3
$\begingroup$

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

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $$ \forall n : \ \max(\{|f_n(-\delta) - f_n(0)| \ \big| \ \delta \in [-1;0]\}) >1/2$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .