= Solution
Set $e_n(x)=|f_n(x)-f(x)|$. The functions $e_n$ are <continuous>, converge pointwise to zero, and decrease with $n$. This remains true even if the direction of monotonicity of $f_n(x)$ differs between points: at each fixed point a monotone convergent real sequence approaches its limit from one side.
Given $\varepsilon>0$, the sets $U_n=\{x:e_n(x)<\varepsilon\}$ are relatively open in $[a,b]$, increase with $n$, and cover the interval. <Compactness> gives a finite subcover; choosing the largest index in it gives $U_N=[a,b]$. For every $n\ge N$ and every $x$, $e_n(x)\le e_N(x)<\varepsilon$. Hence \b[$f_n\to f$ uniformly]. This proves the required version of <Dini's theorem>, rather than assuming a common direction of monotonicity everywhere.
On $[0,1)$, take $f_n(x)=x^n$. Each function and its pointwise limit $f=0$ are <continuous>, and the sequence decreases at every point. Nevertheless
$$
\boxed{\sup_{0\le x<1}|f_n(x)-0|=1\quad\text{for every }n.}
$$
Thus <uniform convergence> fails on the noncompact interval although all the other conditions hold.
Back to article page
