Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.
We shall prove the existence of periodic solutions of (1) provided that either
(43)2cosnτ[g(+∞)-g(-∞)]0000≠∫02πp(t)sin(θ+nt)dt,00∀θ∈[0,2π],
or
(44)2n[F(+∞)-F(-∞)]-2sinnτ[g(+∞)-g(-∞)]0000≠∫02πp(t)sin(θ+nt)dt,00∀θ∈[0,2π]
holds by using (39) and (40). The other cases can be handled similarly by using (43) and (44). We proceed in three steps.

(1) We prove that there exist positive constants M1 and M2 such that, for any 2π-periodic solution x(t) of (29),
(45)∥x∥∞<M1,0000∥x′∥∞<M2.
Assume by contradiction that (45) does not hold. Then there exists a sequence of 2π-periodic solutions {xk(t)}k=1∞ of (29) with λ=λk∈(0,1) such that
(46)∥xk∥∞+∥xk′∥∞⟶+∞, for k→∞.
Write yk(t)=xk′(t)+λkF(xk(t)). Since F is bounded on the interval (-∞,+∞), we have the following:
(47)∥xk∥∞+∥yk∥∞⟶+∞, for k→∞.
Let (rk(t),θk(t)) be the 2π-periodic solution of (32) related to (xk(t),yk(t)). Obviously, rk(t)=n2xk2(t)+yk2(t). Then we have the following:
(48)∥rk∥∞⟶+∞, for k⟶∞.
Without loss of generality, we also assume θk(0)∈[0,2π]. It follows from (39) that
(49)-n∫02πF(xk(t))cos(θk(0)+nt)dt000000+∫02πg(xk(t-τ))sin(θk(0)+nt)dt0000=∫02πp(t)sin(θk(0)+nt)dt.

From (34) and (37) we get that, for t∈[0,2π],
(50)g(xk(t-τ))0000=g(1nrk(t-τ)1/2sinθk(t-τ))0000=g(1nrk(0)1/2sin(θk(0)+n(t-τ))+O(1)).
Therefore,
(51)∫02πg(xk(t-τ))sin(θk(0)+nt)dt0000=∫02πg(1nrk(0)1/2sin(θk(0)+n(t-τ))+O(1))00000000×sin(θk(0)+nt)dt.
Obviously, we have the following:
(52)∫02πg(1nrk(0)1/2sin(θk(0)+n(t-τ))+O(1))00×sin(θk(0)+nt)dt0000=1n∫θk(0)-nτ2nπ+θk(0)-nτg(1nrk(0)1/2sins+O(1))0000000000×sin(s+nτ)ds0000=cosnτn000000×∫θk(0)-nτ2nπ+θk(0)-nτg(1nrk(0)1/2sins+O(1))sinsds000000+sinnτn000000×∫θk(0)-nτ2nπ+θk(0)-nτg(1nrk(0)1/2sins+O(1))cossds.
Since θk(0)∈[0,2π], there exists a subsequence {θkj(0)} such that θkj→θ*, j→∞. By using Lebesgue dominated convergent theorem and the condition (h1), we get the following:
(53)limj→+∞∫θkj(0)-nτ2nπ+θkj(0)-nτg(1nrkj(0)1/2sins+O(1))sinsds0000=2n(g(+∞)-g(-∞)),limj→+∞∫θkj(0)-nτ2nπ+θkj(0)-nτg(1nrkj(0)1/2sins+O(1))cossds=0.
Therefore,
(54)limj→+∞∫02πg(xkj(t-τ))sin(θkj(0)+nt)dt0000=2cosnτ[g(+∞)-g(-∞)].
Similarly, we can get the following:
(55)limj→+∞∫02πF(xkj(t))cos(θkj(0)+nt)dt=0.
Hence, we obtain the following:
(56)2cosnτ[g(+∞)-g(-∞)]=∫02πp(t)sin(θ*+nt)dt,
which contradicts with (43).

On the other hand, it follows from (40) that
(57)n∫02πF(xk(t))sin(θk(0)+nt)dt000000+∫02πg(xk(t-τ))cos(θk(0)+nt)dt0000=∫02πp(t)cos(θk(0)+nt)dt.
According to (50), we have the following:
(58)∫02πg(xk(t-τ))cos(θk(0)+nt)dt0000=∫02πg(1nrk(0)1/2cos(θk(0)+n(t-τ))+O(1))000000000×cos(θk(0)+nt)dt0000=1n∫θk(0)-nτ2nπ+θk(0)-nτg(1nrk(0)1/2sins+O(1))0000000000×cos(s+nτ)ds0000=cosnτn000000×∫θk(0)-nτ2nπ+θk(0)-nτg(1nrk(0)1/2sins+O(1))cossds000000-sinnτn000000×∫θk(0)-nτ2nπ+θk(0)-nτg(1nrk(0)1/2sins+O(1))sinsds.
From (53) we obtain the following:
(59)limj→+∞∫02πg(xkj(t-τ))cos(θkj(0)+nt)dt0000=-2sinnτ[g(+∞)-g(-∞)].
Similarly, we have the following:
(60)limj→+∞∫02πF(xkj(t))sin(θkj(0)+nt)dt0000=2[F(+∞)-F(-∞)].
It follows from (57)–(60) that
(61)2n[F(+∞)-F(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]0000=∫02πp(t)cos(θ*+nt)dt,
which contradicts with (44). Therefore, there exist positive constants M1 and M2 such that (45) holds.

(2) Let x(t)=ϱsin(nt+α), where α is an arbitrary constant. We will prove that there exists M0>0 such that, for ϱ≥M0, Nx∉ImL. Otherwise, there exits a sequence {ϱk} satisfying limk→∞ϱk=+∞ such that Nxk∈ImL with xk(t)=ϱksin(nt+α). We will prove the following:
(62)2cosnτ[g(+∞)-g(-∞)]=∫02πp(t)sin(nt+α)dt,2n[F(+∞)-F(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]0000=∫02πp(t)sin(nt+α)dt.

In fact, since Nxk∈ImL, we have the following:
(63)∫02π[f(xk(t))xk′(t)+g(xk(t-τ))]sin(nt+α)dt0000=∫02πp(t)sin(nt+α)dt,∫02π[f(xk(t))xk′(t)+g(xk(t-τ))]cos(nt+α)dt0000=∫02πp(t)cos(nt+α)dt.

Using the same method as in step 1, we have the following:
(64)limk→∞∫02π[f(xk(t))xk′(t)+g(xk(t-τ))]sin(nt+α)dt0000=2cosnτ[g(+∞)-g(-∞)],limk→∞∫02π[f(xk(t))xk′(t)+g(xk(t-τ))]cos(nt+α)dt0000=2n[F(+∞)-F(-∞)]000000-2sin(nτ)[g(+∞)-g(-∞)].

As a consequence, (62) holds. Thus, we get a contradiction.

(3) Let M>max{nM0,M1,M2} be a sufficiently large constant (if it is necessary, M can be enlarged). Set
(65)Ω={x∈X:∥x∥∞<M, ∥x′∥∞<M}.
From the conclusion in step 1 we know that
(66)Lx≠λNx,00∀x∈∂Ω∩D(L), λ∈(0,1).
From the conclusion in step 2 we know that
(67)Nx∉ImL,00∀x∈∂Ω∩KerL,
which implies QNx≠0 for any x∈∂Ω∩KerL. Since ImQ=KerL, we can take an isomorphism J=identity:ImQ→KerL. In what follows, we will prove the following:
(68)deg{JQN,Ω∩KerL,0}≠0.
To this end, let us define ϕ:KerL→R2, x=asinnt+bcosnt→(a,b), namely,
(69)ϕx=(a,b).
Obviously, ϕ is a linear isomorphism. For any x=asinnt+bcosnt, set
(70)(JQNx)(t)=h1(a,b)sinnt+h2(a,b)cosnt,
where
(71)h1(a,b)=1π∫02π[-f(x(t))x′(t)-g(x(t-τ))+p(t)]0000×sinntdth2(a,b)=1π∫02π[-f(x(t))x′(t)-g(x(t-τ))+p(t)]0000×cosntdt.
Define h:R2→R2 as follows:
(72)h(a,b)=ϕ∘Q∘N∘ϕ-1(a,b)=(h1(a,b),h2(a,b)).
Then we have the following:
(73)deg{JQN,Ω∩KerL,0}=deg{h,ϕ(Ω∩KerL),0}.
To calculate deg{h,ϕ(Ω∩KerL),0}, we first estimate l1 and l2 as follows:
(74)l1(a,b)=1π∫02π[-f(x(t))x′(t)-g(x(t-τ))]sinntdt,l2(a,b)=1π∫02π[-f(x(t))x′(t)-g(x(t-τ))]cosntdt.
Write x=ρsin(nt+ϑ) with ρ=a2+b2, ϑ=arctan(b/a) or ϑ=π+arctan(b/a). Then we have that, for ρ→∞,
(75)∫02πf(x(t))x′(t)sinntdt0000=-n∫02πF(x(t))cosntdt0000=-n∫02πF(ρsin(nt+ϑ))cosntdt0000=-∫ϑ2nπ+ϑF(ρsins)(cosscosϑ+sinssinϑ)ds0000=-cosϑ∫ϑ2nπ+ϑF(ρsins)cossds000000-sinϑ∫ϑ2nπ+ϑF(ρsins)sinsds0000=-2nsinϑ[F(+∞)-F(-∞)]+o(1).
On the other hand, we have that, for ρ→∞,
(76)∫02πg(x(t-τ))sinntdt0000=∫02πg(ρsin(n(t-τ)+ϑ))sinntdt0000=1n∫ϑ-nτ2nπ+ϑ-nτg(ρsins)sin(s+nτ-ϑ)ds0000=cos(nτ-ϑ)n∫ϑ-nτ2nπ+ϑ-nτg(ρsins)sinsds000000+sin(nτ-ϑ)n∫ϑ-nτ2nπ+ϑ-nτg(ρsins)cossds0000=2cos(nτ-ϑ)[g(+∞)-g(-∞)]+o(1).
Therefore, we get the following:
(77)l1(a,b)=2π{cos(nτ-ϑ)[g(+∞)-g(-∞)]nsinϑ[F(+∞)-F(-∞)]000-cos(nτ-ϑ)[g(+∞)-g(-∞)]}+o(1).
To estimate l2, we have that, for ρ→∞,
(78)∫02πf(x(t))x′(t)cosntdt0000=n∫02πF(x(t))sinntdt0000=n∫02πF(ρsin(nt+ϑ))sinntdt0000=cosϑ∫ϑ2nπ+ϑF(ρsins)sinsds000000-sinϑ∫ϑ2nπ+ϑF(ρsins)cossds0000=2ncosϑ[F(+∞)-F(-∞)]+o(1).
Meanwhile, we get that, for ρ→∞,
(79)∫02πg(x(t-τ))cosntdt0000=∫02πg(ρsin(n(t-τ)+ϑ))cosntdt0000=1n∫ϑ-nτ2nπ+ϑ-nτg(ρsins)cos(s+nτ-ϑ)ds0000=cos(nτ-ϑ)n∫ϑ-nτ2nπ+ϑ-nτg(ρsins)cossds000000-sin(nτ-ϑ)n∫ϑ-nτ2nπ+ϑ-nτg(ρsins)sinsds0000=-2sin(nτ-ϑ)[g(+∞)-g(-∞)]+o(1).
Hence, we obtain the following:
(80)l2(a,b)=2π{+sin(nτ-ϑ)[g(+∞)-g(-∞)]-ncosϑ[F(+∞)-F(-∞)]000+sin(nτ-ϑ)[g(+∞)-g(-∞)]}+o(1).
Set
(81)h^1(a,b)=2π{-cos(nτ-ϑ)[g(+∞)-g(-∞)]nsinϑ[F(+∞)-F(-∞)]000-cos(nτ-ϑ)[g(+∞)-g(-∞)]},h^2(a,b)=2π{+sin(nτ-ϑ)[g(+∞)-g(-∞)]-ncosϑ[F(+∞)-F(-∞)]000+sin(nτ-ϑ)[g(+∞)-g(-∞)]},h^(a,b)=(h^1(a,b),h^2(a,b)).
Replacing ϑ in x=ρsin(nt+ϑ) with π+ϑ, we get the following:
(82)h^1(-a,-b)=-2π{-cos(nτ-ϑ)[g(+∞)-g(-∞)]nsinϑ[F(+∞)-F(-∞)]00000-cos(nτ-ϑ)[g(+∞)-g(-∞)]},h^2(-a,-b)=-2π{+sin(nτ-ϑ)[g(+∞)-g(-∞)]-ncosϑ[F(+∞)-F(-∞)]0000+sin(nτ-ϑ)[g(+∞)-g(-∞)]}.
As a consequence,
(83)h^1(-a,-b)=-h^1(a,b),h^2(-a,-b)=-h^2(a,b),h^(-a,-b)=-h^(a,b).
We note that, for a2+b2→+∞,
(84)h1(a,b)=h^1(a,b)+c1+o(1),h2(a,b)=h^2(a,b)+c2+o(1),
where
(85)c1=1π∫02πp(t)sinntdt,c2=1π∫02πp(t)cosntdt.
Let us consider the map H:ϕ(Ω∩KerL)×[0,1]→R2:(a,b,μ)→(h1(a,b,μ),h2(a,b,μ)) with
(86)h1(a,b,μ)=h^1(a,b)+μc1,h2(a,b,μ)=h^2(a,b)+μc2.
Obviously, H is continuous. Next, we shall prove that, for any (a,b,μ)∈∂ϕ(Ω∩KerL)×[0,1],
(87)(h1(a,b,μ),h2(a,b,μ))≠(0,0).
Otherwise, there exists some (a,b,μ)∈∂ϕ(Ω∩KerL)×[0,1] such that
(88)h1(a,b,μ)=0, h2(a,b,μ)=0.
Then we have the following:
(89)2nsinϑ[F(+∞)-F(-∞)]000000-2cos(nτ-ϑ)[g(+∞)-g(-∞)]0000=-μπc1,-2ncosϑ[F(+∞)-F(-∞)]000000+2sin(nτ-ϑ)[g(+∞)-g(-∞)]0000=-μπc2.
Therefore, we get the following:
(90)2cosnτ[g(+∞)-g(-∞)]=μ∫02πp(t)sin(nt+ϑ)dt,2n[F(+∞)-F(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]0000=μ∫02πp(t)cos(nt+ϑ)dt.
Since μ∈[0,1], we know from (90) that
(91)2cosnτ[g(+∞)-g(-∞)]∈[-ϱ,ϱ],(92)2n[F(+∞)-F(-∞)]0000-2sin(nτ)[g(+∞)-g(-∞)]∈[-ϱ,ϱ],
where ϱ is given in Remark 2. From Remark 2 we know that (91) and (92) contradict with (43) and (44).

In particular, we have that, for (a,b)∈∂ϕ(Ω∩KerL),
(93)(h^1(a,b),h^2(a,b))≠(0,0),(94)(h^1(a,b)+c1,h^2(a,b)+c2)≠(0,0).
Since h^:R2→R2 is odd and (93) holds, we know from Borsuk Theorem [12] that
(95)deg(h^,ϕ(Ω∩KerL),0)=2m+1≠0,
where m is an integer.

On the other hand, we know from (94) and the expressions of h^1 and h^2 that there exists a positive constant ν, which is independent of a and b, such that, for (a,b)∈R2(96)|h^1(a,b)+c1|+|h^2(a,b)+c2|≥ν.
Consequently, we infer from the homotopy invariance of degree that, if M>max{M0,M1,M2} is large enough; then
(97)deg(JQN,Ω∩KerL,0)0000=deg(h,ϕ(Ω∩KerL),0)0000=deg(H(·,1),ϕ(Ω∩KerL),0)0000=deg(H(·,0),ϕ(Ω∩KerL),0)0000=deg(h^,ϕ(Ω∩KerL),0)0000=2m+1≠0.
Therefore, all conditions of Lemma 4 are satisfied. Thus, (1) has at least one 2π-periodic solution.