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Suppose we have a sequence {$g_n$} converges uniformly on $(-\infty,0]$ to $g:(-\infty,0]\to \mathbb{R}$, also $g_n:(-\infty,0]\to \mathbb{R}$. Now we assume $\lim\limits_{x\to -\infty}g_n(x)$ exist for all $n$ which is even. Let $a_n=\lim\limits_{x\to -\infty}g_{2n}(x)$

Show that $\lim\limits_{x\to -\infty}g(x)$ and $\lim\limits_{n\to -\infty}a_n$ give same value. Does there exist an positive integer $N$ such that $\lim\limits_{x\to -\infty}g_n(x)$ exist for all $n>N$?

For the first question, we need to prove existence first. From the uniform convergence, we find: Given an $\epsilon>0$, there exists an $N_1$ such that $\forall n>N_1$, $|g(y)-g_n(y)|<\frac{\epsilon}{2}$, $\forall y\in(-\infty,0]$. Form the $\lim\limits_{x\to -\infty}g_n(x)$, suppose it is $L$, then there also exists an $\delta>0$ such that $|x|>\delta$, $|g_n(x)-L|<\frac{\epsilon}{2}$. From $$|g(x)-L|\leq |g(x)-g_n(x)|+|g_n(x)-L|$$ we can show the $\lim\limits_{x\to -\infty}g(x)$ exists. But for $\lim\limits_{n\to -\infty}a_n$, I am confused aboth the final $N$ we should take, how to combine the condition together? For the second question, I cannot find the answer.

Can someone help me with my problems? Thank you in advance.

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  • $\begingroup$ Hint : start by showing that $(a_n)$ is a Cauchy sequence and therefore converges. $\endgroup$ Jun 9, 2021 at 7:29

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For the first question, we first need to show that $(a_n)$ converges, by showing that it is a Cauchy sequence.

Let $\epsilon >0$. Since $g_n \to g$ uniformly, there is $N$ an integer such that : $$\forall n\geq N, \forall y\in (-\infty,0], |g_n(y)-g(y)|<\epsilon/3$$ which implies : $$\forall n,m \geq N/2, \forall y\in (-\infty,0], |g_{2n}(y)-g_{2m}(y)|<2\epsilon/3$$ Taking $y\to -\infty$ we see that : $$\forall n,m \geq N/2, |a_n - a_m|\leq 2\epsilon /3 < \epsilon$$

Therefore, $(a_n)$ is Cauchy and converges to some real number $\ell$.


We can now show that $\lim_{-\infty} g$ exists and is equal to $\lim a_n$. Given $\epsilon >0$, let $N$ be an integer such that : \begin{align} \forall n \geq N, &\forall y\in (-\infty,0], |g(y) - g_n(y)|<\epsilon/3\\ \forall n\geq N, &|a_n - \ell|<\epsilon/3 \end{align} Since $\lim_{y\to -\infty} g_{2N}(y) = a_{2N}$, we have $A<0$ such that :
$$\forall y < A, |g_{2N}(y) - a_{2N}|<\epsilon/3$$ Then, for $y<A$, we have : $$|g(y) - \ell|\leq |g(y) - g_{2N}(y)|+|g_{2N}(y)-a_{2N}|+|a_{2N}-\ell| <\epsilon$$

Therefore $\lim_{-\infty} g$ exists and : $$\lim_{y\to - \infty}g(y) = \ell =\lim_{n\to +\infty} a_n$$


For the second question, consider $g_{2n}(y) = 0$ and $g_{2n+1}(y) = (1/n) \sin y$.

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  • $\begingroup$ Thank you! It is an interesting way to consider the Cauchy sequence. $\endgroup$
    – David
    Jun 9, 2021 at 12:07

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