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This is similar but it's not Dini. Let $f_n$ be a sequence of functions that are monotonic, and $f_n \to f$ pointwise, where $f$ is continuous in a compact interval $I$. Then the convergence it's uniform in every compact subset of $I$. It's not necessary that $f_n$ be continuous. Please help me with this, to see some example of problems with uniform convergence.

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To clarify: Is each $f_n$ a monotonic function, or is the sequence itself monotonic (meaning that $f_n(x) \leq f_{n+1}(x)$ for each $x$)? From the phrasing, I assume you mean the former. – Jesse Madnick Nov 17 '11 at 5:48
If $I$ is compact, it is a compact subset of itself, so you might as well just say the convergence is uniform on $I$. – Robert Israel Nov 17 '11 at 6:42
up vote 2 down vote accepted

Suppose it's not uniform. Then there is $\epsilon > 0$ such that for each positive integer $n$, there is $x_n \in I$ such that $|f_n(x_n) - f(x_n)| > \epsilon$. By compactness, some subsequence of $x_n$ converges: wlog we can assume it's the whole sequence, converging to $x_0$. Now $f$ being continuous, there are $a$ and $b$ with $a < x_0 < b$ (unless $x_0$ is an endpoint of $I$, in which case $a$ or $b$ is $x_0$) such that $|f(x) - f(x_0)| < \epsilon/3$ for $a \le x \le b$. For $n$ sufficiently large, we have $a \le x_n \le b$, $|f_n(a) - f(a)| < \epsilon/3$ and $|f_n(b) - f(b)| < \epsilon/3$. But then $f_n(x_n) \le f_n(b) \le f(b) + \epsilon/3 \le f(x_n) + \epsilon$ and similarly $f_n(x_n) \ge f_n(a) \ge f(a) - \epsilon/3 \ge f(x_n) - \epsilon$, contradiction.

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