In this paper, we establish a general result on spherical maxima sharing the same Lagrange multiplier of which the following is a particular consequence: Let X be a real Hilbert space. For each , let . Let be a sequentially weakly upper semicontinuous functional which is Gâteaux differentiable in . Assume that . Then, for each , there exists an open interval and an increasing function such that, for each , one has .
Here and in what follows, X is a real Hilbert space and is a functional, with . For each , set
A point such that
is called a spherical maximum of J. Assuming that J is , spherical maxima are important in connection with the eigenvalue problem
Actually, if is a spherical maximum of J, by the classical Lagrange multiplier theorem, there exists such that
More specifically, one could be interested in the multiplicity of solutions for (1), in the sense of finding some for which there are more points x satisfying (1). In this connection, however, just because of dependence of on , the existence of more spherical maxima in does not imply automatically the existence of some for which (1) has more solutions. So, in order to the multiplicity of solutions of (1), it is important to know when, at least for some , the spherical maxima in share the same Lagrange multiplier.
The aim of the present note is to give a contribution along such a direction.
Here is our basic result.
Theorem 1For some , assume thatJis Gâteaux differentiable inand that
Assume also that, for some , with
if , the restriction of the functionaltois sequentially weakly lower semicontinuous.
For each , put
Then the following assertions hold:
the functionηis convex and decreasing in , with ;
for each , the setis non-empty and, for every , one has
for each , with , and each , , one has
ifAdenotes the set of allsuch thatis a singleton, then the function () is continuous with respect to the weak topology; if, in addition, Jis sequentially weakly upper semicontinuous in , thenis continuous with respect to the strong topology.
Before proving Theorem 1, let us recall a proposition from  that will be used in the proof.
Proposition 1LetYbe a non-empty set, two functions, anda, btwo real numbers, with . Letbe a global minimum of the functionanda global minimum of the function .
Then one has .
Proof of Theorem 1 By definition, the function η is the upper envelope of a family of functions which are decreasing and convex in . So, η is convex and non-increasing. We also have
for all and so
In turn, this implies that η is decreasing as it never vanishes. Now, fix . So, we have
Consequently, by (3),
Observe that, for each , the restriction to of the functional is sequentially weakly lower semicontinuous. In this connection, it is enough to notice that
Fix a sequence in such that
Up to a subsequence, we can suppose that converges weakly to some . Fix . For each large enough, we have
But then, by sequential weak lower semicontinuity, we have
Hence, since ϵ is arbitrary, we have
that is, . Now, let be any point of . Let us show that . Indeed, since , there exists such that
Clearly, this is equivalent to
and hence, since , we have , as claimed. Clearly, as . Moreover, if , we have
from which we get
Now, let u be any global maximum of . Then we have
for all . Hence, as , the point u is a local minimum of the functional . Consequently, we have
and the proof of (ii) is complete. To prove (iii), observe that
As a consequence, for each , with , and for each , , we have
Therefore, in view of Proposition 1, we have
We claim that
Arguing by contradiction, assume that . In view of (ii), this would imply that and so, at the same time,
In turn, this would imply and hence , a contradiction. So, (iii) holds. Finally, let us prove (iv). For each , continue to denote by the unique point of . Let and let be any sequence in A converging to r. Up to a subsequence, converges weakly to some . Moreover, for each , , one has
From this, after easy manipulations, we get
Since the sequence is bounded above, we have
On the other hand, by sequential weak semicontinuity, we also have
Now, passing in (4) to the lim inf, in view of (5) and (6), we obtain
which is equivalent to
Since this holds for all , we have . So, is continuous at r with respect to the weak topology. Now, assuming also that J is sequentially weakly upper semicontinuous, in view of the continuity of η in , we have
Since X is a Hilbert space and converges weakly to , this implies that
which shows the continuity of at r in the strong topology. □
Remark 1 Clearly, when J is sequentially weakly upper semicontinuous in , the assertions of Theorem 1 hold in the whole interval , since a can be any positive number.
Remark 2 The simplest way to satisfy condition (2) is, of course, to assume that
Another reasonable way is provided by the following proposition.
Proposition 2For some , assume thatJis Gâteaux differentiable inand that there exists a global maximumofsuch that
Then (2) holds with .
Proof For each , set
Clearly, ω is derivable in . In particular, one has
So, by assumption, and hence, in a left neighborhood of 1, we have
which implies the validity of (2) with . □
Also, notice the following consequence of Theorem 1.
Theorem 2For some , let the assumptions of Theorem 1 be satisfied.
Then there exists an open intervaland an increasing functionsuch that, for each , one has
Clearly, I is an open interval since η is continuous and decreasing. Now, for each , pick . Finally, set
for all . Taking (iii) into account, we then realize that the function φ (whose range is contained in ) is the composition of two decreasing functions, and so it is increasing. Clearly, the conclusion follows directly from (ii). □
We conclude deriving from Theorem 1 the following multiplicity result.
Theorem 3For some , assume thatJis sequentially weakly upper semicontinuous in
, Gâteaux differentiable inand satisfies (2). Moreover, assume that there existssatisfying
such thathas either two global maxima or a global maximum at whichvanishes.
Then there existssuch that the equation
has at least two non-zero solutions which are global minima of the restriction of the functionalto .
Proof For each , in view of (7), we can pick (recall Remark 1), so that
Two cases can occur. First, assume that . So, for some . So, by (ii), for each global maximum u of , we have . As a consequence, in this case, has at least two global maxima which, by (ii) again, satisfies the conclusion with . Now, suppose that . In this case, in view of (8), the function ψ is discontinuous and hence, in view of (iv), there exists some such that has at least two elements which, by (ii), satisfy the conclusion with . □
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