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Suppose $f_n$ is a sequence of functions defined a set $K$ with pointwise limit function $f$.

I am confused about the following.

If the following conditions are satisfied:

  1. $f_n$ is continuous on $K$ for all $n$.
  2. The pointwise limit $f$ is continuous on $K$.
  3. $K$ is a compact interval (i.e., a closed and bounded interval in $\mathbb{R}$).
  4. The convergence of $f_n$ to $f$ is increasing or decreasing.

Then does this imply that $f_n$ is uniformly convergent to $f$?

Now a different problem: If one of these conditions is not satisfied, does this imply that $f_n$ is not uniformly convergent to $f$?

If $f_n$ is not uniformly convergent to f, does this mean that one these four conditions doesn't hold?

In general: what is the logical relationship between uniform convergence and these four conditions?

Please, I need answers to all the above questions because this theorem always confuses me when I solve the problems and I don't know how to use it properly. Thanks for your help in advance.

share|cite|improve this question
In 4) you want to say precisely $f_n(x)\le f_{n+1}(x)$ for all $n,x$ (for "increasing"). If one of the conditions doesn't hold, can you construct an example of functions that satisfy the other three conditions but fail to converge uniformly? – David Mitra Nov 16 '11 at 19:04
Dini's theorem provides a sufficient condition for uniform convergence, but it is not necessary. That is, if you are told that all of those 4 hypotheses are met, you can conclude that the convergence is uniform. [Moreover, if any of these conditions is omitted, the conclusion does not follow automatically.] On the other hand, if you are told that the convergence is uniform, you cannot conclude any of the four conditions without additional information. – Srivatsan Nov 16 '11 at 19:20
To address your penultimate question, if the convergence is not uniform, then at least one of the four conditions cannot hold: this is nothing but the contrapositive form of Dini's theorem. – Srivatsan Nov 16 '11 at 19:20
up vote 9 down vote accepted

The theorem as stated is true (see my comment though): If 1), 2), 3), and 4) hold, then $f_n$ converges uniformly to $f$.

The answer to your third question is "yes". This is the contrapositive of the theorem, which is logically equivalent to the theorem.

I think the answer to your second question is no (specifically, 4) does not necessarily have to hold).

Other observations:

Remark 1: The theorem is "tight" (?); that is, each of the hypotheses are needed to insure that the convergence is uniform.

To see this:

Take $K=[0,1]$. The functions $f_n(x)=x^n$ give a sequence satisfying 1), 3), and 4), but not 2).

$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.

Take $K=[0,\infty)$ and $f_n(x)=\begin{cases} 0,\; & 0\leq x\leq n\\x-n , &n< x\leq n+1\\ 1 ,& x>n+1\end{cases}$. This sequence satisfies 1), 2) and 4) but not 3).

$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.

Take $K=[0,1]$ and let $f_n(x)$ be the function whose graph consists of the straight line segment from $(0,0)$ to $({1\over2n},1)$, the straight line segment from $({1\over2n},1)$ to $({1\over n}, 0)$, and the straight line segment from $({1\over n},0)$ to $(1,0)$. Then $f_n$ satisfies 1), 2), and 3), but not 4).

$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.

Take $K=[0,1]$ and $f_n(x)=\begin{cases}1, & 0\le x\le 1-{1\over n}\cr 0,&1-{1\over n}< x<1\\1,\;&x=1\end{cases} $. This sequence satisfies 2), 3) and 4) but not 1).

$f_n$ converges to $f=1$ pointwise but does not converge uniformly to $f$ on $K$.

Remark 2: If $(f_n)$ is uniformly convergent to $f$, you may not conclude that all four conditions hold. For example $f_n(x)=\begin{cases}1/n,& 0\leq x<1/2\\ -1/n& 1/2\leq x\leq1 \end{cases}$ converges uniformly to the zero function. But $(f_n)$ does not satisfy 1) or 4).

This shows that the converse of the theorem is false. It is not an "if and only if" theorem.

I will expand this post later (if that's allowed).

share|cite|improve this answer
But if $f_n$ is a continuous function for all n on K and converges pointwise to $f$ where $f$ is not continuous on K. Does this imply that $f_n$ is not uniformly convergent to $f$ ? What can we say about uniform convergence if $f_n$ is not continuous and converges pointwise to $f$ (f can be continuous or discontinuous)? – M.Krov Nov 16 '11 at 20:33
First question: yes, if $K$ is compact. This is a theorem you should know.. Second question: I think any of the various alternatives could happen. – David Mitra Nov 16 '11 at 20:42
David, I edited your answer since the command \cases{ } seems to be outdated or something of the sort. I replaced it with \begin{cases} \end{cases} – Pedro Tamaroff Jun 2 '13 at 23:25

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