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ЧЕБЫШЕВСКИЙ СБОРНИК Том 14 Выпуск 1 (2013)
ЭКСТРЕМУМЫ ВЕКТОРОЗНАЧНЫХ ФУНКЦИЙ НЕСКОЛЬКИХ ВЕЩЕСТВЕННЫХ ПЕРЕМЕННЫХ
Джела Шушич (г. Подгорица, Черногория)
Аннотация
В данной работе мы попытались обобщить обычное понятие экстремума функции вещественного переменного на векторозначные функции нескольких вещественных переменных. Нашей задачей было построить такое обобщение, чтобы для него остались верными обычные свойства и соотношения для экстремума вещественнозначных функций. Рассматриваемое обобщение также характеризуется эквивалентным обобщением.
Наши определения и связанные с ними результаты проиллюстрированы многочисленными примерами.
EXTREMUMS OF VECTOR-VALUED FUNCTIONS OF SEVERAL REAL VARIABLES
Jela Susic (Podgorica, Montenegro)
Abstract
In this paper we try to give a generalization of the usual notion of extremum of real functions to the vector-valued functions of several real variables. Our aim is that in this generalization remain valid the usual properties and relations for extremum of real functions. A considered generalization is also characterized by an equivalent generalization. Our definitions and related results are illustrated by numerous examples.
1. The notion of extremums of vector-valued functions of several real variables
Let R be the set of all real numbers, and let Rn be the n-dimensional vector space with the usual Euclidean norm || ■ ||, that is, for x = (x1,... ,xn) G Rn,
IWI = (ЕП.1 x?)1/2.
Defintion 1. Let f : U ^ Rm be a function, where U is an open subset of Rn. The point xo G U is said to be the extremum of a function f in U if there holds
||f(xo) - / (a)||2 + ||f(xo) - / (b)||2 > ||/(a) - / (b)||2 for all a,b G U.
Although it seems that Definition 1 is abstract, it can be shown that it is a natural generalization of the notion of the usual local extremum of real functions (see e.g., [1, Lecture 19]). This is shown by the following result.
Theorem 1. Let / : U ^ R be a function, where U is an open subset of Rn. Then the point x0 G U is an extremum of / U in accordance with Definition 1 if and only if x0 the usual local extremum of / in U.
Proof. Let x0 be an extremum of / by Definition 1. Then there holds
|f(a) - /(xo)|2 + |/(b) - /(xo)|2 > |/(a) - /(b)|2 for ah a,b G °(xo)
where O(xo) is a neighbourhood of a point xo. Suppose that xo is not the usual local extremum of a real function /. This means that there exist points a, b G U for which a = /(a) - /(xo) and /3 = /(xo) - /(b) such that a and /3 are positive numbers.
Substituting the previous equalities in above inequality, we obtain
a2 + P2 > (a + P)2 ^ a2 + ^2 > a2 + P2 + 2a^ ^ 0 > a^.
This contradicts the fact that a and P are positive numbers, and hence, xo is the usual local extremal value of a real function /. Conversely, suppose that xo is a usual local extremum of a function /. Without loss of generality, we can suppose that a function / does not attain a local maximum at a point xo. Then there exists a neighbourhood O(xo) of xo such that /(a) < /(xo) and /(b) < /(xo) for all a,b G O(xo). Clearly, at least one of the following inequalities there holds: /(a) < /(b) < /(xo) or /(a) < /(b) < /(xo). If the first inequality is satisfied then |f (a) - /(xo)| > |f (a)-/(b)|. If the second inequality is satisfied then |f (b)-/(xo)| > |f (a)-/(b)|. In both cases we have max{|f (a)-/(xo)|, |f (b)-/(xo)|} > |f (a)-/(b)|. This inequality yields |f (a) - /(xo)|2 + |f (b) - /(xo)|2 > |f (a) - /(b)|2 for all a, b G O(xo), i.e.,
xo is an extreme point of a function / by Definition 1. This completes the proof. ■ Hence, Definition 1 may be considered as a generalization of the notion of a usual local extremum of real functions to vector-valued functions of several variables (for more information on these functions see e.g., [2, Chapter XIV]).
The following result gives a necessary condition for a point to be an extremum of a vector-valued function of several variables, which is analogous to those of a real function.
Theorem 2 (generalized Fermat’s theorem). Let / : U ^ Rm be a function, where U C Rn is an open subset of Rn. Suppose that xo G U is an extreme point of / in accordance with Definition 1. If / is a differentiable function at a point xo then //(xo) = °.
Proof. Let Xo = (xq, •••, xjn) and /(x) = (fi(x),fm(x)) (/ i G {1, 2, ...,m} are coordinate maps of /). Then there exists a neighbourhood O(xo) of xo such that for all a = (aQ,an) G O(xo) and b = (bo,bn) G O(xo) there holds
E™i(/i(a) - fi(xo))2 + E™i(/i(xo) - fi(b))2 > Em=i(/i(a) - fi(b))2. (1)
Let a(£) = (xo + e, ...,xn) and b(£) = (xQ - e, ...,xn) where e > 0 is chosen so that a£,b£ G O(x0)). Replacing in (1) a and b by a£ and b£, respectively, we find that
E™i((/i(xo+exn) - fi(xo>xn))2 + E™i((/i(x0>xn) - fi(xQ -exn))2 >
xn) - fi(x0
Em=i((/i(x0 + e,...,xn) - fi(x0- e,...,xn))
n
o
n2 o)
(2)
By the differentiability of coordinate maps fi with i G {1, ...,m} it follows that for sufficiently small e we have
fi(xl+ x%) - /i(4,-,xo) = |^(xo)e + o(e).
Substituting the above equality in (2), we obtain
Efai(|£-(*o)£ + °M)2 + E" - «-*)? >
E™i(J£(xo)e + °(e) + f^(xo)- - o(-e))2
Dividing the above inequality by e2, we find that
E£i(|fr(*o) + ^)2 + E™ i(g(xo) + ^)2 > E™i(2g(xo) + ^ + ^)2.
Letting e ^ 0 we get
2TZAWi? a4« E™ ,(ft)2 < 0 ft = o for all i € {1..............................in}.
Assuming a(£) = (x0,x0 + e,...,xn) and b(£) = (x0,x0 - e,...,xn), and using the previous considerations, we obtain = 0 for each i G ... etc. a^ =
(*oi ...,Xq + e) and 6(£) = (xq, x'q, ...,Xq — s) ... <=* = 0 for all i G {1,
rM 3/i I 0. .0
* ■ da-,,
9fm 9fm 0. .0
- dxi ‘ ' da-,, -
is a zero-matrix, and the proof is
This shows that the Jacobian matrix [f/(x0)]n completed. ■
EXAMPLES
Example 1. The function / : R2 ^ R2 defined as /(x,y) = (x2 + y3,x+y2) does not
attain an extreme value at the point (0, 0) because of [f/(°, 0)] =
00
10
=
00
00
and so, //(0, 0) = O. It is natural then to propose the question: is / has an extreme point?
Example 2. The function / : R ^ R2 defined as /(x) = (x,x2) does not attain an extreme value at the point 0 because of //(0) = , i.e., //(0) = O. It is
natural to propose the question: whether / has an extreme point?
Example 3. The function / : R ^ R2 defined as /(x) = (x2,x2) has an extreme point, namely, the point 0 G R. Let I(0) be an arbitrary small interval which contains zero and let e,£ G I(0). Then
11/(e) - /(0)H2 + 11/(*) - /(0)H2 = ll(e2,e2)| + ||(£V2)|| = 2e4 + 2£4 11/(e) - /(£)l|2 = |(e2 - £2,e2 - £2||2 = 2(e2 - £2)2 = 2e4 + 2£4 - 4e2^2 From the above equalities it follows that
11/(e) - /(0)|2 + 11/(*) - /(0)ll2 > 11/(e) - /(<f)||2,
i.e., x0 = 0 is an extreme point of the function /. This is the only extreme point of the function / because of if x0 = 0 then f/(x0) = ^2x°^ = , and hence x0 is
not an extreme point of the function /.
Example 4. The function / : R ^ R2 defined as /(v) = (cos v, sin v) has no extreme point because of //(^) = ^SOS^ for all V G R.
However, there exist different examples of functions / : U ^ Rm where U is an open subset of Rn, such that there exists a point x0 G U for which //(x0) = O. In order to verify whether some point is an extreme point of a function / we have only Definition 1, but such a verification is very complicated and non-practical in general. The following considerations solve this problem.
Definition 2. Let / : U ^ Rm be a function, where U C Rn is an open set. / = (f1,...,/m) and let x0 G U .A point x0 is said to be an extreme point of a function / if the function r : U x U G R2n ^ R defined as
r(a1, ...J an, b1 ) bn)
E™1 /i(a1J •••J an)fi(b1J •••J bn) - Em=l(/i(a1J •••J an) + /i(b1J •••J bn))fi(x0)
attains a local minimum at a point (x0,x0) = (xQ, x^,xn, x0, x^,xn) G R2n.
It seems that Definition 2 is abstract, but the following result gives its complete characterization.
Theorem 3. Let / : U ^ Rm be a function, where U C Rn is an open set and let x0 G U. Then x0 is an extreme point in accordance with Definition 1 if and only if x0 is an extreme point of a function / in accordance with Definition 2.
Proof. Let x0 be an extreme point of a function / by Definition 1. Then there exists a neighbourhood O(x0) of a point x0 G U such that for all a, b G O(x0) holds
11/(xo) - / (a) ||2 + Ilf (xo) - / (b)||2 > Ilf (a) - / (b)||2 (1)
For fixed a, b G O(x0) consider the vectors /(x0) - /(a) and /(x0) - /(b) in the space Rm. Then we have
2(/(xo) - f (a),/(xo) - f (b)) =
||f (xo) - f (a)|2 + ||f (xo) - f (b)|2 - ||f (xo) - f (a) - f (xo) + f (b)|2 (iz (1))
^ (/(xo) - f (a)J f (xo) - f (b)) > 0 (2)
Hence, the inequality (2) is equivalent with the inequality (1). Multiplying scalary in (2^ we get ||f (xo)||2 - (/(a) + f (b),/(xo)) + (/(a),f (b)) > ^ i.e.,
E™1 /i(a1J •••, an)/i(b1J •••, bn) -Em=l(/i(a1J •••, an) + /i(b1J •••, bn))/i(x0)+ ||f (x0)|2 >
i.e.,
r(ai, •••, an, bi, ...,bn) + ||f (xo)|2 > 0 ^ r(ai, •••,an,bi, •••,bn) >
-|/(xo)|2 (3)
The above relation is satisfied for all (aQ, ...,an,b1, •••,bn) G O(x0) x O(x0) while for (ai,..., an, bi, ••.,bn) = (xQ, x^,..., x^-, x0, xg, •••,xn)
we have
r(ai,..., an, bi, •••, bn) = -||f (xo)|2. (4)
Conversely, if xo is an extreme point by Definition 2, then the relations (3) and (4) are satisfied, where in (3) a neighbourhood O(x0) x O(x0) of a point x0 is not explicitly given, while this is the case for a neighbourhood O(x0, x0) G R2n (but each neighbourhood of type O(x0,x0) contains a neighbourhood of type O(x0) x O(x0)). The relations (3) and (4) are equivalent with the relation (2) , while the relation (2) is equivalent with the inequality (1), that is, xo is an extreme point by Definition 1. The proof is completed.■
Remark. We have used the fact that the scalar product (■, ■) in a real Hilbert space H may be expressed in terms of related norm by the following identity:
(x, y) = for all x, y G H.
0
Remark. As noticed above, by using Definition 2 it is very complicated and non-practical to verify whether a point xo G U C Rn is an extreme point of a function / : U C Rn ^ Rm where U is an open set.
By Definition 2, it is sufficient to examine whether (x0,x0) G R2n is a local extreme point of a real function r : U x U G R2n ^ R, and the function r can be easily consctructed from a function /. In dependence of the behaviour of a function r in some neighbourhood of the point (x0,x0), we have different investigations related to the question whether this point is a local minimum of the function r. The following considerations gives analytic solution of this problem.
2. The investigation of a function r in a neighbourhood of the point x§(= (x0,x0))
Theorem 4 (Necessary conditions for optimality). Let U be an open subset of Rn and let / : U ^ Rm be a function of class C1 in some neighbourhood of a point
x0 E U such that f is a twice differentiable function at a point x0. If x0 is an extreme point of a function f, then the partial derivatives of the first and the second order of f at a point x0 are equal to zero.
Proof. By the assumptions for a function f, we conclude that there exist the partial derivatives of the function r up to second order at a point x^ Clearly, by Fermat’s theorem, the partial derivatives of the first order of r at a point x0 are equal to zero. For the partial derivatives of the second order we have
sS7,(4) = E™, sSfc(*o)/,(*o) - E™, sSfc(*o)/,(*o) = 0
aSfc(*o) = Efci j0r(-i-o)/.(.i-(,) - E”i s0;(*o)/.(.i-o) = 0 (, ‘2\ _ dfi ( \df,
(*o) = E™i£(xo)tOro) = 0
dakdbs^ 0' Z—/i=l dak
for all k, s E because of all partial derivatives of the first order of the
function f at a point x0 are equal to zero. The proof is completed^
Remark. Theorem 4 shows that the second form r"(x2) of the function r cannot be applied for examining whether a point x0 = (x0,x0) is a local minimum of the function r, i.e., whether x0 is an extreme point of the function f. However, we have the following result which gives sufficient conditions of optimality.
Theorem 5 (Sufficient conditions of optimality). Let U C Rn be an open set and let f : U ^ Rm be a 2p times differentiable function at a point x0 E U (p E N and p > 2). If the forms r(k)(x0)(h, h,h) = 0 for k E {3,..., 2p — 1} and
r(2p)(x2)(h, h,h) > 0 then x0 an extreme point of the function f.
Proof. By the assumptions, we find that
T(xq + h) - T(xq) = + r"^fh'h) + .... + r(>)(^A'"’fe) + r(h)
where ppj —>• 0 as h —> 0. As the forms r'(xo) r"(xo) are trivial by Theorem 4, we obtain
r(xg + h) - r(;rg) = rar'l’°^?‘.....................h) + r(h). (1)
Suppose that r does not have a local minimum at a point x0. Then there exists
a sequence {x|} C R2n such that x| ^ x0 as k ^ to and T(x|) < r(x0) for all
2 2
k E N. x? can be written as xi = Xq + ||x? — x§|| m^J^m • Denoting = ||x| — x§ || i
llxk Xoll
#2- x2 2 2
hk = n‘ j»_‘ °|i the previous equality becomes xk = ■ <uJn. where otk —> 0 as k —> to
llxk Xoll
and || hk || = 1 for all k E N .In view of the fact that {hk} is a subset of a compact sphere in the space R2n it follows that it contains a convergent subsequence hkl, so that hki —> h with \\h\\ = 1. In view of (1) for each I we have
0 > T(xfc2) — r(xg) = r(x2 + Q'A,^,)-r(x2) = r(3P)(gg)(0fctft^/,fc,’-"’0fct/,fc,)+rKfefet)
whence it follows that
‘2pf^2p)(xl)(hkl,hkl,...,hkl) r{aklhkiK n )(hki,hkh...,hki) , r{aklhkl)
akA (2p)l + ^ (2p)! + < U-
Letting / —> oo in above inequality, and after this multiplying by (2p)\ we obtain r^2?^(xg)(h, < 0 where h is a non-zero vector (because of \\h\\ = 1).
This is a contradiction with the assumption of Theorem 5 that r(2p)(x0) is a positive definite form. It follows that x0 is a local minimum of the function r, i.e..
x0 is an extreme point of the function f. This completes the proof. ■
The following theorem gives a good classification of extreme points on an open set U C Rn (i.e., it immediately excludes some points).
Theorem 6. Let U be an open subset of Rn and let f : U ^ Rm be a 2p +1 times differentiable function at a point x0 E U (p E N and p > 1) and T(2p+1)(x0) is a
nontrivial form where the forms r(k)(x2) are trivial for all k E {1,2p}. Then x0
is not an extreme point of a function f.
Proof. By the assumptions of Theorem 6 it follows that for all h E R2n \ {0}
r(;rj + h) - r(4) = rl,,+‘;<g<y.........h) + r{h)
where n^ipp+i —> 0 as h —> 0. Since r^2p+1^(xo) is a nontrivial form, it follows that
there exists a vector h ^ 0 such that r^2p+1^(xo)(/?., h,,..., h) ^ 0. Furthermore, we have
r(4 + ah) - r(*® = °S) + r(ah) r|,°^r':r5) =
rO+1)(.T5)(fe,fe,...,fe) . r(ah) 11 T~ 112'P+1
(2p+l)! _r sgn(a)\\ah\\2P+1 11 11 ^ >
Now suppose that x0 is an extreme point of a function f.
Letting a ^ 0+, we find that the left hand side of (1) is positivna in this case (because r attains a minimum at a point x0 i a > 0) and it converges as
a —> 0+ because of the right hand side of the equality converges to •
Consequently, we have —(2^+1^ ' ' ^ — Letting a, —> 0— and proceeding in a a similar manner as previously, we arrive to the inequality
The last two inequalities yields —^+1^’"'’^ = which contradicts the
assertion that r(2p+1)(x2) is a nontrivial form. Hence, x0 is not an extreme point of a function f. This completes the proof.■
Corollary. If x0 E Rn is a point such that the third form rw(x2) is nontrivial, then x0 is not an extreme point of a function f.
EXAMPLES
Example 1. The function f : R2 ^ R2 defined as f(x,y) = (x2 + y2,x2 + y2) has the extreme point x0 = (0, 0) because of for the function r : R4 ^ R we have r(a1; a2, b1, b2) = 2(a1 + «2)(&1 + &2). Obviously, r atttains a local minimum at a point x2 = (0, 0, 0, 0), and it is easy to verify that rw(x0) is a trivial form (r(x2) and r'(x2) are trivial forms by Theorem 4) and
r(4)(x0)((hi, h2, h-3, h4)(hi, Л.3, h4)(hb Л.3, h4)(hb h2, Л.3, 6,4)) =
8(h!h| + hih4 + Л|Л| + h|h4) > 0,
and therefore, T(xq + h.) — T(xq) = r< ’fooHhXKh) ^ q rp^g means ^2 jg a }ocaj
minimum of the function Г, and hence, x0 = (0, 0) is an extreme point of a function f.
Example 2. The function f : R2 ^ R2 defined as f(x,y) = (x2 + y,x + y2) does not have none extreme point in R2 because of
Г(аі, «2, bi, 62) = (a2 + a2)(b2 + 62) + (a2 + ai)(b2 + bi) — (ai + a2 + b2 + b2)(x2 + yo) —
(a| + ai + tf2 + 6i)(yo + xo)• This shows that Уо, xo, yo) = 2, i.e., Г'"(хо) is
a nontrivial form for all Ж0 Є R4, and by the previous consequence of Theorem 6 it folows that f does not have none extreme point.
Example 3. The function f : R ^ R3 defined as f (x) = (x2, x3, x5) has the extreme point x0 = 0 because of
Г(«, b) = a2b2 + a3b3 + a5b5 = a2b2(1 + ab + a3b3) > 0 = Г(0, 0) for small a and b,
i.e., Г attains a local minimum at the point (0, 0), and therefore, x0 is an extreme point of a function f.
Example 4. The function f : R ^ R2 defined as f ( (/?) = (cos (/?, sin<^) does not have none extreme point because of
r(a, b) = cos a cos b + sin a sin b — (sin a + sin b) sin x0 — (cos a + cosb) cos x0 = cos(a — b) — cos(a — x0) — cos(b — x0).
For a = b = x0 we have r(x0, x0) = 1 — 1 — 1 = —1.
For sequences an = xo + ^ and bn = xo — ^ holds (an, bn) —> (xo, xo) as n —>■ 00, and it folows that each neighbourhood of a point x0 = (x0,x0) contains points of the form (an,bn). We have
/ 7 \ 2п п 2 п п
і (сіп, bn) = cos------2 cos — = 2 cos------2 cos-------1.
n n n n
As z(t) = t2 — 2t — 1 is a decreasing function in a neighbourhood of the point t =1 and cos - < 1 for all n, it follows that Г(а„, bn) < —1 for sufficiently large n, whence it follows that (x0, x0) is not a local minimum of the function Г. This shows that x0
is not an extreme point of a function f.
Example 5. Let f : Rn ^ Rm be a function defined as f(xi,...,xn) = c = (ci,cm) Є Rm (c is a constant). Then r(ai,an, bi,bn) = c2 — 2cjcj =
— I|c||2 (= const). It follows that Г attains a local minimum at every point x0 = (x0,x0) Є R2n, that is, each x0 Є Rn is an extreme point of the function f. Example 6. (The function with countably many extreme points). Let f : U C Rn ^ Rm where U = У U (Uj are open disjoint subsets of the set U). Clearly, such a set U exists. For example, about every point in Z x Z x ... x Z C Rn we describe a ball with the radius Define
f(ub...,u„) = ((ui — aj)2 + ... + (u„ — аП)2,..., (ui — ai)2 + ... + (u„ — a^)2) na
Uj, (ai,..., аП) Є Uj, i Є {1,..., n}.
Before we give another definition of extremum of a function f : U C ^ we will prove the following two lemmas.
Lemma 1. Let K : ^ be a mapping of the space into itself and let
be the orthonormal base of this space. Then the following assertions are
equivalent.
1. K preserves the scalar product.
2. K is a linear isometry, i.e., holds
(a) K is a linear mapping.
(b) For each x G ||Kx|| = ||x|| holds.
3. K is a linear orthogonal operator, i.e., it holds
(a) K is a linear mapping.
(b) For all G {1,...,m} holds (^ei5 Kej) = , where is the Kronecker
delta.
Proof. (1) ^ (2). We have (K0, K0) = (0, 0) = 0, i.e., ||K0||2 = 0 ^ ||K0|| = 0 ^ K0 = 0. Let x G be an arbitrary vector. Then
||9fcr|| = = (x,x)^ = ||x||.
Prove that K is a linear mapping. Let x, y G Rm. Then we have
||K(x + y) — Kx — Ky||2 = (K(x + y) — Kx — Ky, K(x + y) — Kx — Ky) =
(K(x + y), K(x + y)) — 2(k(x + y), Kx) — 2(K(x + y), Ky) + 2(Kx, Ky) + (Kx, Kx) + (Ky, Ky) = (x + y, x + y) — 2(x + y, x) — 2(x + y, y) + 2(x, y) + (x, x) + (y, y) =
(x + y, x + y) — 2(x + y, x + y) = —(x + y, x + y) + 2(x, y) + (x, x) + (y, y) =
— (x x) — 2(x y) — (y, y) + (x, x) + 2(x, y) + ^ y) = 0,
i.e.,
||K(x + y) — Kx — Ky||2 = 0 ^ K(x + y) = Kx + Ky.
By the additivity of the operator K and the fact that K0 = 0, it can be easily seen
that K(qx) = qKx for each rational number q and for each vector x G
Now let a be an irrational number. Then there exists a sequence of rational numbers qn that converge to the irrational number a. We will now prove that K:
is a continuous operator.
||9fcr—9fa/|| = 9fjy, = (3?(x—y), y))^ = (x—y,x—y)^ = ||x—y\\,
i.e., K satisfies the Lipschitz conditition, and thus, K is continuous. It follows that K(anx) = anKx. Letting n ^ to and since anx ^ ax it follows that K(ax) = aKx. (2) (1). Let x,y G Rm. Then
(Kx, Ky)
INI2+IMI2-l№-y)ll2
2
||.T||2+||y||2-||.T—y||2 2
(x,y).
(2) ^ (3). We also see from (2) that K is a linear map. Now assume that K(e1)
11 *** a1m
(a11, ..., am1)------ etc. K(em) (a1mt and R
RT R
(Ke1, Ke1)
(Ke1, Kem)
am1
. We have
_(Kem, Ke1) ... (Kem, Kem)_
It follows (2) from that (Kej, Kej) = (e^, ej) = 5j, and hence, RTR =
Besides (3) it is satisfied 1 = det(RTR) = det RT det R = det R det R = det R2
whence it follows that det R =1 ili det R =1.
(3) ^ (2). Again we have that K is a linear map, and as it is showed previously,
from (2) ^ (3) we have RTR = It follows that (Kx, Kx) = (Rx, Rx) =
(RT Rx,x) = (Emx,x) = (x,x), i.e., (Kx, Kx) = (x,x) ^ ||Kx|
x
||Kx|
||x||. (R denotes the matrix of the linear operator K). The proof is completed.■ Definition. The mapping which satisfies any of the conditions (1), (2) or (3) of Lemma 1 is said to be the rotation of the space Rm.
Corollary (of Lemma 1). LetK be a rotation of the space K2, and let R be a matrix
a2 + 32 = 1
of the operator K, i.e.,
By Lemma 1, we have RTR = E2
aY + 35 = 0 . Y2 + 52 = 1
It follows that there exist real numbers p,9 G [0,n) such that a = cos p, 3 = sin P, Y = cos 9, 5 = sin 9 and from aY + 35 = 0 it follows that cos p cos 9 + sin p sin 9 = 0,
i.e., cos(p — 0) = 0 p — 6,= |=^6l = p — |. Therefore, R
cos sin
sin
cos
for
some p G [0,n).
Lemma 2. Let S be a set of vectors in the space Rm where m G {1, 2}. If S is a set such that (a,b) > 0 for all a,b G S, then there exists an orthonormal base (v1,..., vm) of the space Rm such that a = m=1 a^Vj for all a G S, where a“ > 0 for
all i G {1,..., m}.
Conversely, let m be any positive integer and let S be a set of vectors of Rm such that a = m=1 a“vj for all a G S, where {vi}m=1 is a certain fixed orthonormal base of the space Rm. Then (a, b) > 0 for all a, b G S.
Proof. Let m =1 and S be a subset of R such that (a, b) > 0 for all a, b G S. Let {v1} be a base of the space R. Without loss of generality, we can suppose that ||i'i|| = 1 (otherwise, we assume y^)- If S contains only zero vector, then (0, i'i) = 0 and clearly, the assertion is true.
Now we suppose that for a G S it holds (a,v1) = 0. Then either (a,v1) > 0 or (a,v1) < 0. If (a,v1) < 0 then instead of v1 we assume —v1 ({—v1} is also an orthonormal base of R). It follows that (a,v1) > 0. We will prove the last inequality for all a G S \ {0}. Suppose contrary, i.e., that there exists b G S \ {0} such that (b,v1) < 0. Therefore, 0 < (a,b) = ((a,v1)v1, (b,v1)v1)) = (a,v1)(b,v1) < 0. This
a
2
contradiction implies that the assertion is true for m = 1.
Next suppose that m = 2. If S contains only zero vector, then proceeding as in the previous case, we find that the assertion is true for any orthonormal base. Further, we assume that there exists a G S different from zero. Consider the function
F(x) = (jjfjp^) f°r x £ S. We can assume that each element a in S of length 1 (if
this is not true, then instead of a we consider the vector tAt-). (S is a closure of the
IMI
set S).
The function F is continuous and S' is a compact set as a closed subset of a compact central sphere with the radius 1. By Weierstrass theorem, there exists G S such that infTe^F(x) = (]jfj[, yi)- Since F(x) >0 for all x G S it follows that F(v\) = (jj^j|, i'i). Consider the orthogonal subspace i) to the vector v\. This is one-dimensional vector space, and hence, its base is {t’2}, so that 11^211 = 1- Clearly, we have (t’i, i'2) = 0 and ||i'i|| = 1 because of v\ G S. If (a,v2) > 0 then for such a v2 we have an orthonormal base {v1,v2}, while if this is not true then instead of v2 we assume —v2 such that (a, —v2) > 0. In both cases we have an orthonormal base {i'i, v2} of R2 for which (a, v2) > 0 and (a, Vi) < (py, b) for all b G S. Now let b G S be a non-zero vector. Then we have b = (b, v1)v1 + (b, v2)v2. It follows that
(b,v 1) = ||6||(pj|, i'i) > H&IKpjj-,^i) = jj^j|(<Vl’i) > 0. It remains to prove that (b,v2) > 0. Suppose conversely, i.e., that (b,v2) < 0. Then we have (pj[, ]jfj[) ^ (i'i, -p|-), or equivalently,
(pT^H’^1) + (pj| ’ l'2)(M’l'2) - (l'i’ p[)- >From ^is we find that (pi, t,2)(pp l'2) > (lR[’ t'i)(1 “ (pif’t'1)- Multiplying this by ||a||||b|| we obtain (b,V2)(a,V2) > (a,V1)(|b| — (b,V1)). (*)
Now consider the following three cases.
Case 1. (a,v1) = 0. Then since a = 0, a = (a,v1)v1 + (a,v2)v2 and (a,v2) > 0 it follows that (a,v2) > 0. Then in (*) we obtain (b,v2)(a,v2) > 0 which is impossible.
Case 2. ||b|| = (b,V1) ^ ||(b,v1)v1 + (b,v2)v2| = (b,^) ^ (b,V1)2 + (b,V2)2 = (b,v1)2 (b,v2) = 0, which contradicts the assumption that (b, v2) < 0. From
the previous consideration and since by the Cauchy-Schwarz inequality, (b,v1) > ||b| |v11 = ||b|| we find that (b,v1) < ||b|, i.e., |b| — (b,v1) > 0.
Case 3. (a, v2) = 0. Since a = 0, a = (a, v1)v1 + (a, v2)v2 and (a, v1) > 0 it follows that (a,v1) > 0. In this case the left hand side in the relation (*) is equal to 0 while its right hand side is greater that 0e. A contradiction.
Thus, from the previous cases we conclude that (a,v1) > 0, (a,v2) > 0 and H&ll — (b,v 1) > 0, while from (*) we have (6,t’2) > > 0, we have
(b,v2) > 0, which contradicts the assumption (b,v2) < 0. Hence, we must have (b, v2) > 0. Therefore, b = (b, v1)v1 + (b, v2)v2 where (b, v1) and (b, v2) are nonnegative numbers, and so, the assertion is proved for the case m = 2.
It remains to prove the converse assertion of the lemma. Let m G N and let S be a set of vectors in Rm for which there exists an orthonormal base {v1,..., vm} C Rm such that a = Em=1 a?"^ for all a G S, where a“ > 0 for each i G {1,...,n}. Now
assume that a, b G S. We have a = ™ 1 a“Vj and b = Em=1 a^V where a?, ab > 0
for all i G {1,...,n}. It follows that (a,b) = ™1 a“ab > 0. The proof of lemma is
completed. ■
Example (which shows that the first part of the assertion of Lemma 2 is not true for m > 3).
Firstly, we will show the following auxiliary assertion.
The assertion. There exists a countable set {£^£2, -"^i, --} of vectors in the space Rn(n > 2) such that every its subset consisting of n elements is a linearly independent set.
Proof. Assume that £^ £2,..., £n is a standard base of the space Rn. Choose a vector £n+1 such that it does not belong to the set L({f1, f2,..., fn-1}) (L(A) is a vector space generated by the set A) where {f1, /2,..., /n-1} are arbitrary subsets of the set {£b £2,..., £n} consisting of n — 1 elements. It follows that the set {£^ £2,..., £n+1) possesses the above property. Choose £n+2 such that it does not belong to the subsets L({/1, /2,..., /n-1}) where {/1, /2,..., /n-1} is an arbitrary subset of {£1, £2,..., £n+1} consisting of n — 1 elements, ,... etc, we procced inductively.
In other words, £n+k is chosen so that it does not belong to the the subsets L({/1, /2,..., /n-1}) where {/1, /2,..., /n-1} is any subset of {£1, £2,..., £n+fc-1}(k G N) consisting of n — 1 elements.
Every subset of the set {£1, £2,..., £j,..} consisting of n elements is a linearly independent set. Clearly, if there would be be exist a subset consisting of n linearly dependent vectors, then one of thes vectors should be a linear combination of other vectors (their maximal number is n — 1), which contradicts a construction of this vector. This completes the proof. ■
Now we return to the example. Consider the following subset of Rm(m > 3):
Km = {(xi,x2, Xm = \JYhJi xi } {Km is a conus).
Assume that x,y G Km, i.e.,
x= (a;i,a;2,...,a;m_i, ■sjY.TJi '4) and y = (yu y2, 1, ■sjT.TJi Vi)-
Using Cauchy-Schwarz inequality we have
(x- v) = TZ1' *<№ +
Suppose that
( m
k c a = y
aivi
i = 1
ai > 0 and
{v1, v2,..., vm} is an orthonormal base of the space Rm
Assuming that a,b G A and (a,b) = 0 then (^a^,^’=1 bivi) = 0, whence it follows that im 1 aibi = 0. The last equality shows that at least half of coordinates of the vector a or b is equal to zero. It is necessary that the set K also has this property i.e., that for every pair of vectors in K that are mutually orthogonal, and one of these vectors must lie in a vector subspace of < [yj of the space Rm. The number of such subspaces is finite, because of they are subspaces of the space Rm
and generated by subsets of the set {v1, v2,..., vm} (vi belongs to the subspace if for all x, y G K with (x, y) = 0, assuming that x is such a vector which has greater than half of its coordinates in the base {v1, v2,..., vm} vise that are equal to zero, then ith coordinate of the vector x is = 0).
Now we claim that there exists a countable subset E of K satisfying the property: E = {a1, b1, a2, b2,..., ar, br,...} with (ai, bi) = 0 for all i G N, and every subset of the set {a1,a2, ..^a^ ..} consisting of m — 1 elements is a linearly independent set, and every subset of the set {b1, b2,..., bi,..} consisting of m — 1 elements is also linearly independent set.
Firstly, we will construct the vectors (ai)ieN. Let {£i}iew be a set constructed in the previous assertion, regarded as a subset of the space Rm-1. If we define ai = (£i, |£i|), then ai G Km for all i G N and it is easily seen that the set {a1, a2,..., ai,...} has the property that every its subset consisting of m — 1 elements is linearly independent set. Take bi = (—£i, ||£i|). We also have bi G Km for all i G N and (ai, bi) = —(£i, £i) + |£i|2 = 0 holds. Hence, we have constructed the set E with the mentioned properties.
As the pairs {ai, bi} are mutually orthogonal and they lie in Km, it follows that at least one of vectors ai or bi must belong to the previous constructed vector spaces of dimensions < |_yj for all i G N. It follows that there exists a subsequence of a sequence (ai)ieN or (bi)ieN (we assume that (aij-)jeN C (ai)ieN) that lie in subspaces of dimensions < [yj • Since the number of these subspaces is finite, it follows that there exists infinitely many terms of a sequence (aij-)jeN, namely, its subsequence (aijk)k&N which lies in one of these subspaces of dimension < |_y_|- Assume that aij 1,aij2,...,aij^_ 1 G (aijfc) N C (ai)ieN. It follows that these vectors are linearly independent, their number is m— 1 and they lie in a subspace of the space Rm whose dimension is < [yj < m — 1 for m > 3. A contradiction! Therefore, it is not possible to lie the conus Km in the set A = {J^aivi | ai > 0 for all i G {1, 2,.., m}} where {v1, v2,..., vm} is a orthonormal base of the space Rm.
Remark. The answer to the question why it is not possible to apply the previous proof in the cases when m < 3 is as follows: it is not possible to construct E with the mentioned properties.
Definition 3. Let / : U ^ Rm be a function where U C Rn is an open set, and let xo G U .A point xo is said to be a strong extreme point of a function / if there exist a rotation R of the space Rm such that the mapping F : U ^ Rm defined as F(x) = /(x0) + R(/(x) — /(x0)) has the property that all its coordinate maps Fi : U ^ R with i G {1, ...,m} attain a local minimum.
Theorem 7. Let / : U ^ Rm be a function where U C Rn is an open set, and let x0 G U. If x0 is a strong extreme point of a function /, then x0 is an extreme point of a function /. Conversely, if m < 2 (m G N), then all extreme points are also strong extreme points of a function /.
Proof. Let x0 be a strong extreme point of a function /. Then there exist a rotation R prostora Rm satisfying the following property: There exists a neighbourhood O(x0) of x0 such that the funcction F(x) = /(x0) + R(/(x) — /(x0)) in this neighbourhood
has the property that the coordinate maps attain a local minimum.
Let e1, ...,em be the standard orthonormal base of the space Rm. Then for each x G O(x0) we have F(x) — /(x0) = R(/(x) — /(x0)) G {^m 1 a^e* | a* > 0},
i.e., R(/(O(x0)) — /(x0)) C {Em^ a^ | a* > 0}. It follows from Lemma 2 that (R(/(x) — /(x0)), R(/(y) — /(y0))) > 0 for each (x,y) G O(x0,x0). It follows from Lemma 1 that (/(x) — /(x0), /(y) — /(y0)) > 0. This shows that r(x, y) + ||/(x0)||2 > 0 ^ r(x,y) > —1|/(x0)||2 i r(x0,x0) = —1|/(x0)||2, i.e., r attains a local minimum at the point x^ = (x0,x0) and hence, x0 is an extreme point of the mapping /.
Conversely, suppose that m < 2 and x0 G U C Rn is an extreme point of the function /. Then the function r(x,y) = (/(x) — /(x0),/(y) — /(y0)) — 11/(x0)||2 attains a local minimum at a point x0 = (x0,x0). Since r(x0)2 = —1|/(x0)||2, this is equivalent with (/(x) — /(x0),/(y) — /(y0)) > 0 for all x,y G O^). Hence, there exists a neighbourhood O(x0) of a point x0 which lies in U C Rn such that the above relation is valid for all x,y G O(x0).
It follows by Lemma 2 that there exists an orthonormal base {v1,..., vm} of the space Rm such that /(x) — /(x0) G {J^™1 a*e* | a* > 0} for all x G O(x0). Hence, /(x) — /(x0) = m 1 afv* where af > 0 for each i G {1, ...,m}. By using Lemma 1, there exists a rotation R of the space Rm such that R(v*) = e* for all i G {1, ...,m}, where (e^m 1— is a standard orthonormal base of the space Rm. It follows that R(/(x) — /(x0)) = R£m 1 afvj) = m 1 afR(vj) = JX1 afe* for all x G O(x0).
Therefore, the coordinate maps of the function R(/(x) — /(x0)) attain local minimums at a point x0 which are equal to zero. It follows that this is also true for the function F(x) = /(x0) + R(/(x) — /(x0)) where the local minimums of coordinate maps F* of this function at a point x0 are equal to /*(x0) (/* are the coordinate maps of the function /). The proof is completed. ■
Corollary. If / : U ^ Rm is a function where U C Rn is an open set, and if x0 G U is a strong extreme point of a function /, then /'(x0) = 0.
Proof. By Theorem 7, x0 is an extreme point of a function /, and thus the assertion follows from Theorem 1.
We give here another proof of the corollary. Namely, F'(x) = 0 holds (because of
F^(x0) (h1, ..., hn) = Em 1 //(x0)h* = Em 1 and therefore, (R(/(x) — /(x0)))/f0 = 0, i.e., (R/(x0))' = 0 ^ R/'(x0) = 0 ^ /'(x0) = 0.
Remark. In the case when m =2, i.e., for a function / : U C Rn ^ R2 we have that x0(x0 G U) is an extreme point of a function /, if a function / maps points that are “near"to a point x0 in a rectangular part of the plane R2 with a vertex at a point / (x0).
In order to verify extreme points of these functions, or to determine all extreme points of / there are numerous criteria (necessary and sufficient conditions). EXAMPLES
V3 l 2 2 I \/3
.2 2 _
has a strong extreme point or an extreme point, namely, the point x = 0, because
Example 1. The function / : R ^ R2 defined as / (x)
(1 — cos x, x2)
of R = hence.
'V3 1
2 2
1 Vs
_ 2 2 _
cos g
Sin g — COS g
is a matrix of rotation for the angle f, and
/(x) = R|(l — cosx,x2) =>
=> (1 - >s.r..r2) = R|_1/(x) = R-f (/(x) - /(0)) + /(0) = F(x).
Since the coordinate maps F attain local minimums at the point xo = 0, it
follows that xo = 0 is a strong extreme point of a function f and hence, it is an
extreme point of a function f.
Example 2. We have previously determined the extreme points and strong extreme points of a given function f. For a given point xo £ Rn we will now determine the function f : Rn ^ R2 such that a point x0 is an extreme point of f. We proceed as follows.
There exists a R—rotation of the space R2 for which F(x) = f (xo) + R(f (x) — f (xo)) ^ f (x) = R-1(F(x) — f (xo)) + f (xo), i-e-, f (x) = Ri(F(x) — f (xo)) + f (xo)(R1 = R-1 — is also a rotation).
Now we proceed as follows. Choose an arbitrary function F = (F1,..., Fm) such that the coordinate maps Fj : Rn ^ R (i £ {1, ...,m}) attain a local minimum at a point xo. Let fj(xo) = Fj(xo) za i £ {1,..., m}, and define f (x) = R(F(x) — f (xo)) + f (xo) for an arbitrary rotation R. Clearly, xo is a strong extreme point, and hence, it is an extreme point of a function f.
Example 3. Consider the mapping f : R ^ R2 defined as f (x) = (cosx, sinx). If there would be exist a point xo which is a strong extreme point of f, then would be exist a rotation R of the space R2 such that
R(f(x))
(cos x, sin x)
cos p sin p sin p — cos p
(cos x cos p + sin x sin p, cos x sin p — sin x cos p) = (cos (x — p), sin (x — <^)) ^
F (x) = R(f (x) — f (xo)) + f (xo)) =
(cos (x — <^) — cos (xo — p) + cos xo, sin (x — <^) — sin (xo — p) + sin xo).
Then must be exist p such that F1(xo) and F2(xo) are local minimums of the functions F1 and F2. As F1 and F2 are differentiable functions, it follows that F1 (xo) = 0 and F2(xo) = 0, i.e., sin(xo — p) = 0 and cos(xo — p) = 0, which is impossible because of the functions sin i cos does not vanish at the same point. Example 4. Consider the function / : R2 —> R3 defined as /(x, y) = (x, y, \Jx2 + y2. The point (0, 0) is an extreme point of the function / because of T((ai, a2, b\, b2)) = ai&i+a262+ \Jci\ + ci22 \Jb\ + > 0 by Cauchy-Schwarz inequality, and T(0, 0, 0, 0) =
0, i.e., (0, 0, 0, 0) is a local minimum of the mapping r, and so, (0, 0) is an extreme point of the function f. Now we will prove that the point xo = (0, 0) is not a strong extreme point of the function f.
Suppose contrary, i.e., that there exists a rotation R of the space R3 such that F(x, y) = /(0, 0) + у) - /(0, 0)) = Щх, у, sjx2 + у2) =
= (Fi(x,y),F2(x,y),F3(x,y)) > (Zi(0, 0),/2(0, 0),/з(0, 0)) = (0, 0, 0)
for each (x, y) G 0((0,0)) С R2, i.e., that there exists a rotation R of the space R3 such that the set R(x, y, \Jx2 + y2) for (x, y) in this neighbourhood 0((0,0)), is a subset of the set A = {Ej=1 aiei I ai > 0,i G {1,2,3}}, or equivalently, (x, y, sjx2 + у2) С В = {Ei=i Q'iR_1ej | cti > 0, г G {1,2,3}}. By Lemma 1 it follows that {R-1e1, R-1e2, R-1e3} is an orthonormal base of the space R3.
In the same manner as in example for conus Km we can prove that the the previously mentioned is not possible. The only change consists in the fact that the set E = {a1, b1,..., a, bj,..} (see the example for conus Km) must belong to /(0(0, 0)). This can be made in the manner that instead of a (i G N) we assume
— where /?o(o,o) ^ ^ with — G /(0(0,0)). In a similar manner we proceed for vectors bi (i G N). Hence, the point (0, 0) is not a strong extreme point of the mapping /.
References
1. Arhipov G. I., Sadovnicij V. A., Cubarikov V. N. Lekcii po matematicheskomu analizu, I (Russian). Moscow: MSU, 1995.
2. Arhipov G. I., Sadovnicij V. A., Cubarikov V. N. Lekcii po matematicheskomu analizu, II (Russian). Moscow: MSU, 1997.
University of Montenegro, Podgorica, Montenegro Поступило 9.03.2013
