I have been given the following statement by a professor:

Let $m\leq x\leq M$ and define $K=\max(|m|,|M|)$. Then $|x|\leq K$.

Now I can clearly see that this is true, and working through case-by-case (i.e. $m>0$ and $M>0$, $m<0$ and $M>0$ with $|m|<|M|$ etc) I can also prove it is true for each case. But I feel that there must be a much more elegant proof than the one I have. Can anyone think of one or know one?

Thanks all.


I do not know if this is elegant but I would prove it this way. One has $|m|\leq K$ and $|M|\leq K$ by definition of $K$.

This means $-K\leq m\leq K$ and $-K\leq M\leq K$. Now one has $$-K\leq m\leq x\leq M\leq K$$ i.e $-K\leq x\leq K$ and this is exactly $|x|\leq K$

  • $\begingroup$ Ah I like it. Simple to follow and a lot cleaner than what I have. Thanks. $\endgroup$ – AloneAndConfused Mar 2 '15 at 16:13

Since $x \in [m,M]$, this inequality follows directly from the convexity of $t \mapsto |t|$.


It's a proof by case:

If $x \leq 0$, then $|x| \leq |m| \leq K$. If If $x > 0$, then $|x| \leq |M| \leq K$

  • $\begingroup$ The OP supplied a proof by cases and asked for another proof. $\endgroup$ – user795305 Mar 2 '15 at 19:57
  • $\begingroup$ @benbaer, but this proof has less cases $\endgroup$ – RiaD Mar 2 '15 at 20:48

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