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Suppose $K$ is compact.
(a){$f_n$} is a sequence of continuous functions on $K$
(b){$f_n$} converges pointwise to a continuous function $f$ on $K$
(c) $f_{n+1}(x) \le f_n(x)$ for all $ x \in K$, $n=1,2,3,...$
Then $f_n \rightarrow f$ uniformly on K.

Proof)
Put $g_n=f_n-f$. Then $g_n$ is continuous, $g_n \rightarrow 0$ pointwise, and $g_{n+1} \le g_n$
Let $\epsilon >0$ be given. Let $K_n$ be the set of all $x \in K$ with $\epsilon \le g_n(x)$.
Since $g_n$ is continuous, $K_n$ is closed by theorem(*), hence compact.
.
.
.

Theorem(*)
A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous
iff $f^{-1}(C)$ is closed in $X$ for every closed set $C$ in $Y$.

In here, I have no idea how to apply that theorem to the proof.
$K_n$ is $f^{-1}(C)$ in that theorem?

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Yes, applied to $g_n$ with $C=[\epsilon,\infty)$ (then $K_n=g_n^{-1}(C)$). –  David Mitra Dec 9 '12 at 15:33
    
Then what is $X$ and $Y$ in here? First, I guess both are $K$, but think it's wrong. –  niagara Dec 9 '12 at 15:36
    
$X$ is $K$ and $Y$ (presumeably) is $\Bbb R$. –  David Mitra Dec 9 '12 at 15:39
    
Thank you very much! –  niagara Dec 9 '12 at 15:43

1 Answer 1

up vote 2 down vote accepted

The first theorem you're talking about is known as Dini's theorem. I know the proof for $f:[a,b]\to \Bbb R$, but it can be adjusted for your case.

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. $\blacktriangle$

ADD A little googling brought up this, which is where I learned the proof from. It has a few examples showing the hypotheses

$(1)$ $f_n$ decreases with $n$

$(2)$ The domain is compact.

$(3)$ $f$ is continuous.

are all essential.

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