New Periodic Solutions for the Singular Hamiltonian System

and Applied Analysis 3 The limit (1 + 󵄩󵄩󵄩󵄩xn 󵄩󵄩󵄩󵄩 )f 󸀠 (x n ) → 0 implies f 󸀠 (x n ) x n 󳨀→ 0 (13) and so


Introduction
We are mainly interested in the existence of periodic solutions   → () ∈ Ω, with a prescribed period, of the second-order differential equation ẍ = −  (, ) , (1) with Ω = R  − {0} ( ∈ N,  ≥ 2) and  ∈  1 (R × Ω, R) where   (, ⋅) denotes the gradient of the function (, ⋅) defined on Ω.The study of the periodic solutions of such equations has a substantial literature with the works  of particular importance for our purpose.
In this paper, we prove the following new theorem.
Then the system (1) has a -periodic solution.
then, ∀ > 0, the system (1) has a -periodic solution.From the above example, we see that our potential does not satisfy any of conditions ( 2 ), ( 3 ), and ( 4 ).

A Few Lemmas
In order to prove Theorem 3, we will need to recall the following useful lemmas.
Lemma 6 (Eberlein-Shmulyan [15]).A Banach space  is reflexive if and only if every bounded sequence in  has a weakly convergent subsequence.
Lemma 7 (Ekeland [8]).Let  be a Banach space, and suppose Φ defined on  is Gateaux-differentiable, lower semicontinuous, and bounded from below.Then there is a sequence {  } such that Definition 8 (see [8]).Let  be a Banach space and  ∈  1 (, ).We say  satisfies the ()  condition if whenever then {  } has a strongly convergent subsequence.
Interestingly, Cerami [22] considers a weaker compact condition on a Banach space than the classical ()  condition.Here we introduce a similar condition in an open subset of a Banach space.Definition 9 (see [8]).Let  be a Banach space; Λ is an open subset; and suppose Φ defined on Λ is Gateaux-differentiable.We say that  satisfies the ()  condition if whenever {  } is a sequence such that Φ (  ) →  with (1 +           )      Φ  (  )      → 0, (7) then {  } has a strongly convergent subsequence in Λ.
With this definition, we can deduce a minimizing result in an open subset of a Banach space, the proof of which is similar to the standard one.
then {  } has a strongly convergent subsequence and the limit is in Λ.
Proof.By condition ( 1 ) and Lemma 11, we must have (  ) → +∞ as   → Λ.Since (  ) → , we know that, for any given  > 0, there exists  such that when  > , there holds the inequality The limit (1 +           )  (  ) → 0 implies and so Using condition (3) together with the limits and inequalities (11), (12), and ( 14), we can choose  > 0 such that when  is large enough, there holds which implies weakly convergent subsequence; we still denote it by   , and let the limit be .We can show in a standard fashion that this subsequence is strongly convergent in  1 .To complete the proof, we write it out.Since the sequence   is bounded in  1 , so, by Sobolev's embedding inequality, we know it is also bounded in maximum norm, and, by condition ( 1 ) and Lemma 11, we know that when  is large, By  ∈  1 (R × Ω, R), when  is large,   (  ()) is also uniformly bounded in maximum norm; we have Taking  =  and  =   in the above equation, we get Since   (  ) → 0, hence   (  ) → 0; furthermore, since   is bounded, so   (  )  → 0. Hence, by (21) That is,   →  strongly in Λ.Then by Lemma 10 the proof of Theorem 3 is complete.
is bounded.In the following we prove that |  (0)| is bounded; otherwise, there is a subsequence, still denoted by {  (0)}, such that and the uniformly bounded property for   (  ()), we have lim {  } has a subsequence, still denoted by {  } subject to   (0) → (0).