Download ANOTHER PROOF FOR JACKSON THEOREM Jackson Theorem is

ANOTHER PROOF FOR JACKSON THEOREM
THANKS TO: LUDMIL ZIKATANOV
Jackson Theorem is an example of so called direct theorems of approximation thory.
We will prove two variants of the Jackson theorem: for approximation with trigonometric
polynomials and approximation with algebraic polynomials.
1
Theorem 0.1 (Jackson theorem). If f ∈ W∞
(I) then there exists trigonometric polynomial
T (x) of degree n such that
c
(0.1)
En (f ) ≤ kf ′k∞ .
n
Before we move to the proof of this theorem, let us first introduce some notation and
state and prove some results that are needed in the proof of the theorem. Define Kn (t) to
be the Jackson kernel:
4
sin nt
−1
2
,
Kn (t) := λn
sin 2t
where λn is defined as
4
Z π
Z π
sin nt
−1
2
dt, and hence
Kn (t) dt = 1.
λn =
sin 2t
−π
−π
We now define for a given function f ∈ L∞ (−π, π):
Z π
Jn (f ; x) =
f (x + t)Kn (t) dt.
−π
Below we prove that Jn (f ; x) is in fact a trigonometric polynomial, which gives the bound
stated in the theorem. In that sense, the proof here is a constructive proof, explicitly giving
a trigonometric polynomial T (x) for which
c
kf − T k∞ ≤ kf ′k∞ .
n
If g(t) is an integrable 2π periodic function, and a is a real number, by change of variables t = τ + 2π we have that
Z −π+a
Z π+a
Z π+a
g(τ ) dτ =
g(t − 2π) dt =
g(t) dt
−π
π
π
Date: January 30, 2008.
1
2
THANKS TO: LUDMIL ZIKATANOV
Hence,
Z
π
g(t) dt =
−π
=
−π+a
Z
−π
Z π+a
π
π
Z
g(t) dt +
Z
g(t) dt +
g(t) dt
−π+a
π
g(t) dt =
−π+a
Z
π+a
g(t) dt.
−π+a
In summary,
Z
(0.2)
π+a
g(t) dt =
−π+a
Z
π
g(t) dt
−π
Lemma 0.2. The following idenatities hold
sin2 n2 x
2n − 1
3x
x
+ . . . + sin
x =
sin + sin
2
2
2
sin( x2 )
(0.3)
m−1
X
2m − 1
x
cos(kx)] = sin
x
sin( )[−1 + 2
2
2
k=0
Proof. Denote bl = cos(lx), and recall that for any α and β we have
2 sin α sin β = cos(β − α) − cos(α + β).
Hence
x
2 sin( ) sin
2
2l − 1
x = bl−1 − bl .
2
The first identity is a consequence of the following:
X
n
n
x X
x
2l − 1
2l − 1
sin( )
sin
x =
sin( ) sin
x
2 l=1
2
2
2
l=1
n
1
1X
bl−1 − bl = (b0 − bn )
2 l=1
2
n 1
=
(1 − cos(nx)) = sin2
x .
2
2
To prove the second identity in (0.3), we set al = sin 2l−1
x
, and recall that
2
2l + 1
2l − 1
x
x − sin
x = al+1 − al .
2 cos(lx) sin( ) = sin
2
2
2
=
ANOTHER PROOF FOR JACKSON THEOREM
3
We then have
m−1
X
x
cos(kx)
sin( ) −1 + 2
2
k=0
= a1 +
m−1
X
!
m−1
X
x
x
= sin( ) +
2 sin( ) cos(kx)
2
2
k=1
(ak+1 − ak ) = am ,
k=1
which completes the proof.
We now use the above identities to show that Kn (t) is a trigonometric polynomial.
Lemma 0.3. If f is a 2π-periodic function and f ∈ C(−π, π) then Jn (f ; x) is a trigonometric polynomial of order (2n − 2). If f is an even function than Jn (f ; x) is also even.
Proof. Denote
Zn (t) =
sin(nt/2)
sin(t/2)
2
,
we only need to show that Zn is a trigonometric polynomial of order (n−1). Indeed, since
Kn (t) is proportional to [Zn (t)]2 , this will imply that Kn (t) is a trigonometric polynomial
of order (2n − 2). Applying the first identity from (0.3), for Zn (t) we get
n
X
1
2m − 1
sin2 (nt/2)
=
sin
t .
Zn (t) =
sin(t/2) m=1
2
sin2 (t/2)
We now apply the second identity in (0.3) to each term in the sum, to get that
"
#
n
m−1
X
X
1
− sin(t/2) + 2 sin(t/2)
cos(kt)
Zn (t) =
sin(t/2) m=1
k=0
=
n
X
(−1) + 2
n−1
X
k=0
= −n + 2
sin(t/2)
m=1
m=1
= −n + 2
n
X
n−1
X
cos(kt)
m−1
X
cos(kt)
k=0
n
X
1
m=k+1
(n − k) cos(kt).
k=0
Clearly this shows that Zn (t) is a trigonometric polynomial of order (n − 1) and hence
Kn (t) is a trigonometric polynomial of order (2n − 2). Note also that Kn (t) is even and
4
THANKS TO: LUDMIL ZIKATANOV
is a sum of cosine functions only. This is easy to see by squaring Zn (t) and using simple
trigonometric identities.
Z π
Z π+x
(0.4)
Jn (f ; x) =
f (x + τ )Kn (τ ) dτ =
f (t)Kn (t − x) dt.
−π
−π+x
As a consequence, we obtain that
2n−2
X
Kn (t − x) =
αk cos(k(t − x))
k=0
2n−2
X
=
αk cos(kx) cos(kt) +
k=0
2n−2
X
αk sin(kx) sin(kt).
k=1
We then substitute the above expresion in (0.4) and use the fact that f (·), sin(kt) and
cos(kt) are 2π-periodic functions. From (0.2) we then have:
Z π+x
2n−2
2n−2
X
X
αk ηk sin(kx),
αk ξk cos(kx) +
f (t)Kn (t − x) dt =
−π+x
k=1
k=0
where
ξk =
Z
π+x
Z
π+x
f (t) cos(kt) dt =
ηk =
π
Z
π
f (t) cos(kt) dt,
−π
−π+x
and
Z
f (t) sin(kt) dt =
−π+x
f (t) sin(kt) dt.
−π
Hence, from (0.4) it follows that Jn (f ; x) takes the form
Z π
Jn (f ; x) =
f (t)Kn (t − x) dt
−π
=
2n−2
X
k=0
αk ξk cos(kx) +
2n−2
X
αk ηk sin(kx).
k=1
and clearly is a trigonometric polynomial of the same order as Kn (t).
To conclude the proof of the lemma, it remains to show that if f (·) is even function,
then Jn (f ; x) is also even. Since in such case f (t) sin(kt) is an odd function, we obtain
that ηk = 0, for k = 1, . . . , (2n − 2), and hence Jn (f ; x) is even function.
We now turn our attention to the proof of several inequalities from qhich the Jackson
theorem will follow.
ANOTHER PROOF FOR JACKSON THEOREM
5
Lemma 0.4. The following inequalities hold:
t/π ≤ sin(t/2) ≤ t/2,
(0.5)
for all t ∈ [0, π].
32 3
n.
π3
Proof.
(1) To prove that sin(t/2) ≥ t/π for t ∈ [0, π], we set g(t) = sin(t/2) − t/π.
Clearly we have g ′′ = − 41 sin(t/2) ≤ 0. Hence g(t) can only have maximum
inside the interval [0, π] and the minimum value is achieved at one of the end
points. Direct calculation shows that g(0) = g(π) = 0 and hence g(t) ≥ 0 for all
t ∈ [0, π].
(2) To prove the left side inequality in (0.5), we now set g(t) = t/2 − sin(t/2). It is
easy to see that g ′(t) ≥ 0 for all t ∈ [0, π] and hence g(t) ≥ g(0) = 0, which
concludes the proof of (0.5).
(3) The inequality (0.6) is shown as follows:
4
4
Z π
Z π
n
sin(nt/2)
sin(nt/2)
λn = 2
dt ≥ 2
dt
sin(t/2)
sin(t/2)
0
0
4
Z π
n
32
π 24 n4
(nt/π)
= 3 n3 .
dt = 2
≥ 2
4
(t/2)
n π
π
0
λn ≥
(0.6)
We will also need the following result.
Lemma 0.5. There exists an absolute constant c, such that:
Z π
c
tKn (t) dt ≤
n
0
Proof. Using Lemma 0.4, we obtain
Z π
n−1 Z (k+1)π/n
X
tKn (t) dt
tKn (t) dt =
0
kπ/n
k=0
≤ λ−1
n
"Z
"
π/n
t
0
n−1
nt/2
t/π
4
dt +
n−1 Z
X
k=1
#
(k+1)π/n
kπ/n
n2 π 6 X 2 k + 1
π
+
λn 32
k4
k=1
"
#
∞ 2
6
X
1
1
n π
+ π2
+ 4
≤ cn−1
≤
3
λn 32
k
k
k=1
=
#
(k + 1)π n 4
dt
n
k
6
THANKS TO: LUDMIL ZIKATANOV
We are now ready to prove Theorem 0.1.
Rπ
Proof of Theorem 0.1. Recall that −π Kn (t) dt = 1, and hence, for all x ∈ [−π, π] we
have
Z
π
Jn (f ; x) − f (x) =
(f (x + t) − f (x))Kn (t) dt
−π
Note that Kn (t) is non-negative and even and by Lemma 0.5 we have:
Z π
|Jn (f ; x) − f (x)| = (f (x + t) − f (x))Kn (t) dt
−π
Z π
Z π Z x+t
′
Kn (t) dt
≤
|f (x + t) − f (x))|Kn (t) dt =
f
(τ
)
dτ
−π
−π
x
Z π Z x+t
Z π
≤
|f ′ (τ )| dτ Kn (t) dt ≤ kf ′ k∞
|t|Kn (t) dt
−π
x
−π
Z π
′
= 2kf k∞
tKn (t) dt ≤ 2cn−1 kf ′ k∞ .
0
Setting N = 2(n−1), and recalling that Jn is a trigonometric polynomial of order 2(n−1),
for the error of best approximation we have that:
EN (f ) ≤
4c
kf ′ k∞ ,
N +2
and the proof is complete.
Remark 0.6. We remark that if f is even function than, since Jn (f ; x) is also even, we can
write
(0.7)
inf
T ∈Tn,even
kf − T k∞ ≤ cn−1 kf ′ k∞ .
The inequality just stated in the remark gives the following theorem for approximation
with algebraic polynomials.
1
Theorem 0.7. Let f ∈ W∞
(−1, 1). Then
En (f ) ≤ cn−1 kf ′ k∞
Proof. Set g(t) = f (cos t), for t ∈ [−π, π]. Note that g(t) is even and
g ′(t) = −f ′ (cos t) sin t,
and hence
kg ′k∞,(−π,π) ≤ kf ′ k∞,(−1,1) .
From Lemma 0.3 we may conclude that Jn (g; x) is even. Define now
J˜n (f ; x) := Jn (g; arccos(x)).
ANOTHER PROOF FOR JACKSON THEOREM
7
Note that J˜n (f ; x) is an algebraic polynomial of degree ≤ 2(n − 1). We then have
kf − J˜n (f )k∞,(−1,1) = kg − Jn (g)k∞,(0,π)
= kg − Jn (g)k∞,(−π,π) ≤ cn−1 kg ′k∞,(−π,π)
≤ cn−1 kf ′ k∞,(−1,1) .
If the function f is smoother, then we have the following theorem:
Theorem 0.8. Let f ∈ C k is 2π periodic function. Then, the following estimate holds:
c(k)
En (f ) ≤ k kf (k) k∞ .
n
Proof. We will first prove the following inequality:
c
(0.8)
En (f ) ≤ En (f ′ ).
n
To show (0.8), we fix f and denote with q the trigonometric polynomial of order n for
which we have that
kf ′ − qk∞ = En (f ′ ).
Clearly, we may write q in the form
n
X
q(x) = a0 +
ak cos(kx) + bk sin(kx) = a0 + r(x).
k=1
Define now
s(x) =
Z
x
r(t) dt.
−π
Note that s(x) is a trigonometric polynomial of order n. Hence
c ′
En (f ) = En (f − s) ≤
kf − s′ k∞
n
c ′
=
(0.9)
kf − rk∞
n
c
c
[kf ′ − qk∞ + |a0 |] = [En (f ′ ) + |a0 |].
≤
n
n
Since f is 2π-periodic, we have that
Z π
f ′ (τ ) dτ = f (π) − f (−π) = 0.
−π
Rπ
Direct calculation using the definition of r(x) shows that −π r(τ ) dτ = 0, as well. Thus
Z π
Z π
′
0=
f − r dτ =
f ′ − q + a0 dτ,
−π
−π
8
THANKS TO: LUDMIL ZIKATANOV
and hence
2πa0 =
Z
π
q − f ′ dτ,
−π
Taking abosolute values on both sides of this identity ad by some obvious inequalities, we
get that
2π|a0 | ≤ 2πkf ′ − qk∞ = 2πEn (f ′ ).
The proof is then concluded by estimating |a0 | with En (f ′ ) in (0.9). The statement of the
theorem follows by several applications of the inequality (0.8).
R EFERENCES
[1] Frank Deutsch, Best approximation in inner product spaces, CMS Books in Mathematics/Ouvrages de
Mathématiques de la SMC, 7, Springer-Verlag, New York, 2001. MR MR1823556 (2002c:41001)
[2] Ronald A. DeVore and George G. Lorentz, Constructive approximation, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 303, Springer-Verlag,
Berlin, 1993. MR MR1261635 (95f:41001)
[3] Eugene Isaacson and Herbert Bishop Keller, Analysis of numerical methods, Dover Publications Inc.,
New York, 1994, Corrected reprint of the 1966 original [Wiley, New York; MR0201039 (34 #924)].
MR MR1280462
[4] S. M. Nikol′ skiı̆, Priblizhenie funktsii mnogikh peremennykh i teoremy vlozheniya; Approximation of
multivariate functions and embedding theorems, “Nauka”, Moscow, 1977, (In Russian). MR MR506247
(81f:46046)
[5] P. P. Petrushev and V. A. Popov, Rational approximation of real functions, Encyclopedia of Mathematics and its Applications, vol. 28, Cambridge University Press, Cambridge, 1987. MR MR940242
(89i:41022)