## 2 Almost Anti-Periodic Discrete Functions
\begin{array}{ll}&\left|\sum\limits_{j=a+1}^{a+b}\ln\|M(j)\|-\sum\limits_{j= 1}^{b}\ln\|M(j)\|\right|\\ &=\left|\sum\limits_{j=\tau+1}^{\tau+b}\ln\|M(j)\|-\sum\limits_{j=1}^{b}\ln\|M (j)\|+\sum\limits_{j=a+1}^{\tau}\ln\|M(j)\|-\sum\limits_{j=a+b+1}^{\tau+b}\ln \|M(j)\|\right|\\ &\leq\left.\sum\limits_{j=1}^{b}\left|\ln\|M(j+\tau)\|-\ln\|M(j)\|\right|+\sum \limits_{j=a+1}^{\tau}\left|\ln\|M(j)\|\right|+\sum\limits_{j=a+b+1}^{\tau+b} \left|\ln\|M(j)\|\right|\\ &\leq\left.\frac{\varepsilon b}{4}+2l_{0}\lambda.\end{array} \]  
Moreover, for the case I, through taking \(a=(k-1)n\), \(b=n\), we have  
\[\left|\sum\limits_{j=(k-1)n+1}^{kn}\ln\|M(j)\|-\sum\limits_{j=1}^{n}\ln\|M(j) \|\right|\leq\frac{\varepsilon n}{4};\]  
for the case II, through taking \(a=(k-1)n\), \(b=n\), one has  
\[\left|\sum\limits_{j=(k-1)n+1}^{kn}\ln\|M(j)\|-\sum\limits_{j=1}^{n}\ln\|M(j) \|\right|\leq\frac{\varepsilon n}{4}+2l_{0}\lambda.\]  
Step 4. We will prove that the Limit (1) exists. For any \(m,n\in\mathbb{N}^{+}\), one has  
\[ \begin{array}{ll}\left|\frac{1}{n}\sum\limits_{j=1}^{n}\ln\|M(j)\|-\frac{1 }{m}\sum\limits_{j=1}^{m}\ln\|M(j)\|\right|&\leq\frac{1}{mn}\left|m\sum \limits_{j=1}^{n}\ln\|M(j)\|-\sum\limits_{j=1}^{mn}\ln\|M(j)\|\right|\\ &+\frac{1}{mn}\left|\sum\limits_{j=1}^{mn}\ln\|M(j)\|-n\sum\limits_{j=1}^{m} \ln\|M(j)\|\right|\\ &\leq\frac{1}{mn}\left[\sum\limits_{k=1}^{m}\left|\sum\limits_{j=(k-1)n+1}^{kn }\ln\|M(j)\|-\sum\limits_{j=1}^{n}\ln\|M(j)\|\right|\right|\\ &+\left.\sum\limits_{k=1}^{n}\left|\sum\limits_{j=(k-1)m+1}^{km}\ln\|M(j)\|- \sum\limits_{j=1}^{m}\ln\|M(j)\|\right|\right]\!.\end{array} \]On the other hand, by Step 3, if \(\tau=a=(k-1)n\), then  
\[\left|\frac{1}{n}\sum\limits_{j=1}^{n}\ln\|M(j)\|-\frac{1}{m}\sum\limits_{j=1}^{ m}\ln\|M(j)\|\right|\leq\frac{1}{mn}(m\frac{en}{4}+n\frac{em}{4})=\frac{\epsilon}{2 }<\epsilon;\]  
if \(\tau\in(a,a+l_{0}]_{\mathbb{Z}}=((k-1)n,(k-1n)+l_{0}]_{\mathbb{Z}}\), then  
\[\left|\frac{1}{n}\sum\limits_{j=1}^{n}\ln\|M(j)\|-\frac{1}{m}\sum \limits_{j=1}^{m}\ln\|M(j)\|\right| \leq\frac{1}{mn}\left[m(\frac{cn}{4}+2l_{0}\lambda)+n(\frac{mc+2l _{0}}{4}\lambda)\right]\] \[=\frac{\epsilon}{2}+2l_{0}\lambda(\frac{1}{n}+\frac{1}{m})<\epsilon\]  
for \(m,n\geq[\frac{8l_{0}\lambda}{\epsilon}]+1\), which indicates that the Limit (1) exists. The proof is completed. Proposition 2: _Let \(M(\cdot)\in\mathfrak{G}\). Then, the following limit equalities hold:_  
(i) \[h=\lim\limits_{n\to\infty}\left[\prod\limits_{j=m+1}^{m+n}\|M(j)\| \right]^{\frac{1}{n}}=\lim\limits_{n\to\infty}\left[\prod\limits_{j=1}^{n}\|M (j)\|\right]^{\frac{1}{n}}.\] (ii) \[h=\lim\limits_{n\to-\infty}\left[\prod\limits_{j=n}^{-1}\|M(j)\| \right]^{-\frac{1}{n}}=\lim\limits_{n\to\infty}\left[\prod\limits_{j=-n}^{-1} \|M(j)\|\right]^{\frac{1}{n}}.\]  
Proof: (i) Through using the process of the proof of Proposition 1, we can turn (i) into the following equality:  
\[\lim\limits_{n\to\infty}\frac{1}{n}\bigg{[}\sum\limits_{j=m+1}^{m+n}\ln\|M(j) \|-\sum\limits_{j=1}^{n}\ln\|M(j)\|\bigg{]}=0. \tag{2}\]  
By the proof of Proposition 1 Step 2, for \(\forall\frac{\epsilon}{2}>0\), there exists a positive integer \(l_{0}(\epsilon)\) such that \(M(j)\) satisfies the following condition:  
\[\big{|}\ln\|M(j+\tau)\|-\ln\|M(j)\|\big{|}\leq\frac{\epsilon}{2}\ \ \text{for all}\ \ j\in\mathbb{Z},\]  
where \(\tau\in[m,m+l]_{\mathbb{Z}}\) and \(m\in\mathbb{Z}\). Next, we will consider \(\tau=m\) and \(\tau\in(m,m+l]_{\mathbb{Z}}\) cases. By the proof of Proposition 1 Step 3, taking \(a=m\) and \(b=n\), the desired results can be shown. Cases I. \(\tau=m\). For \(\tau=m\), we have  
\[\frac{1}{n}\bigg{|}\sum\limits_{j=m+1}^{m+n}\ln\|M(j)\|-\sum \limits_{j=1}^{n}\ln\|M(j)\|\bigg{|}\] \[= \frac{1}{n}\bigg{|}\sum\limits_{j=1+1}^{\tau+n}\ln\|M(j)\|-\sum \limits_{j=1}^{n}\ln\|M(j)\|\bigg{|}\] \[\leq \frac{1}{n}\sum\limits_{j=1}^{n}\big{|}\ln\|M(j+\tau)\|-\ln\|M(j) \|\big{|}\leq\frac{1}{n}\frac{en}{2}<\epsilon.\]  
Therefore, Equation (2) holds. Cases II. \(\tau\in(m,m+l]_{\mathbb{Z}}\).