fourier series with linear splines on an interval

Result.Math. 49 (2006), 171–1841422-6383/010171-14, DOI 10.1007/s00025-006-0217-1c© 2006 Birkhauser Verlag Basel/Switzerland Results in Mathematics

Fourier Serieswith Linear Splines on an Interval

Daniela Rosca

Abstract. We construct orthonormal bases of linear splines on a finite interval[a, b] and then we study the Fourier series associated to these orthonormalbases. For continuous functions defined on [a, b], we prove that the associatedFourier series converges pointwisely on (a, b) and also uniformly on [a, b], if itconvergences pointwisely at a and b.

Mathematics Subject Classification (2000). 42A65, 42A20, 41A15.

Keywords. Fourier series, spline approximation.

1. Preliminaries

Given a finite interval [a, b], let L2(a, b) be the space of real square integrablefunctions on [a, b] and let 〈·, ·〉 be the inner product in L2(a, b) defined by

〈f, g〉 =∫ b

a

f(x)g(x)dx, for f, g ∈ L2(a, b). (1.1)

With respect to the induced norm

‖ · ‖2 = 〈·, ·〉1/2, (1.2)

L2(a, b) becomes a Hilbert space. If f ∈ L2(a, b) and {fn}n∈N is a sequence offunctions in L2(a, b), we say that fn converges to f in norm if ‖fn − f‖2 → 0.

For n ∈ N, let ∆n = {x0, . . . , xn} be a partition of [a, b] with x0 = a, x1 = band let S1

n(∆n) be the set of linear splines on ∆n. Then S1n(∆n) is a vector space

of dimension n + 1 which becomes a Hilbert space with respect to the norm (1.2).In the following we will give some elementary properties which will be used in thesequel.

Research supported by the EU Research Training Network HASSIP, HPRN-CT-2002-00285.

172 D. Rosca Result.Math.

Proposition 1.1. Let f, g : [a1, a2] → R be linear functions. If fi = f(ai) andgi = g(ai) for i = 1, 2, then∫ a2

a1

f(x) dx =a2 − a1

2(f1 + f2) , (1.3)

∫ a2

a1

f(x)g(x) dx =a2 − a1

6(f1g1 + f2g2 + (f1 + f2)(g1 + g2)) , (1.4)

∫ a2

a1

f2(x) dx =a2 − a1

3(f21 + f1f2 + f2

2

). (1.5)

2. An orthonormal basis for S1n(∆n)

Let n ∈ N be fixed. In the following, for a given partition ∆n = {x0, . . . , xn} of theinterval [a, b], with x0 = a, x1 = b, we construct an orthonormal basis of S1

n(∆n).We start with two linear orthogonal functions s0, s1 : [a, b] → R, by giving theirvalues at the points a and b. We take their normalized versions e0, e1 : [a, b] →R, e0 = s0/‖s0‖2, e1 = s1/‖s1‖2, their norms being easily calculated using (1.5).Then the construction consists in n − 1 steps. At each step we insert a node andwe construct a new corresponding spline, which is orthogonal to the previous ones.So suppose that the functions e0, e1, . . . , ek−1 : [a, b] → R are already constructedand they form an orthonormal set of linear splines on the nodes {x0, x1, . . . , xk−1}.Let xk be a new node, distinct of them, situated between the nodes xi and xj suchthat xi < xk < xj . We define the piecewise linear function sk : [a, b] → R,

sk(x) =

⎧⎨⎩

x−xi

xk−xi, for x ∈ [xi, xk],

x−xj

xk−xj, for x ∈ [xk, xj ],

0, otherwise,

with the properties sk(xk) = 1 and sk(xl) = 0 for l ∈ {0, 1, . . . , k − 1}. Then wedetermine the constants λ0, . . . , λk−1 such that the function zk : [a, b] → R,

zk = sk +k−1∑p=0

λpep,

is orthogonal to the functions e0, . . . , ek−1. Using formula (1.4), the coefficient λp

is

λp = −〈sk, ep〉= −

∫ xk

xi

sk(x)ep(x)dx −∫ xj

xk

sk(x)ep(x)dx

= −xk − xi

6ep(xi) − xj − xk

6ep(xj) − xj − xi

3ep(xk).

Since ep is a linear function on the interval [xi, xj ] we have

ep(xk) =1

xj − xi((xk − xi) ep(xj) + (xj − xk) ep(xi)) ,

Vol. 49 (2006) Fourier Series with Linear Splines on an Interval 173

so finally λp can be written as

λp =16((xi − 2xj + xk) ep(xi) + (2xi − xj − xk) ep(xj)

),

for p = 0, . . . , k − 1. Then, the norm of the new function zk will be

‖zk‖22 = ‖sk‖2

2 −k−1∑p=0

λ2p =

13(xj − xi) −

k−1∑p=0

λ2p.

The new spline ek : [a, b] → R will be taken as ek = zk/‖zk‖2.After n− 1 steps we obtain the set {e0, . . . , en}, which constitute an orthonormalbasis of S1

n(∆n).

3. Completeness of the orthonormal set of linear splines

If we repeat the above described procedure of node insertion and construction ofsplines, for a given sequence of distinct nodes {xn}n≥0 we obtain an orthonormalset of linear splines {en}n≥0. If f ∈ L2(a, b), then the generalized Fourier series

∞∑n=0

〈f, en〉en (3.1)

converges in norm (not necessarily to the function f) and Bessel inequality∥∥∥∥∥∞∑

n=1

〈f, en〉en

∥∥∥∥∥2

≤ ‖f‖2

holds. The question is how should the nodes be chosen to assure the completenessof the set {en}n≥0 in L2(a, b), more precisely the convergence of the Fourier series(3.1) to the function f, for all f ∈ L2(a, b). The answer is given in the next theorem.

Theorem 3.1. Let {∆n}n≥1, ∆n = {x0, x1, . . . , xn} be a sequence of partitions ofthe interval [a, b] such that x0 = a, x1 = b. Let En = {e0, . . . , en} be an orthonormalbasis of S1

n(∆n), for each n ∈ N, and let

‖∆n‖ = max {|xj − xi|, xi, xj neighbor nodes of ∆n}denote the norm of the partition ∆n. If ‖∆n‖ → 0, n → ∞, then the set {en}n≥0

is complete in L2[a, b].

Proof. The proof consists in two steps:

Step 1. We prove that the Fourier series (3.1) converges in norm to the functionf if f is continuous on [a, b]. So let f : [a, b] → R be a continuous function. Thenfor every ε > 0 there exists δ > 0 such that

|f(x′) − f(x′′)| < ε, whenever |x′ − x′′| < δ.


Let n be such that ‖∆n‖ < δ and let Sn denote the partial Fourier sum

Sn =n∑

k=0

〈f, ek〉ek.

If we denote by Pn the continuous piecewise linear function on the partition ∆n

which satisfies the conditions Pn(xi) = f(xi), for i = 0, 1, . . . , n, then on each ofthe intervals [xi, xj ] with xi, xj consecutive nodes of ∆n, we have

|Pn(x) − f(x)| =∣∣∣∣f(xi) + (x − xi)

f(xj) − f(xi)xj − xi

− f(x)∣∣∣∣

< |f(xi) − f(x)| + |f(xj) − f(xi)| < 2ε,

for all x ∈ [a, b]. Therefore,∫ b

a

(f(x) − Sn(x))2 dx ≤∫ b

a

(f(x) − Pn(x))2 dx < 4(b − a)ε2.

The first inequality holds since Pn belongs to the space generated by the basis{e0, e1, . . . , en}. In conclusion, Sn converges to f in norm.

Step 2. We prove that the set {en}n≥0 is complete in L2(a, b).Let g ∈ L2(a, b). If we denote by T the generalized Fourier series

∞∑n=0

〈g, en〉en, (3.2)

it is known that this series converges in norm. We have to prove that this seriesconverges to g.

Since the space C[a, b] is dense in L2(a, b), there exits a sequence {fn}n≥0 ⊂C[a, b] such that

limn→∞ ‖fn − g‖2 = 0.

This means that for all ε > 0, there exists n0 ∈ N such that

‖fn − g‖2 < ε, whenever n ≥ n0.

Let n ≥ n0 be fixed and let Un denote the Fourier series∞∑

k=0

〈fn, ek〉ek.

From Step 1 we conclude that this series converges to fn in norm, meaning thatUn = fn in L2(a, b). Then we have

‖g − T ‖2 ≤ ‖g − Un‖2,

since the series T realizes the best approximation of the function g in L2(a, b),whence

‖g − Un‖2 ≤ ‖g − fn‖2 + ‖fn − Un‖2 = ‖g − fn‖2 < ε.

Therefore g = T and thus the completeness of the set {en}n≥0 in L2(a, b) isproved. �


In conclusion, the set {en}n≥0 is a complete orthonormal set, which is alsocalled an orthonormal basis of L2(a, b). However, the pointwise convergence of theFourier generalized series (3.2) usually does not hold, so in the following we try tofind results regarding the pointwise convergence.

4. Pointwise convergence of the Fourier series

Let {en}n≥0 be an orthonormal basis of L2(a, b) consisting of linear splines con-structed as we described in Section 2. In the following we study the pointwiseconvergence of the generalized Fourier series

Sn =∞∑

n=0

〈f, en〉en, (4.1)

where f ∈ L2(a, b). Some examples show that the series (4.1) converges point-wisely to f at every point of continuity of f . However, we were not able to givea mathematical proof for the pointwise convergence at the end points a, b of theinterval [a, b]. The main result of this section will be Theorem 4.7, which statesthat the Fourier series (4.1) converges pointwisely at every point x ∈ (a, b), if thefunction f is continuous at x. The proof of this theorem needs some preliminaryresults, which will be given in Lemmas 4.1–4.6.

For these preliminary results, we consider f ∈ L2(a, b) the real-valued func-tion which we wish to approximate. For a fixed n ∈ N, let {e0, . . . , en} be anorthonormal basis of S1

n(∆n), where ∆n = {x0, . . . , xn} is a partition of the inter-val [a, b] with x0 = a, x1 = b. Let Sn : [a, b] → R denote the Fourier sum

Sn =n∑

k=0

〈f, ek〉ek.

Lemma 4.1. If y1 < y2 < y3 are consecutive nodes of ∆n and g : [y1, y3] → R isthe piecewise linear function with g(y1) = g(y3) = 0, g(y2) = 1, then∫ y2

y1

(f(x) − Sn(x)) g(x)dx = 0. (4.2)

Proof. For every t ∈ R we have∫ y2

y1

(f(x) − Sn(x))2 ≤∫ y2

y1

(f(x) − Sn(x) − t g(x))2 dx

=∫ y2

y1

(f(x) − Sn(x))2 − 2t g(x) (f(x) − Sn(x))

+t2g2(x)dx,

so we get the inequality

t2∫ y2

y1

g2(x) dx − 2t

∫ y2

y1

g(x) (f(x) − Sn(x)) dx ≥ 0,


which can be true for all t ∈ R only when (4.2) holds. �

Lemma 4.2. Let y1 < y2 < y3 be consecutive nodes of ∆n and m, M two realnumbers such that

m < f(x) < M, (4.3)for all x ∈ [y1, y3]. Then

min {Sn(y1), Sn(y3)} < 3M − 2Sn(y2), (4.4)max {Sn(y1), Sn(y3)} > 3m − 2Sn(y2). (4.5)

Proof. Let g : [y1, y3] → R be the piecewise linear function with g(y1) = g(y3) =0, g(y2) = 1. Using formula (1.4) we have

0 =∫ y3

y1

(f(x) − Sn(x)) g(x) dx =∫ y3

y1

f(x)g(x) dx −∫ y3

y1

Sn(x)g(x) dx

=∫ y3

y1

f(x)g(x) dx − y3 − y1

3Sn(y2) − y2 − y1

6Sn(y1) − y3 − y2

6Sn(y3),

whence(y2 − y1)Sn(y1) + (y3 − y2)Sn(y3)

y3 − y1=

6y3 − y1

∫ y3

y1

f(x)g(x) dx − 2Sn(y2). (4.6)

Then, from formula (4.3) we obtain

m(y3 − y1)2

<

∫ y3

y1

f(x)g(x) dx <M(y3 − y1)

2,

and further

3m − 2Sn(y2) <6

y3 − y1

∫ y3

y1

f(x)g(x) dx − 2Sn(y2) < 3M − 2Sn(y2).

Replacing the mid-term from relation (4.6), we have

3m − 2Sn(y2) <(y2 − y1)Sn(y1) + (y3 − y2)Sn(y3)

y3 − y1< 3M − 2Sn(y2).

The mid-term can be written asy2 − y1

y3 − y1Sn(y1) +

(1 − y2 − y1

y3 − y1

)Sn(y3),

so it is situated between Sn(y1) and Sn(y3), since y2−y1y3−y1

∈ (0, 1). In conclusion theinequalities (4.4) and (4.5) hold. �

An immediate consequence is the following lemma.

Lemma 4.3. In the hypotheses of Lemma 4.2, the following statements are true.1. If Sn(y2) > 2M − m, then min {Sn(y1), Sn(y3)} < 2m − M,2. If Sn(y2) < 2m − M , then max {Sn(y1), Sn(y3)} > 2M − m.

Another consequence of Lemma 4.2 is the following.


Lemma 4.4. Let y0 < y1 < y2 < y3 < y4 be consecutive nodes of ∆n and m, M realnumbers such that m < f(x) < M, for all x ∈ [y0, y4]. The following statementsare true.1. If Sn(y2) > 2M − m and Sn(y3) ≤ Sn(y1), then Sn(y4) > Sn(y2).2. If Sn(y2) > 2M − m and Sn(y3) ≥ Sn(y1), then Sn(y0) > Sn(y2).3. If Sn(y2) < 2m − M and Sn(y3) ≥ Sn(y1), then Sn(y4) < Sn(y2).4. If Sn(y2) < 2m − M and Sn(y3) ≤ Sn(y1), then Sn(y0) < Sn(y2).

Proof. 1. Suppose that Sn(y4) ≤ Sn(y2). Using Lemma 4.2 for the nodes y1 <y2 < y3, we conclude that

Sn(y3) < 3M − 2Sn(y2).

On the other hand, applying Lemma 4.2 for the nodes y2 < y3 < y4, we obtain

Sn(y2) > 3m − 2Sn(y3) > 3m − 2 (3M − 2Sn(y2)) = 3m− 6M + 4Sn(y2)

and thus we have Sn(y2) < 2M − m, which contradicts the hypothesis.2. 3. 4. The proofs are analogous with the previous case. �

The above lemma can be generalized to the following result.

Lemma 4.5. Let y1 < y2 < . . . < y2k < y2k+1 < . . . < yp be consecutive nodesof ∆n and m, M real numbers such that m < f(x) < M for all x ∈ [y1, yp]. Thefollowing statements are true.1. If Sn(y2k) > 2M − m and Sn(y2k+1) ≤ Sn(y2k−1), then

2m − M > Sn(y2k+1) > Sn(y2k+3) > . . . > Sn(y2·[ p+12 ]−1),

2M − m < Sn(y2k) < Sn(y2k+2) < . . . < Sn(y2·[ p2 ]).

2. If Sn(y2k) > 2M − m and Sn(y2k+1) ≥ Sn(y2k−1), then

2m − M > Sn(y2k−1) > Sn(y2k−3) > . . . > Sn(y1),2M − m < Sn(y2k) < Sn(y2k−2) < . . . < Sn(y2).

3. If Sn(y2k) < 2m − M and Sn(y2k+1) ≥ Sn(y2k−1), then

2M − m < Sn(y2k+1) < Sn(y2k+3) < . . . < Sn(y2·[ p+12 ]−1),

2m − M > Sn(y2k) > Sn(y2k+2) > . . . > Sn(y2·[ p2 ]).

4. If Sn(y2k) < 2m − M and Sn(y2k+1) ≤ Sn(y2k−1), then

2M − m < Sn(y2k−1) < Sn(y2k−3) < . . . < Sn(y1),2m − M > Sn(y2k) > Sn(y2k−2) > . . . > Sn(y2).

Proof. The proof is immediate using Lemmas 4.3, 4.4 and induction. We showhere only the proof of 1, since the other three conclusions can be proved in thesame way. Indeed, applying Lemma 4.3, 1, for the nodes y2k−1 < y2k < y2k+1 weobtain Sn(y2k+1) < 2m−M. Then we apply Lemma 4.4, 1, for the nodes y2k−2 <y2k−1 < y2k < y2k+1 < y2k+2 and we deduce that Sn(y2k+2) > Sn(y2k). Further,applying Lemma 4.4, 3, for the nodes y2k−1 < y2k < y2k+1 < y2k+2 < y2k+3 weconclude that Sn(y2k+3) < Sn(y2k+1). The other inequalities follow by applyingsuccessively Lemma 4.4, 1 and Lemma 4.4, 3. �


The conclusion of Lemma 4.5 is the following: if the value of Sn at a nodey ∈ ∆n becomes large (> 2M − m), respectively small (< 2m − M), then Sn willbe large (resp. small) until one of the end-points of ∆n, depending on the valuesof Sn at the neighbors of y.

Lemma 4.6. Let y1 < y2 be two consecutive nodes of ∆n and let m, M real numberssuch that m < f(x) < M , for all x ∈ [y1, y2]. The following statements are true.1. If Sn(y1) < 2m − M and S(y2) > 2M − m, then∫ y2

y1

(f(x) − Sn(x))2 dx >16(M − m)2(y1 − y2). (4.7)

2. If Sn(y1) > 2M − m and S(y2) < 2m − M, then the inequality (4.7) holds.

Proof. 1. The graph of the spline function on the interval [y1, y2] intersect the linesy = m and y = M at the points of abscises u1 and u2, u1 < u2. Further we canwrite∫ y2

y1

(f(x) − Sn(x))2 dx >

∫ u1

y1

(f(x) − Sn(x))2 dx +∫ y2

u2

(f(x) − Sn(x))2 dx

>

∫ u1

y1

(m − Sn(x))2 dx +∫ y2

u2

(M − Sn(x))2 dx

=13

y2 − y1

Sn(y2) − Sn(y1)((m − Sn(y1))3 + (Sn(y2) − M)3

)

=y2 − y1

3

(1 − M − m

Sn(y2) − Sn(y1)

)×

× ((m − Sn(y1))2 − (m − Sn(y1))(Sn(y2) − M)

+(Sn(y2) − M)2).

From the hypotheses we can easily deduce that m−Sn(y1) > M−m, Sn(y2)−M >M − m, so Sn(y2) − Sn(y1) > 3(M − m) and therefore 1 − M−m

Sn(y2)−Sn(y1)> 2

3 . Onthe other hand,

(m − Sn(y1))2 − (m − Sn(y1))(Sn(y2) − M) + (Sn(y2) − M)2 ≥ 34(Sn(y2) − M)2

>34(M − m)2.

From these inequalities the conclusion is immediate.2. The proof follows the same arguments as in the previous case. �

Theorem 4.7. Let f : [a, b] → R and let {∆n}n≥1, ∆n = {x0, x1, . . . , xn} be asequence of partitions of the interval [a, b] such that x0 = a, x1 = b. For all n ∈ N,let En = {e0, . . . , en} be an orthonormal basis of S1

n(∆n), and let Sn : [a, b] → R

be the Fourier sum

Sn =n∑

k=0

〈f, ek〉ek.


If the function f is continuous at the point x� ∈ (a, b), then

lim‖∆n‖→0

Sn(x�) = f(x�).

Proof. We suppose that Sn(x�) does not converge to f(x�). Then there exists theconstant A > 0 such that for all p ∈ N there exists n > p such that

|f(x�) − Sn(x�)| > A.

From the continuity of f at the point x� we conclude that there exists L > 0 suchthat [x� − L, x� + L] ⊆ [a, b] and

|f(x) − f(x�)| <A

4, whenever |x − x�| < L.

Two cases are possible.

Case 1. There exists n ∈ N such that

∫ b

a

(f(x) − Sn(x))2 dx <A2L

48,

‖∆n‖ <L

4,

Sn(x�) > f(x�) + A. (4.8)

We consider the interval [x� −L, x� + L] and take as bounds of the function f onthis interval the numbers

m = f(x�) − A

4, M = f(x�) +

A

4.

Subcase 1a). x� is a node. Then we denote x� = y1 and let y2 be its neighbor nodein ∆n at which the function Sn takes the smallest value. We can suppose thaty1 < y2. Further, let y1 < y2 < . . . yk < x� + L be all the nodes of the partition∆n which are situated in the interval [x�, x� + L). Then it is easy to see thatyk − y1 > 3L

4 and we also have

Sn(y1) > f(x�) + A > f(x�) +34A = 2M − m. (4.9)

From Lemma 4.5 we further conclude that

Sn(y2) < 2m − M, Sn(y3) > 2M − m, . . .


Then, from Lemma 4.6 we have∫ b

a

(f(x) − Sn(x))2 dx >

∫ yk

y1

(f(x) − Sn(x))2 dx

=k−1∑i=1

∫ yi+1

yi

(f(x) − Sn(x))2 dx

>16(M − m)2(yk − y1) >

A2L

32,

which is a contradiction.

Subcase 1b). x� is not a node. In this case we denote by y1 the neighbor node in∆n where the inequality Sn(y1) > Sn(x�) holds, therefore inequalities (4.9) hold.Again, y2 will be taken as the neighbor of y1 at which the function Sn takes thesmallest value.If y2 > y1, we take as above all the nodes y1 < y2 < . . . < yk of the partition∆n situated in the interval (x�, x� + L). If y2 < y1, we take all the nodes yk <. . . < y3 < y2 in ∆n, situated in the interval (x� − L, x�). In both cases we have|yk − y1| > L/2. Repeating the above calculations we obtain

∫ b

a

(f(x) − Sn(x))2 dx >A2L

48

which is a contradiction.

Case 2. There exists n ∈ N such that∫ b

a

(f(x) − Sn(x))2 dx <A2L

48,

‖∆n‖ <L

4,

Sn(x�) < f(x�) − A. (4.10)

Subcase 2a). x� is a node. We denote by x� = y1 and y2 will be taken as itsneighbor in ∆n at which the function Sn takes the greatest value. Inequalities(4.9) will be replaced by

Sn(x�) < f(x�) − A < f(x�) − 34A = 2m − M. (4.11)

Further, from Lemma 4.5 we have

Sn(y2) > 2M − m, Sn(y3) < 2m − M, . . .

and applying Lemma 4.6 the contradiction follows as in the case 1.


Subcase 2b). x� is not a node. As above, we denote by y1 its neighbor node in ∆n

where the inequality Sn(y1) < Sn(x�) holds, implying the inequalities (4.11) andthus everything reduces to the previous case. �

5. Uniform convergence of the Fourier series

Experience in working with series teaches us that simple pointwise convergenceof a series can be tricky, and a more desirable convergence is the uniform conver-gence. The main result of this section is Theorem 5.2, which proves the uniformconvergence of the Fourier series in the case when it converges pointwisely at aand b and f is continuous on [a, b].

As in the previous sections, let f : [a, b] → R and let {∆n}n≥1, ∆n ={x0, x1, . . . , xn} be a sequence of partitions of the interval [a, b] such that x0 =a, x1 = b. For all n ∈ N, let En = {e0, . . . , en} be an orthonormal basis of S1

n(∆n)and let Sn : [a, b] → R be the Fourier sum

Sn =n∑

k=0

〈f, ek〉ek.

Before we state the main theorem of this section, we give the following lemma,used in the proof of Theorem 5.2.

Lemma 5.1. Let n ∈ N. If xi < xj are two consecutive nodes of the partition ∆n,then for all x ∈ [xi, xj ] the following inequality holds.

max {|Sn(xi) − f(x)|, |Sn(xj) − f(x)|} ≥ |Sn(x) − f(x)|. (5.1)

Proof. Since the function Sn is linear on the interval [xi, xj ], one of the followingsituations is possible.

Sn(xi) ≤ Sn(x) ≤ Sn(xj) or Sn(xj) ≤ Sn(x) ≤ Sn(xi). (5.2)

Then the conclusion follows immediately. �The main result of this section is given in the following theorem.

Theorem 5.2. If f is continuous on [a, b] and the Fourier series∞∑

n=0

〈f, en〉en (5.3)

converges pointwisely to f at the points a and b, then it converges uniformly to fon the interval [a, b], when ‖∆n‖ → 0.

Proof. From Theorem 4.7 and from the hypotheses, we can conclude that theFourier series (5.3) converges pointwisely to the function f , on the interval [a, b].Suppose the series (5.3) is not uniformly convergent to f on [a, b]. Then thereexists A > 0 such that, for all k ∈ N there exists nk ≥ k and ynk

∈ [a, b] for whichthe following inequality holds.

|Snk(ynk

) − f(ynk)| > 5A. (5.4)


For the bounded sequence (ynk)k there exists a cluster point denoted α0. This

point does not belong to {ynk, k ≥ k0} for any k0 ∈ N, since limk→∞ Snk

(α0) −f(α0) = 0. Two cases are possible.

Case 1. There exists n0 ∈ N such that α0 is a node for the partitions ∆n forall n ≥ n0. We distinguish again between two subcases (which could also holdsimultaneously).

Subcase 1a) For each h > 0 the interval [α0, α0 +h] contains an infinity of distinctterms of the sequence (ynk

)k. Since f is uniformly continuous on the interval [a, b],there exists L > 0 such that [α0, α0 + L] ⊆ [a, b] and

|f(x) − f(y)| < A, for all x, y ∈ [α0, α0 + L]. (5.5)

In particular,

f(α0) − A < f(x) < f(α0) + A, for all x ∈ [α0, α0 + L]. (5.6)

If we consider the numbers

M = f(α0) + A,

m = f(α0) − A,

then we will have

m < f(x) < M for all x ∈ [α0, α0 + L]. (5.7)

Since ‖∆n‖ → 0, there exists n1 ≥ n0 such that ‖∆n‖ < L, for all n ≥ n1. Thepartition ∆n1 contains a node, denoted by β0, such that α0 < β0 < α0+L and thisnode belongs to all the partitions ∆n with n ≥ n1. In the following we will focuson the interval [α0, β0], where the inequalities (5.5) and (5.7) also hold. From theconvergence of the sequence (Sn)n to the function f at the nodes α0 and β0, weconclude that there exists n2 ≥ n1 such that

|f(α0) − Sn(α0)| < A, (5.8)

|f(β0) − Sn(β0)| < A, (5.9)

for all n ≥ n2. The interval [α0, β0] contains an infinite number of terms of thesequence (ynk

)k, therefore there exists nk1 ≥ n2 such that α0 < ynk1< β0.

Let xi and xj be two consecutive nodes of ∆nk1such that xi < ynk1

< xj .

Applying Lemma 5.1, from (5.4) we deduce

max{|Snk1

(xi) − f(ynk1)|, |Snk1

(xj) − f(ynk1)|} ≥ |Snk1

(ynk1) − f(ynk1

)|> 5A.

If the above maximum is |Snk1(xi) − f(ynk1

)|, then we have either Snk1(xi) −

f(ynk1) > 5A or Snk1

(xi)−f(ynk1) < −5A. If Snk1

(xi)−f(ynk1) > 5A, then using


(5.6) we obtain

Snk1(xi) > 5A + f(ynk1

)

= 5A + f(α0) + f(ynk1) − f(α0)

> 5A + f(α0) − A > 3A + f(α0)= 2M − m.

If Snk1(xi) − f(ynk1

) < −5A, then as above we get Snk1(xi) < 2m − M. In

both cases we apply Lemma 4.5 and conclude that the corresponding inequality(> 2M −m resp. < 2m−M ) holds until one of the bounds of the interval [α0, β0].

Further, in the case when at α0 we have Snk1(α0) > 2M − m = f(α0) + 3A,

then we obtain a contradiction with (5.8). In the case when Snk1(α0) < 2m−M =

f(α0) − 3A, one contradicts again (5.8).

Subcase 1b) There exists h > 0 such that the interval [α0 − h, α0] contains aninfinity of terms of the sequence (ynk

)k. Using the same arguments as in Subcase1a), we obtain a contradiction.

Case 2. α0 is not a node for any of the partitions ∆n, n ∈ N. Then, as in Case 1we choose L > 0 such that [α0 − L, α0 + L] ⊆ [a, b] and the inequality (5.5) holdsfor all x, y ∈ [α0 − L, α0 + L]. Again, there exists n1 such that ‖∆n‖ < L for alln ≥ n1 and further there exists n2 such that the partition ∆n2 contains two nodesu0, v0 for which the inequalities α0 − L < u0 < α0 < v0 < α0 + L are satisfied.Since limn→∞ = α0, there exists nk1 ≥ n2 such that u0 < ynk1

< v0. Using thesame arguments on the interval [u0, v0], we obtain, as in Case 1, a contradiction.

In conclusion, the assumption of non-uniform convergence is contradictedin all possible cases, therefore the Fourier series (5.3) converges uniformly to fon [a, b].

Remark 5.3. As we have already mentioned at the beginning of Section 4, somenumerical examples show that the pointwise convergence at the points a and bseems to hold if the function f is continuous on [a, b], but a mathematical prooffor this has not be given yet. �

References

[1] G. Folland, Fourier Analysis and its Applications, Brooks/Cole Publishing Company,1992.

Eingegangen am 28. Februar 2006


Daniela RoscaUniversite catholique de Louvain2 chemin du CyclotronB-1348 Louvain-la-NeuveBelgiume-mail: [email protected]

Technical University of Cluj-NapocaDepartment of Mathematicsstr. Daicoviciu 15RO-400 020 Cluj-NapocaRomaniae-mail: [email protected]

fourier series with linear splines on an interval

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