Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

I think your argument works, the only thing that isn't completely justified is why you can conclude that since $\hat{f}_n$ converges uniformly to $f$, so does $f_n$. And where do you use that there are only a finite number of points of discontinuity (the example below shows that this assumption is necessary). You can complete the argument as follows; Given $\epsilon>0$ there exists $\hat{N}$ and $\delta>0$ such that $$ n>\hat{N} \ \Rightarrow \ \forall x \ : \ |\hat{f}_n(x)-f(x)|<\epsilon$$ Label the points of discontinuity $\{y_1,\ldots,y_k\}$. By pointwise convergence we know that for each point of discontinuity $y_j$, there exists and $N_j$ such that $$ n>N_j \ \Rightarrow \ |f_n(x)-\hat{f}_n(x)|<\epsilon$$ But now taking $N=\max(\hat{N},N_1,\ldots,N_k)$ should give $$ n>N \ \Rightarrow \ |f_n(x)-f(x)|<2\epsilon$$

EDIT:The following answers a misunderstood version of the question, but since reference to the example is made above, I will leave it up. Missed the assumption that points of discontinuity are assumed independent of $n$ and finite in number.

I am not convinced that this is necessarily true. How about the sequence of functions $$ f_n(x) = \begin{cases} 0 & \text{ if } x\in ]-\infty,-1/n[\cup [0;\infty[\\ 1/2+1/n & \text{ if }x\in [-1/n;0[ \\ \end{cases}$$ seems to me like this sequence of functions will converge pointwise to $0$ and be stictly decreasing, but it is clearly not uniformly convergent, since $$ \forall n : \ \max(\{|f_n(-\delta) - f_n(0)| \ \big| \ \delta \in [-1;0]\}) >1/2$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.