I'm stuck on an analysis problem to which I've reduced to the following, so some assumptions may be superfluous.
Let $\{ f_n(x) \} \subset C(X,\mathbb{R}^{\geq0})$ (i.e. $f_n$ is continuous and nonnegative) where $X$ is a compact subset of $\mathbb{R}^n$.
Suppose further that for all $n,x$, $f_{n+1}(x) \leq f_n(x)$ and that $f_n$ converges pointwise to the continuous function $f$.
Show that the sequence $f_n$ is equicontinuous.
Any help on this problem is appreciated. Thanks.
By Dini's theorem http://en.wikipedia.org/wiki/Dini's_theorem, the convergence is uniform.
The sequence is actually uniformly equicontinuous.
Take $\epsilon>0$.
There exists $N$ such that $\sup_X|f_n-f|\leq \epsilon/3$ for all $n\geq N$.
Now $f$ is uniformly continuous on the compact $X$, so there exists $\delta_0>0$ such that $|f(x)-f(y)|\leq \epsilon/3 $ for all $|x-y|\leq \delta_0$.
For all $n\geq N$ and for $|x-y|\leq \delta_0$, we thus have, by triangular inequality $$ |f_n(x)-f_n(y)|\leq |f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)| $$ $$ \leq 2\sup_X|f_n-f|+|f(x)-f(y)| \leq 3\frac{\epsilon}{3}=\epsilon. $$
Finally, the functions $f_1,\ldots,f_{N-1}$ are all uniformly continuous and in finite number, so we can clearly find $\delta_1>0$ such that $|f_n(x)-f_n(y)|\leq \epsilon$ for all $|x-y|\leq \delta_1$ and all $n=1,\ldots,N-1$.
I remains to take $\delta=\min\{\delta_0,\delta_1\}$ to get $|f_n(x)-f_n(y)|\leq \epsilon$ for all $|x-y|\leq \delta$ and all $n$.
So the functions $(f_n)$ are uniformly equicontinuous.
Since that $f$ is a continuous function, for $\varepsilon>0$ and $x\in X$ exist $\delta>0$ such that $$|f(y)-f(x)|<\varepsilon, \ \ \ \mbox{if} \ \ \ |y-x|<\delta.$$ Since that the sequence $(f_n)_n$ converge pointwise, $$|f_n(x)-f(x)|\longrightarrow0, \ \ \ \ |f_n(y)-f(y)|\longrightarrow0,$$ for $n$ sufficiently huge. Using the monotonicity of the sequence, we obtain $$ \begin{array}{rcl} |f_1(x)-f_1(y)| & \leq & |f_1(x)-f(x)|+|f_1(y)-f(y)|+|f(y)-f(x)|\\ & \leq & |f_n(x)-f(x)|+|f_n(y)-f(y)|+|f(y)-f(x)|\longrightarrow0, \end{array} $$ when $n\in\mathbb{N}$ is sufficiently huge and $|y-x|<\delta$. This same argument can be used for $f_2, f_3,...,f_n...$. Then, $$|f_n(x)-f_n(y)|<\varepsilon, \ \ \ \ \mbox{if} \ \ \ |y-x|<\delta,$$ for all $n\in\mathbb{N}$, like desired.
$f_n(x)$ is decreasing in n, so $|f_1(x)-f(x)| \geq |f_n(x)-f(x)|$
$$ $$ I don't think it matters, but for the problem I'm working on, it suffices to show that the sequence is eventually equicontinuous, i.e. for a given $\epsilon>0$, the $\delta>0$ required for continuity need only be true for the family ${f_n : n\geq N}$ for some large $N$.
– student Feb 19 '13 at 03:20By continuity of f ; for every $\varepsilon > 0$ (given $x\in $X$)$ , there exist $\Delta>0$ such that $$|f(y)-f(x)|<\varepsilon/3 \ \ \ \mbox{if} \ \ \ |y-x|<\Delta.$$ Since the sequence $(f_n)$ converges pointwise,
$|f_N(x)-f(x)|<\varepsilon/3$ and $|f_N(y)-f(y)|<\varepsilon/3$ for sufficiently large $N$.
And for $n\geq$ $N $ , we have $|f_n(x)-f_n(y)|\leq \ $$|f_N(x)-f(x)|$+$|f_N(y)-f(y)|$+$|f(y)-f(x)|$ $<\epsilon$
Then just take $\delta= min $$\{\delta_1,\delta_2,..,\delta_N,\Delta\}$ where $\delta_i$ satisfies $:|f_i(x)-f_i(y)| < \varepsilon $ if $|x-y|<\delta_i $ .
So , by definition , $f_n$ is equicontinuous.
When you choose $N$, you choose it depending on $x$ and $y$. This in turn makes $\delta$ depend on $x,y$, and so equicontinuity doesn't follow.
– student Feb 19 '13 at 20:21
One minor note, continuity of $f$ is required: just consider ${x^n}$ for $x\in [0,1]$ and $n \in \mathbb{N}$.
– student Feb 19 '13 at 06:38