Dini's Theorem. Uniform convergence and Bolzano Weierstrass.

Assume that we can find $\delta>0$, a subsequence $\{f_{n_k}\}$ and a sequence $\{x_{n_k}\}$ such that $f_{n_k}(x_{n_k})\geqslant \delta$ for all $k$. Bolzano-Weierstrass theorem ($[a,b]$ is compact) allows us to extract of $\{x_{n_k}\}$ a subsequence, denoted $\{t_j\}$, converging to some $t$. We have for all integers $n$ and $m$, denoting $g_k$ the sequence indexed by the integers appearing in $t_k$, $$\delta\leqslant g_{m+n}(t_{m+n})\leqslant g_n(t_{m+n}).$$ Now we fix $n$, and take the $\limsup_{m\to +\infty}$. This gives for all integer $n$, $$\delta\leqslant\limsup_{k\to +\infty}g_k(t_k)\leqslant g_n(t),$$ a contradiction.

In the second theorem, take $a=0, b=1$.

• If we don't assume $f$ continuous, consider $f_n(x):=x^n$.
• The same counter-example considering $(0,1)$.

Here is a direct proof:

THEOREM Suppose $\{f_n\}$ is a sequence of continuous functions from $[a,b]$ to $\Bbb R$ that converge pointwise to a continuous function $f$ over $[a,b]$. If $f_{n+1}\leq f_n$, then convergence is uniform.

PROOF We set $g_n=f_n-f$ and note that $g_n\geq g_{n+1}$ and the $g_n$ are continuous, converging pointwise to $0$. For a given $\epsilon>0$. Consdier the (relatively) open sets (because of continuity of the $g_n$) $O_n=\{x\in [a,b] :g(x)<\epsilon\}$. Note that since $g_n\geq g_{n+1}$ we have $O_n\subset O_{n+1}$. Given $x\in[a,b]$ there is an $n$ such that $g_n(x)<\epsilon$; whence $\bigcup_{n\in\Bbb N}O_n=[a,b]$. But since $[a,b]$ is compact there exists a finite set $K=\{1,\dots,m\}$ such that $\bigcup_{k=1}^m O_{n_k}=[a,b]$. But since $O_n\subset O_{n+1}$ the greatest element of $K$, call it $\ell$, is such that $O_\ell =[a,b]$. And we're done: for every $n\geq \ell$ we have $g_n(x)<\epsilon$; as desired. $\blacktriangle$