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Let $f$ be continuous on [a,b].
Then $f$ is uniformly continuous on [a,b] and there exists $\delta >0$ such that
$|f(s)-f(t)|<\epsilon$ if $|s-t|<\delta$.
Let P={$x_0,x_1,...,x_n$} is a partition of [a,b] with $x_i-x_{i-1}<\delta$ for all i.
If $x_{i-1} \le t \le x_i$ then
$$|f(t)|\le |f(x_i)|+\epsilon$$

Where does this last inequality come from?

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up vote 1 down vote accepted

If $x_{i-1}\le t\le x_i$ and $x_i-x_{i-1}\lt\delta$, then $|x_i-t|\lt\delta$, so $|f(x_i)-f(t)|\lt\epsilon$, and thus $|f(t)|\le|f(x_i)|+\epsilon$.

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Thanks! I wonder when you conclude $|f(t)|\le |f(x_i)|+\epsilon$ from $|f(x_i)-f(t)|<\epsilon$, which theorem is applied? That inequality is absolutely true but I just want to make things clear. – niagara Nov 28 '12 at 16:17
@niagara: $|a|-|b|\le|a-b|$; thus $|f(t)|-|f(x_i)|\le|f(t)-f(x_i)|$. – joriki Nov 28 '12 at 16:22
Thank you very much! – niagara Nov 28 '12 at 16:24
@niagara: You're welcome! – joriki Nov 28 '12 at 16:25

It is clear that $(x_{i-1},t)\subset (x_{i-1},x_{i})$ so that $|x_{i}-t|\leq |x_{i-1}-x_{i}|<\delta$ so that, by the uniform continuity of $f$, $|f(x_{i})-f(t)|<\epsilon$. The desired inequality follows with some manipulation.

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Continuity on a compact set does imply uniform continuity, since $\delta(x)$ can be chosen as a continuous function of $x$ and a continuous function on a compact set attains its maximum, which can then be taken as the $\delta$ independent of $x$. – joriki Nov 28 '12 at 16:08
Oh sorry, yes, I was not paying attention to the fact that it was on a closed interval. – Jebruho Nov 28 '12 at 16:09

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