# Show the following limit exist.

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.

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

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 $$y, we have : $$|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 $$g_{2n+1}(y) = (1/n) \sin y$$.

• Thank you! It is an interesting way to consider the Cauchy sequence. Jun 9, 2021 at 12:07