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This is a very common proof, but there is one point, which many just seem to take for granted.

At one point I'll use the mean value theorem to say

$$ w.l.o.g.\quad x > y \quad x,y \in (a,b) $$

And from the mean value theorem I get a term like this

$$\frac{f(x)-f(y)}{x-y} = f'(\xi) \leq L $$

But people often conclude $$|f(x)-f(y)| \leq L|x-y|$$

Is that step always true? i.e. is it always true that

$$ x = y \rightarrow |x|=|y| $$

Jonathan
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    Well yes, that's true for all functions, not just $| \space |$. – Emax Jan 27 '17 at 14:19
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    It is true, but exactly where is that used in the proof? (Note also that should should take the absolute values on both sides before upper bounding $f'(\xi)$ by $L$. You want to use $\lvert f'(\xi)\rvert \leq L$, which is the stronger statement needed.) – Clement C. Jan 27 '17 at 14:24

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The mapping $x\mapsto |x|$ is a function like any other, and for any function $f$, it is true that if $a=b$, then $f(a)=f(b)$.

5xum
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