Yang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Gavrea and Ivan2010obtained an inequality for a continuous linear functional which annihilates all polynomials of degree at mostk−1 for some positive integerk. In this paper, a new functional proof by Riesz representation theorem is provided. Related results and further applications of the inequality are also brought together.
1. Introduction
Letk ≥ 1 be an integer andf ∈ Cka, b. Denote by
Èkthe set of all polynomials of degree
not exceedingk. LetL:Ca, b → Êbe a continuous linear functional which annihilates all
polynomials of degree at mostk−1; that is,
Lf0, ∀f∈Èk−1. 1.1
It is well known that a continuous linear functional is bounded, and finding the bound or norm of a continuous linear functional is a fundamental task in functional analysis. Recently, in light of the Taylor formula and the Cauchy-Schwarz inequality, Gavrea and Ivan in1obtained an inequality for the continuous linear functionalLsatisfying1.1. In order to state their result, we need some more symbols. Recall that theL2norm of a square integrable
functionfona, bis defined by
fL2a,b
b
a
fx2dx
1/2
and denote byt :max{t,0}the truncated power function. The notationLtft, smeans
that the functionalLis applied tofconsidered as a function oft. The main result of1can now be stated as follows.
Theorem 1.1. The functionalLsatisfies the following inequality:
L
f≤Mkfk
L2a,b, 1.3
where
Mk −
1k
2k−1!LsLtt−s
2k−1
1.4
are the best possible constants. The equality is attained if and only iffis of the form
fs CLtt−sk−1
−k
, s∈a, b, 1.5
whereCis an arbitrary constant and the symbol−kdenotes akth antiderivative off.
Remark 1.2. Usually, the functional L is allowed an interchange with the integral this is silently assumed throughout this paper. This is true in most interesting cases when, for example,Lis an integral or a derivative or a linear combination of them. If the interchange is permitted, then it is easily verified
Ltt−sk−1
−k
−1kk−1!
2k−1! Ltt−s
2k−1
ps, p∈Èk−1. 1.6
It should be pointed out that the inequality1.3can be found in Wang and Han2, Lemma 1 see also3. In this note, we will give a short account of historical background on inequality1.3. A new functional proof based on the Riesz representation theorem4,5is also given. Furthermore, some related results are brought together, and further applications are also included.
2. Historical Background
It is well known that a Hilbert space can be given a Gaussian measure. LetH be a Hilbert space equipped with Gaussian measure andLa continuous linear functional acting onH. Smale in6 a pioneering work on continuous complexity theorydefined an averagewith respect to the Gaussian measureerror for quadrature rules. A result of Smale6says that the average error is proportional toL. More precisely,
Av
f∈HL
f 2
Using 2.1, Smale was able to compute the average error for right rectangle rule, the trapezoidal rule, and Simpson’s rulesee6, Theorem D.
Later on, Wang and Han in2extended and unified results in6, Theorem D, and they also simplified the corresponding analysis given in6. The main observations in2are
iany quadrature rule has its algebraic precision, or equivalently, the corresponding quadrature error functional annihilates some polynomials,
iiand hence the Peano kernel theorem applies.
In fact, more can be stated. The quadrature rule in the above observations can be replaced by any continuous linear one. The main result and its elegant proof deserve to be better known. For reader’s convenience, they are recorded here. To do this, we need more notations. For brevity, let0,1 a, b. Denote byL2the square integrable functions on0,1, and
Hk
f∈C0,1|fk−1 is absolutely continuous andfk∈L2
. 2.2
The inner product onHkis defined by
f, gH
k
k−1
j0
fj0gj0 fk, gk
L2
. 2.3
Then,Hkis a Hilbert space of functions. The result in2can be now stated as follows.
Theorem 2.1. Let L : Hk → Ê be a continuous linear functional satisfying Lf 0, for all
f∈Èk−1. Then,
L2 −1k
2k−1!LsLtt−s
2k−1
. 2.4
It is easily seen thatTheorem 1.1is a rediscovery ofTheorem 2.1. For completeness, we record the original short but beautiful proof of2.4in2.
Proof. We have by the Peano kernel theorem
Lf
1
0
Gksfksds, 2.5
where
Gks 1
k−1!Ltt−s
k−1
. 2.6
And hence,
Obviously,
Gk2L2 L
f1
, 2.8
wheref1∈ Hksatisfying
f1kGk, almost everywhere. 2.9
Solving the above equality, we have
f1s −1 k
2k−1!Ltt−s
2k−1
ps, p∈Èk−1. 2.10
Applying the functionalLto both sides of2.10and noting2.7and2.8, we obtain2.4
as required.
3. A Functional Proof
It seems that the original proof of Wang and Han recorded in the previous section does not fully utilize the spaceHk. We now provide an alternative functional proof.
First, we define an equivalence relation∼onHkwith respect to its subspaceÈk−1since
Lvanishes onÈk−1. We say thatf ∼giff−g∈Èk−1. It is easy to check that the quotient space
Hk/Èk−1 is still a Hilbert space. For anyf ∈ Hk, there must exist a functionF ∼fsuch that
Fj0 0,j 0,1, . . . , k−1. For example,
Fx fx−f0−
k−1
j1
fj0
j! x
j 3.1
may serve this purpose. So, we may assume thatfj0 0,j 0,1, . . . , k−1, for anyf ∈
Hk/Èk−1and, the inner product onHk/Èk−1is
f, gH
k/
k−1
fk, gk
L2
. 3.2
The functionalLcan be viewed as acting onHk/Èk−1, since it vanishes onÈk−1. The Peano
kernel theorem can be rewritten as
Lffk, G k
L2
, 3.3
whereGkis defined by2.6. By the Riesz representation theoremsee, e.g.,4or5, there
exists a uniquef0∈ Hk/Èk−1such that
Lff, f0
Hk/k−1
, 3.4
Lf0Hk/
k−1
Lf0
From3.2and3.4
Lff, f0
Hk/k−1
fk, f0k
L2
. 3.6
From3.3and3.6, we have
f0kGk, almost everywhere, 3.7
which gives
f0s
−1k
2k−1!Ltt−s
2k−1
. 3.8
Applying the linear functionalLto both sides of the above equality gives
Lf0
−1k
2k−1!LsLtt−s
2k−1
, 3.9
which together with3.5yields
L2 −1k
2k−1!LsLtt−s
2k−1
, 3.10
as desired.
Remark 3.1. From the above proof, we see thatf0 given by 3.8 is the representer of the
Hilbert spaceHk/Èk−1.
4. Related Results and Further Applications
Numerical integration and quadrature rules are classical topics in numerical analysis while quadrature error functionals are typical continuous linear functionals on function spaces. It was quadrature error functionals that stimulated study of Smale6and Wang and Han2. So it is natural to consider the applications of2.4to quadrature error estimates.
Example 4.1. Let n be a positive integer,f ∈ Hk and x ∈ 0,1. Let the Euler-Maclaurin
remainder functionalLEMbe defined by
LEMf
1
0
ftdt− 1 n
n−1
i0
f
ix n
k−1
ν1
fν−11−fν−10
whereBνtis theνth Bernoulli polynomial. It is not hard to verify thatLEMvanishes onÈk−1.
So the norm ofLEMcan be calculated according to2.4. It can be found in2 cf.7which
gave a bound in terms of Bernoulli numberB2kB2k0; that is,
LEM
−1k−1
2k! B2k
Bkx
k!
21/2 1
nk. 4.2
Example 4.2. Letm,nbe positive integers andf∈ Hk. Suppose that the following quadrature
rule
1
0
ftdt
m−1
j0
pjf
xj
4.3
is exact for any polynomial of degree≤k−1 for some positive integerk. Then,
LCQf
1
0
ftdt− 1 n
n−1
i0 m−1
j0
pjf
ix
j
n
4.4
defines a composite quadrature error functional which annihilates anyf∈Èk−1. So Theorem
3 applies. The expression for the norm ofLCQcan be found in2. A different but easy-to-use
expression can also be found in7
LCQ 1
k!nk
⎧ ⎨ ⎩
m−1
i,j0
pipj
−1k−1k!2
2k! B2kxi−xj BkxiBkxj
⎫⎬
⎭
1/2
, 4.5
where Bk is the Bernoulli function, defined by Bkt Bk{t}. Here {t} stands for the
fractional part oft.
Example 4.3. The error functionals LM, LT, and LS for the midpoint rule, trapezoidal
quadrature and Simpson’s rule are, respectively,
LMf
1
0
ftdt−f
1 2
,
LTf
1
0
ftdt−1
2
f0 f1,
LSf
1
0
ftdt−1
6
f0 4f
1 2
f1 .
They vanishes onÈ1, È2, andÈ3, respectively. So, 4.5 appliessee7 for details. It is a
routine computation to find their norms and they can be found in2 some of them can also be found in6–8. In the following,H∗kstands for the dual space ofHk
LM H∗ 1 LT H∗ 1 1
2√3,
LM
H∗
2 1
8√5,
LT
H∗
2 1
2√30,
LS H∗ 1 1 6, LS H∗ 2 1
12√30,
LS
H∗
3 1
48√105,
LS
H∗
4 1
576√14.
4.7
From these and1.4, or equivalently2.4, we immediately obtain
LMf
1
0
ftdt−f
1 2 ≤ ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 1 2√3
f
L2, iff∈ H1,
1 8√5
f
L2, iff∈ H2.
LTf
1
0
ftdt−1
2
f0 f1≤
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 1 2√3
f
L2, if f∈ H1,
1 2√30f
L2, if f∈ H2.
LSf
1
0
ftdt−1
6
f0 4f
1 2
f1 ≤
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 6f
L2, if f∈ H1,
1 12√30f
L2, if f∈ H2,
1 48√105f
L2, if f∈ H3,
1 576√14
f4
L2, if f∈ H4. 4.8
Note that there is a mistake in Example 9 in1. The constant 1/576√14 in the last inequality is mistaken to be 1/1152√14 there.
Recently, there is a flurry of interest in the so-called Ostrowski-Gr ¨uss-type inequalities. Some authors, for example, see Ujevi´c9, consider to bound a quadrature error functional in terms of the Chebyshev functional, that is,fk2
L2− "1
0fktdt
2, for some appropriate
Proposition 4.4. Suppose that a continuous linear functionalL:Hk → Êvanishes onÈk−1. Then
for any nonnegativej < k, we have
L
f≤
# $ $ % −1j
2j−1!LsLtt−s
2j−1
⎛ ⎝fj2
L2 −
1
0
fjtdt
2⎞
⎠
1/2
. 4.9
Proof. Letpbe a polynomial inÈk−1such thatp
jt 1. Let
Ft ft−pt
1
0
fjtdt, f∈ Hk. 4.10
Then,F∈ Hkand
LF L
f−p
1
0
fjtdt
Lf, 4.11
sincep∈Èk−1andLvanishes onÈk−1. Obviously,
L
f|LF| ≤ LFj
L2
. 4.12
Moreover, by notingpjt 1, we have
Fj
L2
fj−
1
0
fjtdt
L2
. 4.13
It is trivial to check that
fj−
1
0
fjtdt
2
L2
fj2
L2 −
1
0
fjtdt
2
. 4.14
From4.12–4.14and2.4, follows4.9. This completes the proof.
Note that Proposition 4.4 shows that we have a corresponding inequality 4.9 for everyj < kwhenever we have inequality2.4. It should be mentioned, however,4.9does not hold forkin general especially when the kernel ofLis exactlyÈk−1.
Proposition 4.4can be reformulated in a slightly different language as follows.
Corollary 4.5. Suppose thatLis a continuous linear functional acting on Hk andkerL {f |
Lf 0}Èk−1. Then for any nonnegativej < k, both2.4and4.9hold while only2.4is also
valid fork.
Example 4.6see also7. FromExample 4.3andProposition 4.4, we have
1
0
ftdt−1
6
f0 4f
1 2
f1 ≤ 1 6
⎛ ⎝f2
L2−
1
0
ftdt
2⎞
⎠
1/2
. 4.15
In view ofProposition 4.4orCorollary 4.5, the above inequality is still valid withfreplaced byfandf, respectively, and with obvious change in the coefficients. We omit the details.
Acknowledgment
The work is supported by Zhejiang Provincial Natural Science Foundation of China
Y6100126.
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