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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?

2 Answers 2

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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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    @niagara: You're welcome!2012-11-28
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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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    Oh sorry, yes, I was not paying attention to the fact that it was on a closed interval.2012-11-28