Dini's Theorem and tests for uniform convergence 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:
  
  
*
  
*$f_n$  is continuous on $K$ for all $n$.
  
*The pointwise limit $f$ is continuous on $K$.
  
*$K$ is a compact interval (i.e., a closed and bounded interval in $\mathbb{R}$).
  
*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.
 A: 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).
