Prove that $f_n$ converges uniformly to $f$ on $[a,b)$

Question

Let $-\infty < a < b \leq +\infty$. Suppose that $f_n$ converges uniformly to $f$ on $[a,r]$ for each $r\in(a,b)$. Prove that $f_n$ converges uniformly to $f$ on $[a,b)$ if, and only if, for each sequence ${b_n}$ that $b_n \leq b_{n+1}$ for all n $\in \mathbb{N}$, $b_n \to b$, and $f_n(b_n) -f(b_n) \to 0$.

($\Rightarrow$) Already proved.

($\Leftarrow$) I'm having problems with this part. I think that the way is to construct a sequence. Can you guys give me an ideia of how to prove this?

Answer 1

Suppose that $f_n$ does not converge uniformly to $f$ on $[a,b)$. We must then have the existence of some $\delta>0$ such that there exists an infinite subsequence $(n_k) \subset \mathbb N$, so that $\|f_{n_k}-f\|_\infty\geq\delta$ for every $k\in \mathbb N$. This means that for each $k\in \mathbb N$ there exists a $b_k\in [a,b)$ such that $|f_{n_k}(b_k)-f(b_k)|\geq \delta/2$.

However, our first assumption must mean that $b_k\to b$, else this sequence would contradict our uniform convergence on all closed subintervals (I leave the details of this to you). With some possible reordering and choosing of our $b_k$, we can ensure that $b_{k+1}\geq b_k$, and still have $b_k\to b$. From this we can construct a sequence $(b_n)$ so that $b_{n_k}=b_k$ for all $k$ and still have $b_{n+1}\geq b_n$ and $b_n\to b$ (just by setting $b_n$ to be equal to the greatest $b_{n_k}$ with $n_k<n$ whenever $n\neq n_j$ for some $j$.) We thus have $f_n(b_n)-f(b_n)\to 0$, but $|f_{n_k}(b_{n_k})-f(b_{n_k})|\geq\delta/2$, a contradiction.

Answer 2

Suppose that the convergence is not uniform. Then there is a $\varepsilon>0$ such that$$(\forall N\in\mathbb N)(\exists n\in\mathbb N)\bigl(\exists x\in[a,b)\bigr):n\geqslant N\text{ and }\bigl\lvert f(x)-f_n(x)\bigr\rvert\geqslant\varepsilon.$$So, there is a sequence $(b_n)_{n\in\mathbb N}$ of elements of $[a,b)$ such that the inequality $\bigl\lvert f(b_n)-f_n(b_n)\bigr\rvert\geqslant\varepsilon$ occurs infinitely often. Since $[a,b)$ is bounded, you can assume without loss of generality the sequence $(b_n)_{n\in\mathbb N}$ converges to some $c\in[a,b]$. If $c\in[a,b)$, take some $r\in(c,b)$. Then you know that the convergence is uniform on $[a,r]$. This is impossible, because then we should have $\lim_{n\to\infty}f_n(b_n)=\lim_{n\to\infty}f(b_n)$. So, you can't have $c<b$. Therefore, $c=b$.