Theorem: Let $(f_n)$ be a sequence of real valued functions defined on metric space $E$. Suppose that $(f_n)$ converges uniformly on $E$ to the limit function $f$. Let $x$ be a limit point of $E$. Suppose that $ \lim_{t\to x}f_n(t)=:A_n$ exists for every $n\in \mathbb N$.
Then $\lim A_n$ exists and $\lim_nA_n= \lim_{t\to x} f(t)$
Proof: Let $\epsilon\gt 0$ be given. By Cauchy criterion of uniform convergence, there exists $N$ such that $m\ge N, n\ge N\implies |f_m(t)-f_n(t)|\lt\epsilon$ for all $t\in E$. Taking limit $t\to x$ on both sides to get: $|A_m-A_n|\le\epsilon$. It follows that $(A_n)$ is a Cauchy sequence and hence must converge.
To show that $(A_n)$ converges to $\lim_{t\to x} f(t)$, let's note that
since $(f_n)$ converges to $f$ uniformly on $E$, it follows that there exists $N'$ such that $$n\ge N'\implies |f_n(t)-f(t)|<\epsilon\:\: \forall t\in E\tag 1$$
It follows by $(1)$ that $\lim_{t\to x}|f_n(t)-f(t)|\le\epsilon\implies |A_n-\lim_{t\to x}f(t)|\le\epsilon$, whence it follows that $\lim_n A_n=\lim_{t\to x}f(t)$.
Is my proof correct? Thanks.
Best Answer
The major gap in the proof is your failure to justify why $\lim\limits_{t\to x} f(t)$ even exists.