For convenience, we denote
$$ R_{1}= \Biggl(\sum_{n=1}^{T}f^{2}(n) \Biggr)^{1/2},\qquad R_{2}=\sum_{n=1}^{T}f(n), \quad\mbox{and}\quad R_{3}=\sum_{n=1}^{T}g(n). $$
Proof of Theorem 1.1
According to (3), there exists \(a_{1}>\frac{1}{4\sin^{2}\frac{\pi}{T}}\) satisfying
$$\liminf_{x\longrightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)>\frac{a_{1}}{2}R_{1}^{2}. $$
From (1) and Lemma 2.1, for any \(u\in H_{T}\), one has
$$\begin{aligned} &\Biggl\vert \sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u(n)\bigr)-F_{1}(n,\bar{u})\bigr]\Biggr\vert \\ &\quad=\Biggl\vert \sum_{n=1}^{T} \int_{0}^{1}\bigl(\nabla F_{1}\bigl(n,\bar {u}+s\tilde{u}(n)\bigr),\tilde{u}(n)\bigr)\,ds\Biggr\vert \\ &\quad\leq\sum_{n=1}^{T} \int_{0}^{1}f(n)\bigl|\bar{u}+s\tilde {u}(n)\bigr|^{\alpha}\bigl|\tilde{u}(n)\bigr|\,ds+ \sum_{n=1}^{T} \int_{0}^{1}g(n)\bigl|\tilde{u}(n)\bigr|\,ds \\ &\quad\leq\sum_{n=1}^{T}f(n) \bigl(|\bar{u}|+\bigl| \tilde{u}(n)\bigr|\bigr)^{\alpha }\bigl|\tilde{u}(n)\bigr|+\sum _{n=1}^{T}g(n)\bigl|\tilde{u}(n)\bigr| \\ &\quad\leq\sum_{n=1}^{T}f(n)| \bar{u}|^{\alpha}\bigl|\tilde {u}(n)\bigr|+\sum_{n=1}^{T}f(n)\bigl| \tilde{u}(n)\bigr|^{\alpha+1}+\sum_{n=1}^{T}g(n)\bigl| \tilde{u}(n)\bigr| \\ &\quad\leq|\bar{u}|^{\alpha} \Biggl(\sum_{n=1}^{T}f^{2}(n) \Biggr)^{1/2} \Biggl(\sum_{n=1}^{T}\bigl| \tilde{u}(n)\bigr|^{2} \Biggr)^{1/2}+\|\tilde{u}\| _{\infty}^{\alpha+1}\sum_{n=1}^{T}f(n) +\|\tilde{u}\|_{\infty}\sum_{n=1}^{T}g(n) \\ &\quad\leq\frac{1}{2a_{1}}\sum_{n=1}^{T}\bigl| \tilde{u}(n)\bigr|^{2}+\frac {a_{1}}{2}R_{1}^{2}| \bar{u}|^{2\alpha}+R_{2}\|\tilde{u}\|_{\infty }^{\alpha+1}+R_{3} \|\tilde{u}\|_{\infty} \\ &\quad\leq\frac{1}{8a_{1}\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}+\frac{a_{1}}{2}R_{1}^{2}| \bar{u}|^{2\alpha }+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &\qquad{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2}. \end{aligned}$$
(3.1)
From (2) and Lemma 2.1, for any \(u\in H_{T}\), we have
$$\begin{aligned} &\sum_{n=1}^{T} \bigl[F_{2}\bigl(u(n)\bigr)-F_{2}(\bar{u})\bigr] \\ &\quad=\sum _{n=1}^{T} \int_{0}^{1}\frac{1}{s}\bigl(\nabla F_{2}\bigl(\bar{u}+s\tilde {u}(n)\bigr)-\nabla F_{2}( \bar{u}),s\tilde{u}(n)\bigr)\,ds \\ &\quad\geq-\sum_{n=1}^{T} \int_{0}^{1}rs^{\gamma-1}\bigl|\tilde {u}(n)\bigr|^{\gamma}\,ds \\ &\quad\geq-\frac{rT}{\gamma}\|\tilde{u}\|_{\infty}^{\gamma} \\ &\quad\geq-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma/2} \Biggl( \sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{\gamma/2}. \end{aligned}$$
(3.2)
Combining (3.1) with (3.2), for all \(u\in H_{T}^{1}\) one has
$$\begin{aligned} \varphi(u) ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u(n)\bigr)-F_{1}(n,\bar{u})\bigr] \\ &{}+\sum_{n=1}^{T}\bigl[F_{2} \bigl(u(n)\bigr)-F_{2}(\bar{u})\bigr]+\sum_{n=1}^{T}F(n, \bar{u}) \\ \geq{}& \biggl(\frac{1}{2}-\frac{1}{8a_{1}\sin^{2}\frac{\pi }{T}} \biggr)\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2}- \biggl( \frac {T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2}-\frac{rT}{\gamma} \biggl(\frac {T^{2}-1}{6T} \biggr)^{\gamma/2} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2} \Biggr)^{\gamma/2} \\ &{}+|\bar{u}|^{2\alpha} \Biggl(|\bar{u}|^{-2\alpha}\sum _{n=1}^{T}F(n,\bar{u})-\frac{a_{1}}{2}R_{1}^{2} \Biggr). \end{aligned}$$
Hence, \(\varphi(u)\rightarrow\infty\) as \(\|u\|\rightarrow\infty\). From this result, if \(\{u_{k}\}\subset H_{T}\) is a minimizing sequence for φ, i.e., \(\varphi(u_{k})\rightarrow\inf\varphi \), \(k\rightarrow\infty\), then \(\{u_{k}\}\) is bounded. Since \(H_{T}\) is finite dimensional, going if necessary to a subsequence, we can assume that \(\{u_{k}\}\) converges to some \(u_{0}\in H_{T}\). Because of φ is continuously differentiable on \(H_{T}\), one has
$$ \varphi(u_{0})=\inf\varphi \quad\mbox{and}\quad \varphi'(u_{0}). $$
Obviously, \(u_{0}\in H_{T}\) is a T-periodic solution of system (1.1). □
Proof of Theorem 1.2
Step 1. To prove φ satisfies the (PS) condition. Suppose that \({u_{k}}\) is a (PS) sequence, that is, \(\varphi'(u_{k})\rightarrow0\) as \(k\rightarrow\infty \) and \({\varphi(u_{k})}\) is bounded. According to (5), there exists \(a_{2}>\frac{1}{4\sin^{2}\frac{\pi}{T}}\) satisfying
$$\limsup_{x\rightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)< - \biggl(a_{2}+ \frac{1}{8\sin^{2}\frac{\pi}{T}} \biggr)R_{1}^{2}. $$
In the same way as (3.1), for any \(u\in H_{T}\), one has
$$\begin{aligned} \Biggl\vert \sum_{n=1}^{T} \bigl(\nabla F_{1}\bigl(n,u_{k}(n)\bigr), \tilde{u}_{k}(n)\bigr)\Biggr\vert \leq{}&\frac{1}{8a_{2}\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2}+\frac{a_{2}}{2}R_{1}^{2}| \bar {u}_{k}|^{2\alpha} \\ &{} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2} \end{aligned}$$
(3.3)
and
$$\begin{aligned} \sum_{n=1}^{T}\bigl(\nabla F_{2} \bigl(u_{k}(n)\bigr),\tilde{u}_{k}(n)\bigr)\geq- \frac {rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma/2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\gamma/2}. \end{aligned}$$
Hence, we have
$$\begin{aligned} \|\tilde{u}_{k}\| \geq{}&\bigl\langle \varphi'(u_{k}),\tilde {u}_{k}\bigr\rangle \\ ={}&\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl(\nabla F\bigl(n,u_{k}(n)\bigr),\tilde{u}_{k}(n)\bigr) \\ \geq{}& \biggl(1-\frac{1}{8a_{2}\sin^{2}\frac{\pi}{T}} \biggr)\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}-\frac {a_{2}}{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha} \\ &{}-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma /2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\gamma/2} \end{aligned}$$
(3.4)
for all large k.
By Lemma 2.1, one has
$$ \|\tilde{u}_{k}\|\leq\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin\frac {\pi}{T}} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2} . $$
(3.5)
By (3.4) and (3.5), for all \(u\in H_{T}^{1}\) one has
$$\begin{aligned} \frac{a_{2}}{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha} \geq{}& \biggl(1-\frac{1}{8a_{2}\sin^{2}\frac{\pi}{T}} \biggr) \sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}- \biggl[\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin\frac {\pi}{T}} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \biggr] \Biggl(\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2} \\ &{}-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma /2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\gamma/2} \\ \geq{}&\frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+C_{1}, \end{aligned}$$
(3.6)
where
$$\begin{aligned} C_{1} ={}&\min_{s\in[0,+\infty)}\biggl\{ {\frac{4a_{2}\sin^{2}\frac {\pi}{T}-1}{8a_{2}\sin^{2}\frac{\pi}{T}}s^{2}- \biggl(\frac {T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2}s^{\alpha+1}} -\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma/2}s^{\gamma} \\ &{}- \biggl[\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin\frac {\pi}{T}}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \biggr]s\biggr\} . \end{aligned}$$
By the choice of \(a_{2}>\frac{1}{4\sin^{2}\frac{\pi}{T}}\), \(-\infty < C_{1}<0\). Hence
$$ \sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \leq a_{2}R_{1}^{2}|\bar {u}_{k}|^{2\alpha}-2C_{1}, $$
(3.7)
and then
$$ \Biggl(\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}\leq\sqrt {a_{2}}R_{1}|\bar{u}_{k}|^{\alpha}+C_{2}, $$
(3.8)
where \(0< C_{2}<+\infty\).
From Theorem 1.1, one has
$$\begin{aligned} \Biggl\vert \sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u_{k}(n)\bigr)-F_{1}(n, \bar{u}_{k})\bigr]\Biggr\vert \leq{}&\frac{1}{8a_{2}\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2}+\frac{a_{2}}{2}R_{1}^{2}| \bar {u}_{k}|^{2\alpha} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}. \end{aligned}$$
(3.9)
By (4), we obtain
$$\begin{aligned} &\sum_{n=1}^{T}\bigl[F_{2} \bigl(u_{k}(n)\bigr)-F_{2}(\bar{u}_{k})\bigr] \\ &\quad=\sum_{n=1}^{T} \int_{0}^{1}\frac{1}{s}\bigl(\nabla F_{2}\bigl(\bar {u}_{k}+s\tilde{u}_{k}(n)\bigr)- \nabla F_{2}(\bar{u}_{k}),s\tilde {u}_{k}(n) \bigr)\,ds \\ &\quad\leq\sum_{n=1}^{T} \int_{0}^{1}Cs^{\delta-1}\bigl|\tilde {u}_{k}(t)\bigr|^{\delta}\,ds\leq\frac{CT}{\delta}\| \tilde{u}_{k}\|_{\infty }^{\delta} \\ &\quad\leq\frac{CT}{\delta} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\delta/2} \Biggl( \sum_{=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\delta/2}. \end{aligned}$$
Combining the boundedness of \(\{\varphi(u_{k})\}\) and (3.7)-(3.9), one has
$$\begin{aligned} C_{3} \leq{}&\varphi(u_{k}) \\ ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u_{k}(n)\bigr)-F_{1}(n, \bar{u}_{k})\bigr] \\ &{}+\sum_{n=1}^{T}\bigl[F_{2} \bigl(u_{k}(n)\bigr)-F_{2}(\bar{u}_{k})\bigr]+\sum _{n=1}^{T}F(n,\bar {u}_{k}) \\ \leq{}& \biggl(\frac{1}{2}+\frac{1}{8a_{2}\sin^{2}\frac{\pi }{T}} \biggr)\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+ \biggl(\frac {T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}+\frac {a_{2}}{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha} \\ &{}+\frac{CT}{\delta} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\delta /2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{\delta/2}+\sum_{n=1}^{T}F(n, \bar{u}_{k}) \\ \leq{}& \biggl(\frac{1}{2}+\frac{1}{8a_{2}\sin^{2}\frac{\pi }{T}} \biggr) \bigl(a_{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha}-2C_{1} \bigr) + \frac{a_{2}}{2}R_{1}^{2}|\bar{u}_{k}|^{2\alpha}+ \sum_{n=1}^{T}F(n,\bar {u}_{k}) \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2}\bigl(\sqrt {a_{2}}R_{1}|\bar{u}_{k}|^{\alpha}+C_{2} \bigr)^{\alpha+1} \\ &{}+\frac{CT}{\delta} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\delta/2}\bigl(\sqrt {a_{2}}R_{1}| \bar{u}_{k}|^{\alpha}+C_{2}\bigr)^{\delta} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3}\bigl(\sqrt {a_{2}}R_{1}|\bar{u}_{k}|^{\alpha}+C_{2} \bigr) \\ \leq{}&|\bar{u}_{k}|^{2\alpha} \Biggl[|\bar{u}_{k}|^{-2\alpha } \sum_{n=1}^{T}F(n,\bar{u}_{k})+ \biggl(a_{2}+\frac{1}{8\sin^{2}\frac{\pi}{T}} \biggr)R_{1}^{2} \Biggr] \\ &{}+C_{4}|\bar{u}_{k}|^{\alpha(\alpha+1)}+C_{5}|\bar {u}_{k}|^{\alpha}+C_{6}|\bar{u}_{k}|^{\alpha\delta}+C_{7} \end{aligned}$$
for large k. By the choice of \(a_{2}\), \(\{\bar{u}_{k}\}\) is bounded. From (3.7), \(\{u_{k}\}\) is bounded. In view of \(H_{T}\) is finite dimensional Hilbert space, φ satisfies the (PS) condition.
Step 2. Let \(\tilde{H}_{T}=\{u\in H_{T}:\bar{u}=0 \}\). We show that, for \(u\in\tilde{H}_{T}\),
$$ \varphi(u)\rightarrow+\infty,\quad \|u\|\rightarrow\infty. $$
(3.10)
From (1) and Lemma 2.1, one has
$$\begin{aligned} \Biggl\vert \sum_{n=1}^{T} \bigl[F_{1}\bigl(n,u(n)\bigr)-F(n,0)\bigr]\Biggr\vert ={}&\Biggl\vert \sum_{n=1}^{T} \int_{0}^{1}\bigl(\nabla F_{1}\bigl(n,s u(n)\bigr),u(n)\bigr)\,ds\Biggr\vert \\ \leq{}&\sum_{n=1}^{T}f(n)\bigl|u(n)\bigr|^{\alpha+1}+ \sum_{n=1}^{T}g(n)\bigl|u(n)\bigr| \\ \leq{}& R_{2}\|u\|_{\infty}^{\alpha+1}+R_{3}\|u \|_{\infty} \\ \leq{}& \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha +1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2} \end{aligned}$$
for all \(u\in\tilde{H}_{T}\). It follows from (2) that
$$\begin{aligned} &\sum_{n=1}^{T}\bigl[F_{2} \bigl(u(n)\bigr)-F_{2}(0)\bigr] \\ &\quad=\sum_{n=1}^{T} \int _{0}^{1}\bigl(\nabla F_{2}\bigl(su(n) \bigr)-\nabla F_{2}(0),u(n)\bigr)\,ds \\ &\quad\geq-\sum_{n=1}^{T} \int_{0}^{1}rs^{\gamma-1}\bigl|u(n)\bigr|^{\gamma }\,ds \geq-\frac{rT}{\gamma}\|u\|_{\infty}^{\gamma} \\ &\quad\geq-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma/2} \Biggl( \sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{\gamma/2}. \end{aligned}$$
Hence, we have
$$\begin{aligned} \varphi(u) ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F\bigl(n,u(n)\bigr)-F(n,0)\bigr]+\sum_{n=1}^{T}F(n,0) \\ \geq{}&\frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2}- \biggl(\frac {T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2}+\sum_{n=1}^{T}F(n,0) \\ &{}-\frac{rT}{\gamma} \biggl(\frac{T^{2}-1}{6T} \biggr)^{\gamma /2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{\gamma/2}. \end{aligned}$$
In view of Lemma 2.1, \(\|u\|\rightarrow+\infty\) in \(\tilde{H}_{T}\) if and only if \((\sum_{n=1}^{T}|\Delta u(n)|^{2})^{1/2}\rightarrow\infty\). Hence (3.10) is satisfied.
Step 3. By (5), for all \(u\in(\tilde{H}_{T})^{\bot }=\mathbb{R}^{N}\), one has
$$\varphi(u)=- \sum_{n=1}^{T}F\bigl(n,u(n) \bigr)\rightarrow-\infty, \quad\| u\| \rightarrow\infty. $$
Above all, all conditions of Lemma 2.2 are satisfied. So, by Lemma 2.2, system (1.1) has at least one T-periodic solution. □
Proof of Theorem 1.3
By (7), there exists \(a_{3}>\frac {3}{(12-2(T^{2}-1)r)\sin^{2}\frac{\pi}{T}}\) satisfying
$$\liminf_{|x|\longrightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)>\frac{a_{3}}{2}R_{1}^{2}. $$
Similar to (3.1), we have
$$\begin{aligned} &\sum_{n=1}^{T}\bigl[F_{1} \bigl(n,u(n)\bigr)-F_{1}(n,\bar{u})\bigr] \\ &\quad\geq-\frac{1}{8a_{3}\sin^{2}\frac{\pi}{T}}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}-\frac{a_{3}}{2}R_{1}^{2}| \bar{u}|^{2\alpha }- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &\qquad{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2}. \end{aligned}$$
By (6) and Lemma 2.1, one has
$$\begin{aligned} \sum_{n=1}^{T}\bigl[F_{2} \bigl(u(n)\bigr)-F_{2}(\bar{u})\bigr] &=\sum _{n=1}^{T} \int_{0}^{1}\frac{1}{s}\bigl(\nabla F_{2}\bigl(\bar {u}+s\tilde{u}(n)\bigr)-\nabla F_{2}( \bar{u}),s\tilde{u}(n)\bigr)\,ds \\ &\geq-\sum_{n=1}^{T} \int_{0}^{1}rs\bigl|\tilde{u}(n)\bigr|^{2}\,ds\geq - \frac{(T^{2}-1)r}{12}\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2}. \end{aligned}$$
So, for any \(u\in H_{T}\), we have
$$\begin{aligned} \varphi(u) ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl| \Delta u(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F\bigl(n,u(n)\bigr)-F(n,\bar{u})\bigr]+\sum_{n=1}^{T}F(n, \bar{u}) \\ \geq{}& \biggl(\frac{1}{2}-\frac{1}{8a_{3}\sin^{2}\frac{\pi }{T}}-\frac{(T^{2}-1)r}{12} \biggr)\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2}- \biggl(\frac {T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum_{n=1}^{T}\bigl|\Delta u(n)\bigr|^{2} \Biggr)^{1/2} \\ &{}+|\bar{u}|^{2\alpha} \Biggl(|\bar{u}|^{-2\alpha}\sum _{n=1}^{T}F(n,\bar{u})-\frac{a_{3}}{2}R_{1}^{2} \Biggr). \end{aligned}$$
Therefore, \(\varphi(u)\rightarrow+\infty\) as \(\|u\|\rightarrow+\infty\) due to the choice of \(a_{3}\) and \(r<\frac{6}{T^{2}-1}\). The rest is similar to the proof of Theorem 1.1. □
Proof of Theorem 1.4
First, we prove that φ satisfies the (PS) condition. Suppose that \(\{u_{k}\}\subset H_{T}\) is a (PS) sequence of φ, that is, \(\varphi'(u_{k})\rightarrow0\) as \(k\rightarrow\infty\) and \(\{\varphi(u_{k})\}\) is bounded. By (9), there exists \(a_{4}>\frac{1}{4\sin^{2}\frac{\pi}{T}}\) satisfying
$$ \limsup_{|x|\rightarrow+\infty}|x|^{-2\alpha}\sum _{n=1}^{T}F(n,x)< - \biggl(a_{4}+ \frac{1}{8\sin^{2}\frac{\pi}{T}} \biggr)R_{1}^{2}. $$
(3.11)
By the \((\lambda,\mu)\)-subconvexity of \(G(x)\), we have
$$ G(x)\leq \bigl(2\mu|x|^{\beta}+1 \bigr)G_{0} $$
(3.12)
for all \(x\in\mathbb{R}^{N}\), where \(G_{0}=\max_{|s|\leq1}G(s)\), \(\beta =\log_{2\lambda}(2\mu)<2\).
Then
$$\begin{aligned} \sum_{n=1}^{T}\bigl(\nabla F_{2}\bigl(n,u_{k}(n)\bigr),\tilde{u}_{k}(n) \bigr) &\geq-\sum_{n=1}^{T}h(n)G( \bar{u}_{k}) \\ &\geq-\sum_{n=1}^{T}h(n) \bigl(2\mu| \bar{u}_{k}|^{\beta }+1\bigr)G_{0} \\ &=-2\mu R_{4}|\bar{u}_{k}|^{\beta}-R_{4}, \end{aligned}$$
(3.13)
where \(R_{4}=G_{0}\sum_{n=1}^{T}h(n)\). For large k, according to (3.3) and (3.13) we have
$$\begin{aligned} \|\tilde{u}_{k}\| \geq{}&\bigl\langle \varphi'(u_{k}),\tilde {u}_{k}\bigr\rangle \\ ={}&\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl(\nabla F\bigl(n,u_{k}(n)\bigr),\tilde{u}_{k}(n)\bigr) \\ \geq{}& \biggl(1-\frac{1}{8a_{4}\sin^{2}\frac{\pi}{T}} \biggr)\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2}-\frac{a_{4}}{2}R_{1}^{2}| \bar {u}_{k}|^{2\alpha} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}-2\mu R_{4}|\bar {u}_{k}|^{\beta}-R_{4}. \end{aligned}$$
(3.14)
By (3.5) and (3.14), one has
$$\begin{aligned} &\frac{a_{4}}{2}R_{1}^{2}| \bar{u}_{k}|^{2\alpha}+2\mu R_{4}|\bar {u}_{k}|^{\beta} \\ &\quad\geq \biggl(1-\frac{1}{8a_{4}\sin^{2}\frac {\pi}{T}} \biggr)\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \\ &\qquad{}- \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2}-R_{4} \\ &\qquad{}- \biggl[\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin\frac {\pi}{T}}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \biggr] \Biggl(\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2} \\ &\quad\geq \frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+C_{8}, \end{aligned}$$
(3.15)
where
$$\begin{aligned} C_{8} ={}&\min_{s\in[0,+\infty)}\biggl\{ \biggl( \frac{1}{2}-\frac {1}{8a_{4}\sin^{2}\frac{\pi}{T}} \biggr)s^{2} - \biggl( \frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2}s^{\alpha+1} \\ &{}-R_{4}- \biggl[\frac{(4\sin^{2}\frac{\pi}{T}+1)^{1/2}}{2\sin \frac{\pi}{T}}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \biggr]s\biggr\} . \end{aligned}$$
By the choice of \(a_{4}\), \(-\infty< C_{8}<0\). By (3.15), we have
$$ \sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \leq a_{4}R_{1}^{2}|\bar {u}_{k}|^{2\alpha}+4 \mu R_{4}|\bar{u}_{k}|^{\beta}-2C_{8}, $$
(3.16)
and then
$$ \Biggl(\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}\leq\sqrt {a_{4}}R_{1}|\bar{u}_{k}|^{\alpha}+2\sqrt{ \mu R_{4}}|\bar{u}_{k}|^{\beta /2}+C_{9}, $$
(3.17)
where \(C_{9}>0\). By (8) and (3.12), for any \(u\in H_{T}\), we get
$$\begin{aligned} &\sum_{n=1}^{T} \bigl[F_{2}\bigl(n,u(n)\bigr)-F_{2}(n,\bar{u})\bigr] \\ &\quad=-\sum_{n=1}^{T} \int_{0}^{1}\bigl(\nabla F_{2}\bigl(n,\bar {u}_{k}+s\tilde{u}_{k}(n)\bigr),\tilde{u}_{k}(n) \bigr)\,ds \\ &\quad\leq\sum_{n=1}^{T} \int_{0}^{1}h(n)G\bigl(\bar {u}_{k}+(s+1) \tilde{u}_{k}(n)\bigr)\,ds \\ &\quad\leq\sum_{n=1}^{T} \int_{0}^{1}h(n) \bigl(2\mu\bigl|\bar {u}_{k}+(s+1)\tilde{u}_{k}(n)\bigr|^{\beta}+1 \bigr)G_{0}\,ds \\ &\quad\leq4\mu\sum_{n=1}^{T}h(n) \bigl(| \bar{u}_{k}|^{\beta}+2^{\beta }\bigl|\tilde{u}_{k}(n)\bigr|^{\beta} \bigr)G_{0}+R_{4} \\ &\quad\leq2^{\beta+2}\mu R_{4}\|\tilde{u}_{k} \|_{\infty}^{\beta }+4\mu R_{4}|\bar{u}_{k}|^{\beta}+R_{4} \\ &\quad\leq \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{\beta +2}\mu R_{4} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{\beta /2}+4\mu R_{4}|\bar{u}_{k}|^{\beta}+R_{4}. \end{aligned}$$
(3.18)
Combining the boundedness of \(\{\varphi(u_{k})\}\) and (3.16)-(3.18), one has
$$\begin{aligned} C_{10} \leq{}&\varphi(u_{k}) \\ ={}&\frac{1}{2}\sum_{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+\sum_{n=1}^{T} \bigl[F\bigl(n,u_{k}(n)\bigr)-F(n,\bar{u}_{k})\bigr]+\sum _{n=1}^{T}F(n,\bar {u}_{k}) \\ \leq{}& \biggl(\frac{1}{2}+\frac{1}{8a_{4}\sin^{2}\frac{\pi }{T}} \biggr)\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2}+ \frac {a_{4}}{2}R_{1}^{2}|\bar{u}_{k}|^{2}\\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{(\alpha+1)/2} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \Biggl(\sum _{n=1}^{T}\bigl|\Delta u_{k}(n)\bigr|^{2} \Biggr)^{1/2}\\ &{}+ \biggl(\frac {T^{2}-1}{6T} \biggr)^{\beta/2}2^{\beta+2} \mu R_{4} \Biggl(\sum_{n=1}^{T}\bigl| \Delta u_{k}(n)\bigr|^{2} \Biggr)^{\beta/2} +4\mu R_{4}|\bar{u}_{k}|^{\beta}+R_{4}+ \sum_{n=1}^{T}F(n,\bar{u}_{k}) \\ \leq{}& \biggl(\frac{1}{2}+\frac{1}{8a_{4}\sin^{2}\frac{\pi }{T}} \biggr) \bigl(a_{4}R_{1}^{2}| \bar{u}_{k}|^{2\alpha}+4\mu R_{4}|\bar {u}_{k}|^{\beta}-2C_{8} \bigr)+\frac{a_{4}}{2}R_{1}^{2}| \bar {u}_{k}|^{2\alpha} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \bigl( \sqrt{a_{4}}R_{1}|\bar{u}_{k}|^{\alpha}+2 \sqrt{\mu R_{4}}|\bar {u}_{k}|^{\beta/2}+C_{9} \bigr)^{\alpha+1} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \bigl(\sqrt {a_{4}}R_{1}|\bar{u}_{k}|^{\alpha}+2\sqrt{ \mu R_{4}}|\bar{u}_{k}|^{\beta /2}+C_{9} \bigr) \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{\beta+2}\mu R_{4} \bigl(\sqrt{a_{4}}R_{1}| \bar{u}_{k}|^{\alpha}+2\sqrt{\mu R_{4}}| \bar{u}_{k}|^{\beta/2}+C_{9} \bigr)^{\beta}+4\mu R_{4}|\bar {u}_{k}|^{\beta}\\ &{}+R_{4}+\sum _{n=1}^{T}F(n,\bar{u}_{k}) \\ \leq{}& \biggl(1+\frac{1}{8a_{4}\sin^{2}\frac{\pi}{T}} \biggr)a_{4}R_{1}^{2}| \bar{u}_{k}|^{2\alpha}+ \biggl(6+\frac{1}{2a_{4}\sin ^{2}\frac{\pi}{T}} \biggr)\mu R_{4}|\bar{u}_{k}|^{\beta}\\ &{}- \biggl(1+ \frac {1}{4a_{4}\sin^{2}\frac{\pi}{T}} \biggr)C_{8} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}R_{2} \bigl(2^{\alpha}a_{4}^{\frac{\alpha+1}{2}}R_{1}^{\alpha+1}| \bar {u}_{k}|^{\alpha(\alpha+1)} +2^{3\alpha+1}\mu^{\frac{\alpha+1}{2}}R_{4}^{\frac{\alpha+1}{2}}| \bar {u}_{k}|^{\frac{\beta(\alpha+1)}{2}} +2^{2\alpha}C_{9}^{\alpha+1} \bigr) \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3} \bigl(\sqrt {a_{4}}R_{1}|\bar{u}_{k}|^{\alpha}+2\sqrt{ \mu R_{4}}|\bar{u}_{k}|^{\beta /2}+C_{9} \bigr) \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{\beta+2}\mu R_{4} \bigl(2^{\beta-1}a_{4}^{\frac{\beta}{2}}R_{1}^{\beta}| \bar {u}_{k}|^{\alpha\beta}+2^{3\beta-2}\mu^{\frac{\beta}{2}}R_{4}^{\frac {\beta}{2}}| \bar{u}_{k}|^{\frac{\beta^{2}}{2}} +2^{2(\beta-1)}C_{9}^{\beta} \bigr)\\ &{}+R_{4}+\sum_{n=1}^{T}F(n, \bar {u}_{k}) \\ ={}&|\bar{u}_{k}|^{2\alpha}\Biggl[|\bar{u}_{k}|^{-2\alpha} \sum_{n=1}^{T}F_{1}(n, \bar{u}_{k})+ \biggl(a_{4}+\frac{1}{8\sin^{2}\frac{\pi }{T}} \biggr)R_{1}^{2} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}\sqrt{a_{4}}R_{1}R_{3}|\bar {u}_{k}|^{-\alpha} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}2^{\alpha }a_{4}^{\frac{\alpha+1}{2}}R_{1}^{\alpha+1}| \bar{u}_{k}|^{\alpha(\alpha-1)} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{2\beta+1}\mu a_{4}^{\frac {\beta}{2}}R_{1}^{\beta}R_{4}| \bar{u}_{k}|^{\alpha(\beta-2)}\Biggr] \\ &{}+|\bar{u}_{k}|^{\beta}\Biggl[|\bar{u}_{k}|^{-\beta} \sum_{n=1}^{T}F_{2}(n, \bar{u}_{k})+ \biggl(6+\frac{1}{2a_{4}\sin^{2}\frac{\pi }{T}} \biggr)\mu R_{4}\\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{4\beta } \mu^{\frac{\beta+2}{2}}R_{4}^{\frac{\beta+2}{2}}|\bar{u}_{k}|^{\frac {1}{2}\beta^{2}-\beta} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}2^{3\alpha +1} \mu^{\frac{\alpha+2}{2}}R_{2}R_{4}^{\frac{\alpha+1}{2}}|\bar {u}_{k}|^{\frac{\beta(\alpha-1)}{2}} + \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}2R_{3}\sqrt{\mu R_{4}}|\bar {u}_{k}|^{-\beta/2}\Biggr] \\ &{}- \biggl(1+\frac{1}{4a_{4}\sin^{2}\frac{\pi}{T}} \biggr)C_{8}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{(\alpha+1)/2}2^{2\alpha }R_{2}C_{9}^{\alpha+1}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{1/2}R_{3}C_{9} \\ &{}+ \biggl(\frac{T^{2}-1}{6T} \biggr)^{\beta/2}2^{3\beta}\mu R_{4}C_{9}^{\beta}+R_{4}. \end{aligned}$$
Combining (3.11) and the above inequality, we see that \(\{|\bar {u}|\}\) is bounded. By (3.16), \(\{u_{k}\}\) is bounded. Since \(H_{T}\) is a finite dimensional Hilbert space, φ satisfies the (PS) condition.
Similar to the proof of Theorem 1.2, all conditions of Lemma 2.2 are satisfied. So, the proof of Theorem 1.4 is completed. □