CC BY
25Probl. Anal. Issues Anal. Vol. 8 (26), No3, 2019, pp. 3-15
DOI: 10.15393/j3.art.2019.7110
3
UDC 517.521.2
G. G. Akniyev
APPROXIMATION PROPERTIES OF SOME DISCRETE FOURIER SUMS FOR PIECEWISE SMOOTH DISCONTINUOUS FUNCTIONS
Abstract. Denote by Ln, n(f,x) a trigonometric polynomial of order at most n possessing the least quadratic deviation from f with respect to the system [tk = u + 2N^}fc=0 , where u e R and n ^ N/2. Let D1 be the space of 2n-periodic piecewise continuously differentiable functions f with a finite number of jump discontinuity points —n = < ... < = n and with absolutely continuous derivatives on each interval ((i,(i+1). In the present article, we consider the problem of approximation of functions f e D1 by the trigonometric polynomials Ln, N(f,x). We have found the exact order estimate |f(x) — Ln,N(f,x)| ^ c(f,e)/n, |x — {i| ^ e. The proofs of these estimations are based on comparing of approximating properties of discrete and continuous finite Fourier series.
Key words: function approximation, trigonometric polynomials, Fourier series
2010 Mathematical Subject Classification: 41A25
1. Introduction. Let D1 be the space of 2n-periodic functions f, each of which has a finite number of jump discontinuity points Q(f) = = l£i}I=o, where —n = £o < £1 < ... < = n, f (&) = (f (£ — 0) + + f (£i + 0))/2 and has an absolutely continuous derivative f on each interval (£i,£i+1) (0 ^ i ^ m) (here we say that a function f is absolutely continuous on an interval (a,b) if the function f is absolutely continuous on the segment [a, b], where f (x) = f (x) for x e (a, b), f (a) = f (a + 0), and f (b) = f (b — 0)). One of such functions is f (x) = sign(sinx).
Denote by Ln,N (f, x) (1 ^ n ^ |_N/2_|) the trigonometric polynomial of order at most n that possesses the least quadratic deviation from the function f with respect to the system {tk}N—1, where tk = u + 2nk/N (u e R). In other words, the minimum of the sums Y1 fc=o1 If (tk) — Tn(tk)|2
© Petrozavodsk State University, 2019
on the set of trigonometric polynomials Tn of order n is attained when Tn = Ln,N(f)• In particular, L|_n/2J,n(f,tk) = f(tk). It is easy to show (see [13]) that for n < N/2 the polynomial Ln,N(f,x) can be represented as follows:
n 1 N — 1
Ln,N(/,x) = £ cVN)(/)eivx, cVN)(/) = - £ f (tk)e
v=—n k=0
-ivtk .
and for n = —/2:
(N ) , —
Ln/2,N (f,x) = Ln/2- 1, N (f,x) + aN/2 (/) cos —(x - u)
where
n- i
an2n)(f) = aNN/2(f) = - £ /(tk) cos ^(tk - u).
k=0
By Sn(f, x) we denote the partial Fourier sum of order n of /:
n
Sn(/, x) = — + ^^ (ak cos kx + bk sin kx), k=i
where
ak = — /(t) cos ktdt, bk = — /(t) sin ktdt. n J n J
—n —n
To read more about approximation of functions by trigonometric polynomials, see [4-7], [9-12], [14].
Also, later we will need the function
hp(x)
and the well-known inequalities
cos x, p = 0, sin x, p =1
£
k=i
sin kx
k
^ -, 2'
£hP(kx)
k=i
Isin 2 I
x = 2in, i = 0, ±1, ±2,
ro 1 o
Lko=£ • (5)
k2 6 fc=i
It is easy to show, that the Fourier series converges pointwise for any function f e D1 and, therefore, the function can be represented as follows:
f (x) = — + ^^ (ak cos kx + sin kx) • fc=i
In the previous works, the author found estimates for the value |f(x) — Ln,N(f,x)| for 2n-periodic piecewise-linear and piecewise-smooth continuous functions (see [1], [2]). Also, two particular cases of such functions - 2n-periodic functions f (x) = |x| and f (x) = sign x, x e [—- were considered in [3]. The goal of this work is to estimate |f (x) — Ln,N(f,x)| for f e D1 as n, N ^ x>. We obtained the following result:
Theorem 1. For a function f e D1, the following estimate holds:
If(x) — (f,x)| ^ , 1 ^ n ^ LN/2J, |x — >e, (6)
n
where i = 0,1,..., m. The order of this estimate cannot be improved.
To prove this theorem, we use a lemma from [13]:
Lemma 1. [13] If the Fourier series of f converges at the points tk = u + 2kn/N, then the representation
Ln,N (f,x) = Sn(f,x) + R„, N (f,x),
where
2 A n
Rn,N(f,x) = -> f(t)Dn(x — t) cos^N(u — t)dt, (7)
n i=1 {
^ —n
1 n
D„(x) = 2^£cos kx, (8)
fc=i
holds true when 2n < N.
From this lemma, we have the following estimate:
|f(x) — Ln,N(f,x)| ^ If(x) — Sn(f,x)| + |Rn,N(f,x)|, n < N/2. (9)
In the case 2n = N, from (1) and (9) we have
If (x) — Ln, N (f,x)| ^
^ If(x) — Sn-1(f,x)| + |Rn-1,N(f,x)| + |anN^(f)|, n = N/2. (10 The estimate for |f(x) — Sn(f, x)|, where f e D1, were obtained in [8]:
|f (x) — Sn(f,x)| ^ , |x — £i| ^ e.
'11)
n
Now we have to estimate the values |Rn,N(f,x)| and |an2n)(f)|, which is done in the following sections.
2. The estimate for |Rn,N(f,x)|. From (7) and (8), we can get the representation
1
Rn,N (f,x) = -£ f (t)cos ^N (u — t)dt+
+— f (t) cos k(x — t) cos ^N(u — t)dt
m=1_
fc=1
Rn, N (f, x) + Rn, N (f, x).
Lemma 2. For a e (0, 2], the following inequality holds:
^^ sin kx
k
k — S)
^ c.
Proof. Performing the Abel transformation (summation by parts), we get
sin kx k — $)
e
e
k=1
1 — — - _
1 k2 1
e
(k+1)2 / j=1
sin jx j
e
a
(2 + k
=1k2 (1 +1)2 (1 — U) (1 —
(k+1)2
k
1 sin jx j
j=1 J
Using (5) and the inequalities
a2
(2 + k
(1 + k)2 (1 — H) (1 —
16
^ —, ■2 N 15'
k
1 sin j x
k^ j j=1 J
1
1
2
a
1
2
a
we have
ro . ,
sin kx k= k i1 — )
A 1 a2 (2 + k
k2 (1 +1)2 (1 — fi) (1 — H+F)
k . .
1 sin jx j
j=i J
C c.
This completes the proof. □
Lemma 3. For f E D1, the following holds:
n
f (t)hp(k(t — x))hq (uN (t — u))dt =
—n
^ff^2 £ (f (6 — 0) — f (6 + 0)) hp(k(6 — x))hi—q(uN(6 — u))
i=0
(—1)q uN
(UN)2 — k2 fo
(uNf—^ I f (t)hp(k(t — x))hi—q(uN(t — u))dt+
—n
m— 1
+ (( J» _ (f (6i — 0) — f (6i + 0)) hi—p(fc(6 — x))hq (uN (6i — u))
) k i=o
( 1)21+Pk if (t)hi— p(k(t — x))hq (uN (t — u))dt. (12)
(UN )2 — k2
—n
Proof. Perform integration by parts: / f (t)hp(k(t — x))hq(uN(t — u))dt
(_1)q m—1
E (f (6 — 0) — f (6 + 0)) hp(k(6i — x))hi—q(uN(6i — u)) — u i=0
n
)q
(-i)q
J f' (t)hp(k(t — x))hi—q (uN (t — u))dt+
(— 1)p+® k
+ — f (t)h1-p(k(t — x))h1-q (uN (t —u))dt. (13)
Repeat integration by parts for the last integral in (13) f (t)hp(k(t — x))h® (uN (t — u))dt =
—n
m— 1
(_1)q m—1
^ E (f (£i — 0) — f (£i + 0)) hp(k(£i — x))h1—q(uN(£i — u)) —
i=o
(-1)® f ,
v ' f (t)hp(k(t — x))h1—q(uN(t — u))dt+
UN J
—n
m— 1
(_1)1+pk ^1
+ ( , In2 E (f (£i — 0) — f (£i + 0)) h1— p(k(£i — x))h®(uN(£i — u))— (UN) 1=0
( ( UNf^ i f' (t)h1—p(k(t —x))hq(uN(t —u))dt+
—n
n
k2
+ ^7^2 / f (t)hp(k(t — x))h® (uN (t —u))dt.
—n
By moving the last integral to the left-hand side and dividing both sides by 1, we get (12). □
Lemma 4. The value |R,N (f, x) |, where f e D1, can be estimated as follows:
|<N(f,x)| ^ f. Proof. Performing integration by parts twice, we get
Rn, n (f,x) =
1 ro n 1 ro m—1
= 1 E f (t) cos uN (t — u)dt =1 EE f (t) cos uN (t — u)dt = n n ^=1 i=0 &
œ -, m—1
nN e " e (f (Ci - 0) - f (Ci + 0)) sin(Ci - u) + n ¿=1 ß i=0
+
1 A 1
nN2 E ß ¿=1 r
m— 1
e (f'(Ci - 0) - f (Ci + 0)) cosßN(Ci - u)- / f ''(t) cos ßN(t - u)dt
Applying some simple transformations and using (3), we have
m— 1
Rn(f, x) I ^ — E |f (Ci - 0) - f (Ci + 0)|
+
1 ^ 1
e
nN^ ß2
¿=1 p
i=0 m— 1
e
¿=1
sin ßN(Ci - u)
Ef' (Ci - 0) - f (Ci + 0) + / f'' (t)
i=0
ß
dt
+
cf)
N '
This completes the proof. □
Lemma 5. The value \RnN(f,x)|, where f E D1, can be estimated as follows:
(f, x) I ^
f) N
|x - Ci| ^ e.
Proof. Using Lemma 3, we have
Rn, N(f, x) = - I f (t) cos k(t - x) cos ßN(t - u)dt
¿=1 fc=1_n
m1
2 v—„ss sinßN(Ci - u) ^^ cosk(Ci - x)
nN
E(f (Ci - 0) - f (Ci + 0))£
i=0
¿=1
ß
e
k=1 1 - (¿N
—2 ~ 1 ™ 1
LßL
-2 f (t) cos k(t - x) sin ßN(t - u)dt+
nN ^ß ti 1 - (A) —n
m1
+je (f (ci - 0) - / (ci+0)) e cos ßn'ci - u) ;k sin ^ -x)+
n i=0 ß k=1 1 - (¿n)
2
n
2
n
+
2 ^ 1 n
£
k
nN2 ^ ß2 (k \2
k=1 1 - (^v) -n
f (t) sin k(t — x) cos ßN(t — u)dt
= Cn (f, x) + R;2n (f, x) + Rn,3N (f, x) + <4n (f, x).
Here we estimate only the values IR^n(f,x)| and IR/n(f,x)|, because |Rn;3w(f,x)| and | R'4N (f, x)| can be estimated in the similar way. Begin with | Rn,1N(f, x)|. Consider the expression
A = £cos k(& — x)£
sin ßN (C — u)
k=1
M=1
1 ß ( 1 —
Applying the Abel transformation, we get
^ sinßN(& — u) ^ .ic X.
A = ~f—;—t^yz^ cos j — x) +
m=1 ß 1
_ i jn_
un
j=1
n-1 ro +£ £
k=1 ^=1
sin ßN (£j — u) ß
1
1
V1 — 1 — (ft1) /
£cos j (& — x). j=1
Using (4), Lemma 2 and the fact that
1
1
k
2 + k
1— (£) 1— (»)'
(ßNH1 — (ä)2 1 — 0+1
we get
|A| ^
I sin
gi-x i '
From this, we get the estimate for IR^n(f,x)|:
KN(f,x)| ^
c
N
m-1
£
j=0
f & — 0) — f (C + 0)
sin
gi-x
c(f,e)
N :
|x — ^ e. (14)
n
2
2
c
In the similar way, we get the estimate
|<3n (/,*)|^ /, |x - Cil ^ e.
Now we estimate |R'2N(/, x) |. Consider the integral
:i5)
B
/ (t) cos k(t — x) sin ßN(t — u)dt.
Using Lemma 3, we estimate the value |B| as follows:
|B| ^
m— 1
E|/' (Ci — 0) — / (Ci + 0) + f (t)
i=0
dt
c(f)
ßN '
Now we have
Rn, N(/,x)|
B
„ ^ ^ n
_ V-V_
nN ßf^ 1 /fc \2 k=1 1 — (^^NJ
c(/) N '
:i6)
The value \ Rn_4/v(f, x) \ can be estimated in the similar way:
R4n(/,x)|^ / ■
:i7)
From (14)-(17) we have
(/,x)| ^
c(/,e)
N
|x — Ci| ^ e.
Lemma is proved. □
Finally, from Lemmas 4 and 5, we have
|Rn,N(/,x)| ^ , |x — Cil ^ e.
:is)
3. The estimate for |ai2n)(f )|. From (2), using that tj = u + 2nk/N, we have
'(N) (/) = N e (—1)'/(tk) = N e (/(t2k) — /(t2k+l))
2n— 1
n— 1
N
fc=0
N
fc=0
n
n
C
n
and
n— 1
|aN(fN e |f (t2k) — f(t2k+1)|.
N k=o
Denote by G the subset of indexes {k}n=0, such that for k e G the segment [t2k, t2k+1] does not contain any point £i, 0 ^ i ^ m. Denote G = {k}n—0\G. Now write
|aN(f)| ^ N e |f (t2k) — f (t2k+1)| + N e |f (t2k) — f (t2k+1)| . (19)
kec
For each k e G, the segment [t2k,t2k+1] lies entirely inside some interval (£i,£i+1) and, therefore, the function f is differentiable on it, which allows us to use the mean-value theorem and get the following inequality:
c(f)
|f(t2k) — f(t2k+1 )| ^ C(f) 112k — t2k+1| ^ f. (20)
For a k e G, there are s(k) points 1 < 2 < ... < £ifc s(fc) inside the segment [t2k,t2k+1]. Now we estimate the value |f(t2k) — f(t2k+1)| for k e G. First, we need the following lemma:
Lemma 6. For f e D1 and the segment [a,b], where [a, b] C [—n,n], the following holds:
|f(a) — f (b)| ^ c(f)(s + |a — b|),
where s is the number of jump discontinuity points x 1, x 1, . . . , x s of the function f on the segment [a,b].
Proof. Here we consider only the case a < xi < ... < xs < b. The proof for the cases a = x1 or b = xs is similar. Consider the following inequality:
s
|f (a) — f (b)| ^ |f (a) — f (x1 — 0)| + e |f (xi — 0) — f (xi + 0)| +
i=1
s—1
+ e |f (xi + 0) — f (xi+1 — 0)| + |f (xs + 0) — f (b)|.
i=1
Function f is differentiable on each of the intervals (a,x1), (x1,x2), ..., (xs—1,xs), (xs,b). Using the mean-value theorem, we can write
|f(a) - f (b)| ^ c(f)|a - b| + L |/(xi - 0) - f (x + 0)| ^
i=1
^ c(f)|a - b| + sM ^ c(f)|a - b| + c(f)s,
where M = max |f'(x, - 0) - f'(x, + 0)|. □ From this lemma
L |f (t2k) - f (t2k+1 )| ^
^ c(f w s(k) + |) ^ c(f) L s(k) + L2n
fcec
N / - v / ^ N
fcec fcec fceG
Each point ^1,C2,... ,£m-1 may be included in one or two segments [t2k,t2k+1 ], k G G, therefore, ^ s(k) < 2m. Using this and the fact that
fceG
|G| ^ m, we have
l |f (t2k) - f (t2k+1)| ^ c(f). (21)
fceG
From (19), (20), and (21) inequality
K(f)l^ (22)
follows.
4. The proof of Theorem 1. The proof of estimate (6) from Theorem 1 immediately follows from inequalities (9), (10), (11), (18), (22), and n ^ N/2. To prove that the order of this estimate cannot be improved, consider the value |f1(2) - L4n,N(f1, f)|, where 4n < N/2 and f1(x) = = sign(sin x). From Lemma 1, get the inequality
|f (x) - Ln, N (f,x)| ^ |f (x) - Sn(f,x)| - (f,x)| .
It is easy to show that the following representation takes place:
f (x) = 2 y (1 - (-1)k)sin kx = 4 ^ sin(2k - 1)n (23)
n k n 2k — 1 '
fc=1 fc=1
4 sin(2fc - l)x
S2n(fl,x) = - >
IT ' '
2k - 1
k=l
Using this, we can estimate the value |f (2) — S4n f1, | from below:
fl ( 2)- S4n (fi'2
4
e
k=2n+l
(-1)
k+l
2k1
e
i ^ \4k - 3
k=n+l
1
4k 1
8 ^ 1 K k=+i k2 (4 - D (4 - I)
>
From this and (23) we have
i
f l ( 2 ) - L4n, N
(fi'i )
1/4
1/4 4n '
4n
R
4 n,N
(fi-i )
In the previous sections we showed that |R4n,N (/1, 2) | ^ c/N. Denote by N(n) a number such that for each N ^ N(n) inequality | R4n, N (/1, n)| ^ 14~ holds. Now, we have
fl (2) - ^4n,N(n) (fl> 2)
1/8 4n
c
4n'
From this we see that the order of estimate (6) cannot be improved. Theorem 1 is proved.
References
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Received November 21, 2018.
In revised form, September 24, 2019.
Accepted September 24, 2019.
Published online October 15, 2019.
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