/compile/output.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 7, No. 3, 2014, 230-245
ISSN 1307-5543 – www.ejpam.com

Total Least Squares Fitting the Three-Parameter Inverse Weibull

Density

Dragan Jukić, Darija Marković∗

Department of Mathematics, J.J. Strossmayer University of Osijek, Trg Ljudevita Gaja 6, HR-31 000

Osijek, Croatia

Abstract. The focus of this paper is on a nonlinear weighted total least squares fitting problem for
the three-parameter inverse Weibull density which is frequently employed as a model in reliability and
lifetime studies. As a main result, a theorem on the existence of the total least squares estimator is
obtained, as well as its generalization in the lq norm (1≤ q <∞).

2010 Mathematics Subject Classifications: 65D10, 62J02, 62G07, 62N05

Key Words and Phrases: inverse Weibull density, total least squares, total least squares estimate,
existence problem, data fitting

1. Introduction

The probability density function of the random variable T having a three-parameter inverse
Weibull distribution (IWD) with location parameter α ≥ 0, scale parameter η > 0 and shape
parameter β > 0 is given by

f (t;α,β ,η) =

(

β
η

� η
t−α
�β+1

e−(
η

t−α )
β

t > α

0 t ≤ α.
(1)

If α = 0, the resulting distribution is called the two-parameter inverse Weibull distribution.
This model was developed by Erto [6].

The IWD is very flexible and by an appropriate choice of the shape parameter β the den-
sity curve can assume a wide variety of shapes (see Fig. 1). The density function is strictly
increasing on (α, tm] and strictly decreasing on [tm,∞), where tm = α + η(1 + 1/β)−1/β .
This implies that the density function is unimodal with the maximum value at tm. This is in
contrast to the standard Weibull model where the shape is either decreasing (for β ≤ 1) or
unimodal (for β > 1). When β = 1, the IWD becomes an inverse exponential distribution;

∗Corresponding author.

Email addresses: jukicd@mathos.hr (D. Jukić),darija@mathos.hr (D. Marković)

http://www.ejpam.com 230 c© 2014 EJPAM All rights reserved.



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 231

when β = 2, it is identical to the inverse Rayleigh distribution; when β = 0.5, it approximates
the inverse Gamma distribution. That is the reason why the IWD is a frequently used model
in reliability and lifetime studies (see e.g. Cohen and Whitten [5], Lawles [18], Murthy et

al. [21], Nelson [22]).

t

f(t) ✻

✲

β=0.5

❄

β=3✛

β=2✛
β=1
�

�✠

Figure 1: Plots of the inverse Weibull density for some values of β and by assuming α= 0 and
η= 1.2

In practice, the unknown parameters α, β and η of the three-parameter inverse Weibull
density (1) are not known in advance and must be estimated from a random sample t1, . . . , tn

consisting of n observations of the three-parameter inverse Weibull random variable T . There
is no unique way to estimate the unknown parameters and many different methods have been
proposed in the literature (see e.g. Abbasi et al. [1], Lawless [18], Marušić et al. [20], Murthy et

al. [21], Nelson [22], Silverman [26], Smith and Naylor [27, 28], Tapia and Thompson [29]).
A very popular method for parameter estimation is the least squares method. The non-

linear weighted ordinary least squares (OLS) fitting problem for the three-parameter inverse
Weibull density is considered by Marušić et al. [20]. In this paper we consider the nonlin-
ear weighted total least squares (TLS) fitting problem for the three-parameter inverse Weibull
density function. The structure of the paper is as follows. In Section 2 we briefly describe
the TLS method and present our main result (Theorem 1) which guarantees the existence of
the TLS estimator for the three-parametric inverse Weibull density. Its generalization in the lq
norm (1≤ q <∞) is given in Theorem 2. All proofs are given in Section 3.

2. The TLS Fitting Problem for the Three-parameter Inverse Weibull Density

Both the OLS and the TLS method require the initial nonparametric density estimates f̂

which need to be as good as possible (see e.g. Silverman [26], Marušić et al. [20]). Suppose
we are given the points (t i , yi), i = 1, . . . , n, n> 3, where

0< t1 < t2 < . . .< tn

are observations of the nonnegative three-parameter inverse Weibull random variable T and
yi := f̂ (t i) are the respective density estimates.



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 232

The goal of the OLS method (see e.g. [2, 3, 8, 11, 13, 14, 19, 25]) is to choose the unknown
parameters of density function (1) such that the weighted sum of squared distances between
the model and the data is as small as possible. To be more precise, let wi > 0, i = 1, . . . , n,
be the data weights which describe the assumed relative accuracy of the data. The unknown
parameters α,β and η have to be estimated by minimizing the functional

S(α,β ,η) =
n
∑

i=1

wi[ f (t i;α,β ,η)− yi]
2

on the set
P :=
�

(α,β ,η) ∈ R3 : α≥ 0;β ,η > 0
	

.

A point (α⋆,β⋆,η⋆) ∈ P such that S(α⋆,β⋆,η⋆) = inf(α,β ,η)∈P S(α,β ,η) is called the OLS
estimator, if it exists. As we have already mentioned, this problem has been solved by Marušić
et al. [20].

In the OLS approach the observations t i of the independent variable are assumed to be
exact and only the estimates yi of the density (dependent variable) are subject to random
errors. Unfortunately, this assumption does not seem to be very realistic in practice, and many
errors (sampling errors, human errors, modeling errors and instrument errors) prevent us from
knowing t i exactly. In such situation, when also the observations of the independent variable
contains errors, it seems reasonable to estimate the unknown parameters so that the weighted
sum of squares of all errors is minimized. This approach, known as the total least squares

(TLS) method, is a natural generalization of the OLS method (see e.g. [7]). In the statistics
literature, the TLS approach is known as errors-in-variables regression or orthogonal distance

regression, and in numerical analysis it was first considered by Golub and Van Loan [9].
The TLS method can be described as follows. Let wi , pi > 0, i = 1, . . . , n, be some weights.

If we assume that yi contains unknown additive error ǫi and that t i has unknown additive
error δi , then the mathematical model becomes

yi = f (t i +δi;α,β ,η) + ǫi, i = 1, . . . , n.

The unknown parameters α,β and η of density function (1) have to be estimated by mini-
mizing the weighted sum of squares of all errors, i.e. by minimizing the functional (see e.g.
[4, 7, 10, 17, 24])

T (α,β ,η,δ) =
n
∑

i=1

wi[ f (t i +δi;α,β ,η)− yi]
2 +

n
∑

i=1

piδ
2
i (2)

on the set P ×Rn. A point (α⋆,β⋆,η⋆) in P is called the total least squares estimator (TLS es-
timator) of the unknown parameters (α,β ,η) for the three-parameter inverse Weibull density,
if there exists δ⋆ ∈ Rn such that

T (α⋆,β⋆,η⋆,δ⋆) = inf
(α,β ,η,δ)∈P ×Rn

T (α,β ,η,δ).

Numerical methods for solving the nonlinear TLS problem are described in Boggs et al.
[4] and Schwetlick and Tiller [24]. As in the case of the OLS approach, before the iterative



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 233

minimization of the sum of squares it is still necessary to ask whether the TLS estimator exists.
In the case of nonlinear TLS problems it is still extremely difficult to answer this question (see
e.g. [3, 7, 12, 15–17]).

The difference between the OLS and the TLS approach is illustrated in Fig. 2. Geometri-
cally, if wi = pi for all i = 1, . . . , n, minimization of functional T corresponds to minimization
of the weighted sum of squares of distances from data points to the model curve.

(ti, yi)

❛

❛

❛

❛

t

y ✻

✲

(a) OLS

(ti, yi)

❛

❛

❛

❛

t

y ✻

✲

(ti+δi, f(ti+δi;α, β, η))

(b) TLS

Figure 2: The difference between the OLS and TLS approaches

Our main existence result for the TLS problem for the three-parameter inverse Weibull
density is given in the next theorem.

Theorem 1. Let the points (t i , yi), i = 1, . . . , n, n> 3, be given, such that 0< t1 < t2 < . . .< tn

and yi > 0, i = 1, . . . , n. Furthermore, let wi , pi > 0, i = 1, . . . , n, be some weights. Then there

exists a point (α⋆,β⋆,η⋆,δ⋆) ∈ P ×Rn such that

T (α⋆,β⋆,η⋆,δ⋆) = inf
(α,β ,η,δ)∈P ×Rn

T (α,β ,η,δ),

i.e. the TLS estimator exists.

The proof is given in Section 3. The following total lq norm (q ≥ 1) generalization of
Theorem 1 holds true.

Theorem 2. Suppose 1 ≤ q <∞. Let the points and weights be the same as in Theorem 1.

Define

Tq(α,β ,η,δ) :=
n
∑

i=1

wi

�

� f (t i +δi;α,β ,η)− yi

�

�
q
+

n
∑

i=1

pi |δi|q. (3)

Then there exists a point (α⋆q,β⋆q ,η⋆q,δ⋆q) ∈ P ×Rn such that

Tq(α
⋆
q,β⋆q ,η⋆q,δ⋆q) = inf

(α,β ,η,δ)∈P ×Rn
Tq(α,β ,η,δ).

The proof of this theorem is omitted as it is similar to that of Theorem 1. It suffices to
replace the l2 norm with the lq norm. Thereby all parts of the proof remain the same.



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 234

3. Proof of Theorem 1

Before starting the proof of Theorem 1, we need some preliminary results.

Lemma 1. Suppose we are given data (wi, t i , yi), i ∈ I := {1, . . . , n}, n> 3, such that

0 < t1 < t2 < . . . < tn and yi > 0, i ∈ I . Let wi , pi > 0, i ∈ I , be some weights. Given any real

number q, 1≤ q <∞, and any nonempty subset I0 of I , let

ΣI0
:=
∑

i∈I\I0

wi y
q

i
+
∑

i∈I0

pi |t i −τ0|q,

where

τ0 ∈ argmin
x

n
∑

i=1

pi |t i − x |q.

Then there exists a point inP ×Rn at which functional Tq defined by (3) attains a value less than

ΣI0
.

Summation
∑

i∈I0

is to be understood as follows: The sum over those indices i ≤ n for which

i ∈ I0. If there are no such indices, the sum is empty; following the usual convention, we
define it to be zero. Summation

∑

i∈I\I0

has similar meanings.

It is easy to verify that t1 ≤mini∈I0
t i ≤ τ0 ≤maxi∈I0

t i ≤ tn. Note that for the case when
q = 2, τ0 is a well known weighted arithmetic mean, and for the case when q = 1, τ0 is a
weighted median of the data (see e.g. Sabo and Scitovski [23]).

Proof. Since τ0 is an element of the closed interval [t1, tn], there exists r ∈ {1, . . . , n} such
that

τ0 ∈ (tr−1, tr],

where t0 = 0 by definition. Let us first choose real y0 such that

0< y0 <min
i∈I

yi (4)

and then define functions α,β ,η : (0,1)→ R by:

β(b) :=τ0 y0
eb

b

η(b) :=τ0 b1/β(b),

α(b) :=τ0 −η(b)b−1/2β(b) = η(b)
�

b−1/β(b) − b−1/2β(b)
�

.

Clearly, functions β and η are positive. Furthermore, by using the inequality
b−1/β(b) − b−1/2β(b) > 0, which holds for every b ∈ (0,1), it is easy to show that function α is
also positive. Thus, we have showed that (α(b),β(b),η(b)) ∈ P for all b ∈ (0,1). Let us now
associate with each real b ∈ (0,1) a three-parametric inverse Weibull density function

f (t;α(β),β(b),η(b)) =







β(b)

t−α(b)
� η(b)

t−α(b)
�β(b)

e−
�

η(b)
t−α(b)
�β(b)

t > α(b)

0 t ≤ α(b).
(5)



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 235

This function has maximum at the point

α(b) +η(b)(1+ 1/β(b))−1/β(b) = τ0 − ǫ(b),

where

ǫ(b) := η(b)
�

b−1/2β(b) −
�

1+
1
β(b)

�−1/β(b)�
.

It is strictly increasing on (α(b),τ0−ǫ(b)] and strictly decreasing on [τ0−ǫ(b),∞). Further-
more, by a straightforward calculation, it can be verified that

f (τ0 +α(b);α(b),β(b),η(b)) = y0, (6)

lim
b→0
β(b) =∞, (7)

lim
b→0
η(b) = τ0, (8)

lim
b→0
α(b) = 0. (9)

Now we are going to show that

lim
b→0

f (t;α(b),β(b),η(b)) = 0, t 6= τ0. (10)

First, in view of (8) and (9), we obtain

lim
b→0

� η(b)

t −α(b)
�

=
τ0

t
.

If τ0 < t, then from (7) and (9) it follows readily that limb→0 e−
�

η(b)
t−α(b)
�β(b)

= 1 and

limb→0 β(b)
�

η(b)

t−α(b)
�β(b)

= 0, and therefore

lim
b→0

f (t;α(b),β(b),η(b)) = lim
b→0

� β(b)

t −α(b)
� η(b)

t −α(b)
�β(b)

e−
�

η(b)
t−α(b)
�β(b) �

= 0.

If τ0 > t, then there exists a sufficiently great k0 ∈ N such that

e<
� η(b)

t −α(b)
�k0

for every sufficiently small b > 0. Now, by using the inequality x < ex (x ≥ 0) we obtain

β(b)< eβ(b) <
� η(b)

t −α(b)
�k0β(b)

, b ≈ 0,

and therefore, for any b ≈ 0 we have

0< f (t;α(b),β(b),η(b)) =
β(b)

t −α(b)
� η(b)

t −α(b)
�β(b)

e−
�

η(b)
t−α(b)
�β(b)



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 236

<
1

t −α(b)
� η(b)

t −α(b)
�(k0+1)β(b)

e−
�

η(b)
t−α(b)
�β(b)

.

Since

lim
b→0

� η(b)

t −α(b)
�(k0+1)β(b)

e−
�

η(b)
t−α(b)
�β(b)

= 0,

then from the above-mentioned inequality it follows that

lim
b→0

f (t;α(b),β(b),η(b)) = 0, t > τ0.

Thus, we proved the desired limits (10).
Note that

f (τ0;α(b),β(b),η(b)) =
β(b)

τ0 −α(b)
� η(b)

τ0 −α(b)
�β(b)

e−
�

η(b)
τ0−α(b)
�β(b)

=
β(b)

τ0 −α(b)
p

b e−
p

b =
τ0 y0 eb−pb

(τ0 −α(b))
p

b
,

from where taking the limit as b→ 0 it follows that

lim
b→0

f (τ0;α(b),β(b),η(b)) =∞. (11)

Due to (9), (10) and (11), we may suppose that b is sufficiently small, so that

0< α(b)< t1 (12)

0< f (t i;α(b),β(b),η(b))< yi , if t i 6= τ0 (13)

f (τ0;α(b),β(b),η(b))>max
i∈I

yi . (14)

Let us now show that for each i ∈ I0 and for every b ∈ (0,1) there exists a unique number
τi(b) such that (see Figure 3)









t i < τi(b)< τ0 − ǫ(b)< τ0, if t i < τ0

t i < τi(b)< t i + ǫ(b), if t i = τ0

τ0 < τi(b)< t i , if t i > τ0

(15)

and
f (τi(b);α(b),β(b),η(b)) = yi . (16)

First, since the function t 7→ f (t i;α(b),β(b),η(b)) has maximum at the point τ0 − ǫ(b) and
it is strictly increasing on (α(b),τ0−ǫ(b)] and strictly decreasing on [τ0−ǫ(b),∞), by using
(4), (6), (13) and (14) we obtain








f (t i;α(b),β(b),η(b))< yi < f (τ0 − ǫ(b);α(b),β(b),η(b)), if t i < τ0

f (t i;α(b),β(b),η(b))< yi < f (τ0;α(b),β(b),η(b)), if t i > τ0

f (t i + ǫ(b);α(b),β(b),η(b))< yi < f (t i;α(b),β(b),η(b)), if t i = τ0.



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 237

The existence of the desired numbers τi(b), i ∈ I0, follows from the well-known Intermediate
Value Theorem which states that a continuous real function assumes all intermediate values
on a closed interval, while uniqueness follows from monotonicity.

t

f(t) ✻

✲
τ0−ε(b) τ0 τ0+ε(b)

❝

s

s

(tj , yj)(τj(b), yj)
❝s

s ❝

(ti, yi) (τi(b), yi)

s

Figure 3: i, j ∈ I0, t i < τ0, t j > τ0; 0 < δi(b) = τi(b) − t i < τ0 − ǫ(b) − t i < τ0 − t i;
τ0 − t j < τ j(b)− t j = δ j(b)< 0

Setting

δi(b) :=

¨

τi(b)− t i , if i ∈ I0

0, if i ∈ I\I0
, (17)

(16) becomes
f (t i +δi(b);α(b),β(b),η(b)) = yi , i ∈ I0. (18)

Note that only one of the following two cases can occur:

(i) |I0|= 1, or

(ii) |I0|> 1.

Case (i): |I0| = 1. In this case we have τ0 = tr . It follows from (15) that 0 < δr(b) < ǫ(b).
Without loss of generality, in addition to (12)-(14) we may suppose that b is sufficiently small,
so that

tr−1 + ǫ(b)<
tr−1 + tr

2
< tr − ǫ(b)

and
f ((tr−1 + tr)/2;α(b),β(b),η(b))<min

i∈I
yi .

Due to these two additional assumptions and the fact that the function t 7→ f (t i;α(b),β(b),η(b))
is strictly increasing on (α(b), tr −ǫ(b)] and strictly decreasing on [tr −ǫ(b),∞), we deduce:



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 238

If t i < tr , then

0< f (t i;α(b)−δr(b),β(b),η(b)) = f (t i +δr(b);α(b),β(b),η(b))

< f (t i + ǫ(b);α(b),β(b),η(b))≤ f (tr−1 + ǫ(b);α(b),β(b),η(b))

< f ((tr−1 + tr)/2;α(b),β(b),η(b))<min
i∈I

yi ≤ yi , (19)

whereas if t i > tr , then

0< f (t i;α(b)−δr(b),β(b),η(b)) = f (t i +δr(b);α(b),β(b),η(b))

< f (t i;α(b),β(b),η(b))< yi . (20)

Thus, it follows from (18), (19) and (20) that, for every b ∈ (0,1),

Tq(α(b)−δr(b),β(b),η(b),0) =
n
∑

i=1

wi

�

� f (t i;α(b)−δr(b),β(b),η(b))− yi

�

�
q

<

n
∑

i=1
i 6=r

wi y
q

i
= ΣI0

Case |I0|> 1. Note that only one of the following two subcases can occur:

(i) τ0 6= t i for all i ∈ I0, or

(ii) τ0 = tr for some r ∈ I0.

Subcase (i): In this subcase, it follows from (13), (15), (17) and (18) that, for every b ∈ (0,1),

Tq(α(b),β(b),η(b),δ(b)) =
∑

i∈I\I0

wi

�

� f (t i;α(b),β(b),η(b))− yi

�

�
q
+
∑

i∈I0

pi |δi(b)|q

<
∑

i∈I\I0

wi y
q

i
+
∑

i∈I0

pi |t i −τ0|q = ΣI0
.

Subcase (ii): Assume that τ0 = tr for some r ∈ I0. Let index s ∈ I0 be such that ts < τ0. Then
by (15), for every b ∈ (0,1),

0< δr(b)< ǫ(b) and 0< δs(b)< tr − ǫ(b)− ts

and therefore
ps|δs(b)|q + pr |δr(b)|q < ps|tr − ǫ(b)− ts|q + prǫ

q(b).

It can be easily shown that the above right-hand side is less than

ps|tr − ts|q

whenever b is small enough. Therefore, for every small enough b we have

Tq(α(b),β(b),η(b),δ(b)) =
∑

i∈I\I0

wi

�

� f (t i;α(b),β(b),η(b))− yi

�

�
q



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 239

+ pr |δr(b)|q + ps|δs(b)|q +
∑

i∈I0\{r,s}
pi |δi(b)|q

<
∑

i∈I\I0

wi y
q

i
+
∑

i∈I0

pi |t i −τ0|q = ΣI0
.

This completes the proof of the lemma.

Proof of Theorem 1.

Proof. Since functional T is nonnegative, there exists

T ⋆ := inf
(α,β ,η,δ)∈P ×Rn

T (α,β ,η,δ).

To complete the proof it should be shown that there exists a point (α⋆,β⋆,η⋆,δ⋆) ∈ P ×Rn

such that T (α⋆,β⋆,η⋆,δ⋆) = T ⋆.
Let (αk,βk,ηk,δk) be a sequence in P ×Rn, such that

T ⋆ = lim
k→∞

T (αk,βk,ηk,δk) = lim
k→∞

�∑

i∈I

wi

�

f (t i +δ
k
i ;αk,βk,ηk)− yi

�2
+
∑

i∈I

pi(δ
k
i )

2
�

= lim
k→∞

¦ ∑

t i+δ
k
i
≤αk

wi y2
i +
∑

t i+δ
k
i
>αk

wi

�βk

ηk

� ηk

t i + δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk

−yi

�2

+
∑

i∈I

pi(δ
k
i )

2
©

. (21)

where I = {1, . . . , n}. The summation
∑

t i+δ
k
i
≤αk

(or
∑

t i+δ
k
i
>αk

) is to be understood as follows: The

sum over those indices i ≤ n for which t i + δ
k
i
≤ αk (or t i + δ

k
i
> αk). If there are no such

points t i , the sum is empty; following the usual convention, we define it to be zero.
There is no loss of generality in assuming that all sequences (αk), (βk), (ηk), (δ

k
1), . . . , (δk

n)

are monotone. This is possible because the sequence (αk,βk,ηk,δk
1, . . . ,δk

n) has a subse-

quence (αlk
,βlk

,ηlk
,δlk

1 , . . . ,δlk
n ), such that all its component sequences are monotone; and

since limk→∞ T (αlk
,βlk

,ηlk
,δlk) = limk→∞ T (αk,βk,ηk,δk) = T ⋆.

Since each monotone sequence of real numbers converges in the extended real number
system R̄, define

α⋆ := lim
k→∞

αk, β⋆ := lim
k→∞

βk, η⋆ := lim
k→∞

ηk, δ
⋆ := lim

k→∞
δ

k = (δ⋆1, . . . ,δ⋆n).

Note that 0 ≤ α⋆,β⋆,η⋆ ≤ ∞, because (αk,βk,ηk) ∈ P . Also note that δ⋆
i
∈ R for each

i = 1, . . . , n. Indeed, if |δ⋆
i
| =∞ for some i, then it would follow from (21) that T ⋆ =∞,

which is impossible.
To complete the proof it is enough to show that (α⋆,β⋆,η⋆) ∈ P , i.e. that

0 ≤ α⋆ <∞ and β⋆,η⋆ ∈ (0,∞). The continuity of the functional T will then imply that
T ⋆ = limk→∞ T (αk,βk,ηk,δk) = T (α⋆,β⋆,η⋆,δ⋆).



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 240

It remains to show that (α⋆,β⋆,η⋆) ∈ P . The proof will be done in five steps. In step 1 we
will show that α⋆ < tn. In step 2 we will show that β⋆ 6= 0. The proof that η⋆ 6=∞ will be
done in step 3. In step 4 we prove that η⋆ 6= 0. Finally, in step 5 we show that β⋆ 6=∞.

Step 1. If α⋆ ≥ tn, from (21) it follows that T ⋆ =
∑n

i=1 wi y2
i
+
∑

i∈I piδ
⋆2

i
. Since according to

Lemma 1 (for q = 2 and I0 = {1}) there exists a point in P ×Rn at which functional T attains
a value smaller than ΣI0

and since ΣI0
<
∑n

i=1 wi y2
i
+
∑

i∈I piδ
⋆2

i
, this means that in this way

(α⋆ ≥ tn) functional T cannot attain its infimum. Thus, we have proved that α⋆ < tn.
Before continuing the proof, let us introduce some notation and make one remark. First

let us define

I0 :=

¨

Iα⋆ , if Iα⋆ 6= ;
{1}, otherwise

where Iα⋆ := {i ∈ I : t i +δ
⋆
i
= α⋆}. Let us note that Lemma 1 with q = 2 implies that

T ⋆ <
∑

i∈I\I0

wi y2
i +
∑

i∈I0

pi(t i −τI0
)2 =: ΣI0

, (22)

where τI0
=

∑

i∈I0
pi t i
∑

i∈I0
pi

.

By taking an appropriate subsequence of (αk,βk,ηk,δk), if necessary, we may assume that
if t i + δ

⋆
i
< α⋆, then t i + δ

k
i
< αk for every k ∈ N. Similarly, if t i + δ

⋆
i
> α⋆, we may assume

that t i +δ
k
i
> αk for every k ∈ N. Due to this, now it is easy to show that from (21) it follows

that

T ⋆ ≥
∑

t i+δ
⋆
i
<α⋆

wi y2
i + lim

k→∞

¦ ∑

t i+δ
⋆
i
>α⋆

wi

�βk

ηk

� ηk

t i +δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk

−yi

�2©

+
∑

i∈I

piδ
⋆2

i . (23)

Step 2. If β⋆ = 0, then by using the inequality x < ex (x ≥ 0) we obtain

0<
βk

ηk

� ηk

t i + δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk

<
βk

t i +δ
k
i
−αk

, if t i +δ
⋆
i > α

⋆,

wherefrom it follows readily that

lim
k→∞

�βk

ηk

� ηk

t i +δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk
�

= 0, if t i +δ
⋆
i > α

⋆.

Now, from (23) it follows that

T ⋆ ≥
∑

i∈I\I0

wi y2
i +
∑

i∈I0

piδ
⋆2

i



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 241

=
∑

i∈I\I0

wi y2
i +
∑

i∈I0

pi(t i −α⋆)2

≥
∑

i∈I\I0

wi y2
i +
∑

i∈I0

pi(t i −τI0
)2 = ΣI0

, (24)

which contradicts (22). Therefore, in this way (β⋆ = 0) functional T cannot attain its infimum.
Thus, we have proved that β⋆ 6= 0.

The last inequality in (24) follows directly from a well-known fact that the quadratic func-
tion x 7→∑i∈I0

pi(t i − x)2 attains its minimum
∑

i∈I0
pi(t i −τI0

)2 at point τI0
.

Step 3. Let us show that η⋆ 6=∞. We prove this by contradiction. Suppose on the contrary
that η⋆ = ∞. Without loss of generality, we may then assume that if t i + δ

⋆
i
> α⋆, then

e< ηk

t i+δ
k
i
−αk

for all k ∈ N. Then from the inequality x < ex (x ≥ 0) it follows that if t i+δ
⋆
i
> α⋆,

then

βk < eβk <
� ηk

t i +δ
k
i
−αk

�βk

, k ∈ N.

Thus, if t i +δ
⋆
i
> α⋆, then

0<
βk

ηk

� ηk

t i +δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk

=
βk

t i +δ
k
i
−αk

� ηk

t i +δ
k
i
−αk

�βk
e
−
�

ηk

ti+δ
k
i
−αk

�βk

<
1

t i +δ
k
i
−αk

� ηk

t i +δ
k
i
−αk

�2βk
e
−
�

ηk

ti+δ
k
i
−αk

�βk

. (25)

Furthermore, since limk→∞
� ηk

t i+δ
k
i
−αk

�

=∞ and β⋆ 6= 0, we have limk→∞
� ηk

t i+δ
k
i
−αk

�βk
=∞

and therefore limk→∞
� ηk

t i+δ
k
i
−αk

�2βk e
−
�

ηk

ti+δ
k
i
−αk

�βk

= 0, so that from (25) it follows that

lim
k→∞

�βk

ηk

� ηk

t i +δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk
�

= 0, if t i +δ
⋆
i > α

⋆.

Putting the above limits into (23), we immediately obtain

T ⋆ ≥
∑

i∈I\I0

wi y2
i +
∑

i∈I

piδ
⋆2

i ≥ ΣI0
,

which contradicts (22). Hence we proved that η⋆ 6=∞.
So far we have shown that α⋆ < tn, β⋆ 6= 0 and η⋆ 6=∞. By using this, in the next step

we will show that η⋆ 6= 0.

Step 4. Let us show that η⋆ 6= 0. To see this, suppose on the contrary that η⋆ = 0. Then only
one of the following two cases can occur:



D. Jukić, D. Marković / Eur. J. Pure Appl. Math, 7 (2014), 230-245 242

(i) η⋆ = 0 and β⋆ ∈ (0,∞), or

(ii) η⋆ = 0 and β⋆ =∞.

Now, we are going to show that functional T cannot attain its infimum in either of these two
cases, which will prove that η⋆ 6= 0.
Case (i): η⋆ = 0 and β⋆ ∈ (0,∞). In this case we would have

lim
k→∞

βk

ηk

� ηk

t i +δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk

= lim
k→∞

βk

t i +δ
k
i
−αk

� ηk

t i +δ
k
i
−αk

�βk
e
−
�

ηk

ti+δ
k
i
−αk

�βk

= 0, if t i +δ
⋆
i > α

⋆

and hence from (23) it would follow that

T ⋆ ≥
∑

t i+δ
⋆
i
6=α⋆

wi y2
i +
∑

i∈I

piδ
⋆2

i ≥ ΣI0

which contradicts assumption (22).
Case (ii): η⋆ = 0 and β⋆ =∞. Since ηk→ 0, there exists a real number L > 1 and sufficiently
great k0 ∈ N such that if t i + δ

⋆
i
> α⋆ and k > k0, then ηk/(t i + δ

k
i
−αk) < 1/L. Without loss

of generality, we may assume that k0 = 1. Thus, if t i +δ
⋆
i
> α⋆, then

0<
βk

ηk

� ηk

t i +δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk

=
βk

t i +δ
k
i
−αk

� ηk

t i +δ
k
i
−αk

�βk
e
−
�

ηk

ti+δ
k
i
−αk

�βk

<
1

t i +δ
k
i
−αk

� βk

Lβk

�

e
−
�

ηk

ti+δ
k
i
−αk

�βk

. (26)

Furthermore, since

lim
k→∞

� βk

Lβk

�

= 0 and lim
k→∞

e
−
�

ηk

ti+δ
k
i
−αk

�βk

= 1,

from (26) it follows that

lim
k→∞

�βk

ηk

� ηk

t i +δ
k
i
−αk

�βk+1
e
−
�

ηk

ti+δ
k
i
−αk

�βk
�

= 0, if t i +δ
⋆
i > α

⋆.

Finally, from (23) we obtain T ⋆ ≥ ∑t i+δ
⋆
i
6=α⋆ wi y2

i
+
∑

i∈I piδ
⋆2

i
≥ ΣI0

, which contradicts as-
sumption (22). This means that in this case functional T cannot attain its infimum.

Thus, we have proved that η⋆ 6= 0.



REFERENCES 243

Step 5. It remains to show that β⋆ 6= ∞. We prove this by contradiction. Suppose that
β⋆ =∞. Arguing as in case (ii) from step 4, it can be shown that

lim
k→∞

� βk

t i +δ
k
i
−αk

� ηk

t i +δ
k
i
−αk

�βk
e
−
�

ηk

ti+δ
k
i
−αk

�βk
�

= 0, if 0<
η⋆

t i +δ
⋆
i
−α⋆ < 1. (27)

If η⋆

t i+δ
⋆
i
−α⋆ > 1, then there exists a sufficiently great k0 ∈ N such that e <

� ηk

t i+δ
k
i
−αk

�k0 . Now,

by using the inequality x < ex (x ≥ 0) we obtain

βk < eβk <
� ηk

t i +δ
k
i
−αk

�k0βk

, k ∈ N,

and therefore

0<
βk

t i + δ
k
i
−αk

� ηk

t i +δ
k
i
−αk

�βk
e
−
�

ηk

ti+δ
k
i
−αk

�βk

<
1

t i + δ
k
i
−αk

� ηk

t i +δ
k
i
−αk

�(k0+1)βk
e
−
�

ηk

ti+δ
k
i
−αk

�βk

. (28)

Since limk→∞
� ηk

t i+δ
k
i
−αk

�βk
=∞, we have that

lim
k→∞
� ηk

t i +δ
k
i
−αk

�(k0+1)βk e
−
�

ηk

ti+δ
k
i
−αk

�βk

= 0

and therefore from (28) it follows that

lim
k→∞

� βk

t i +δ
k
i
−αk

� ηk

t i +δ
k
i
−αk

�βk
e
−
�

ηk

ti+δ
k
i
−αk

�βk
�

= 0, if
η⋆

t i +δ
⋆
i
−α⋆ > 1. (29)

From (23), (27) and (29) we would obtain T ⋆ ≥ ∑t i+δ
⋆
i
6=α⋆ wi y2

i
+
∑

i∈I piδ
⋆2

i
≥ ΣI0

, which
contradicts (22). Thus, we have proved that β⋆ 6=∞ and completed the proof.

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