Let $f_n$ be monotonically decreasing sequence of functions on a compact interval $I\subset\mathbb{R}$ that converges pointwise to a continuous function $F$.
If all $f_n$ were continuous, Dini would give uniform convergence to $F$. But all $f_n$ are discontinuous in finitely many points $x^d$, $x^d$ independent of $n$. However, if $f_n$ is discontinuous in $x^d$ from the left [right] then $f_n(x^d)>f_n(x^d-\epsilon)$ $\ \ \ \ [f_n(x^d)>f_n(x^d+\epsilon)]$.
It seems reasonable that $f_n$ converges also uniformly to $F$.$\ \ \ \ $ (1)
I would try to prove (1) by partitioning $I$ into compact subsets $a_i = [x^d_i, x^d_{i+1}]$ ($x_0^d=\min(I), x_m^d=\max(I))$ and proving uniform convergence on each subset for $\hat{f}_{n,i}$, where $\hat{f}_{n,i}(x)=f_n(x)$ for all $x\neq x^d$ and $\hat{f}_{n,i}(x^d_i) = \lim_{\epsilon\rightarrow 0}f_n(x^d_i+\epsilon)$ , $\hat{f}_{n,i}(x^d_{i+1}) = \lim_{\epsilon\rightarrow 0}f_n(x^d_{i+1}-\epsilon)$.
Each interval is compact. All $\hat{f}_{n,i}$ should be monotonously decreasing and continuous converging pointwise and hence (Dini) uniformly to $F|_i$. Uniform convergence of $f_n\rightarrow F$ should follow directly.
- Is my assumption (1) true?
- Or did I miss something and there is even a simple counterexample?
- Any ideas for a more elegant proof?
Thank you very much for your help!
Bernd