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Prove that if $|f'(x)| \leq M$, for all $x \in [a,b]$, then $$f(a)-M(b-a)\leq f(b) \leq f(a)+M(b-a)$$


What I tried so far:

Proof. Let $x \in [a,b]$

Assume $|f'(x)|\leq M$

Show $f(a)-(Mb-Ma)\leq f(b) \leq f(a)+(Mb-Ma)$

By assumption that $f'$ defined on $[a,b]$, we have following:

$f$ is cts on [a,b]

Also differentiable on (a,b)

By MVT have $\exists c \in (a,b) s.t.$

$f'(c)=\frac{f(b)-f(a)}{b-a}$

by assumption also have $-M\leq f'(x) \leq M$

That $c \in (a,b), have -M \leq f'(c) \leq M$

Have $-M \leq \frac{f(b)-f(a)}{b-a} \leq M$

Therefore $f(a)-M(b-a)\leq f(b) \leq f(a)+M(b-a)$

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  • $\begingroup$ You are not using the hypothesis at all. How do you get $|f(x)| \leq M$? $\endgroup$ – Kavi Rama Murthy Jun 9 at 0:31
  • $\begingroup$ @Kavi Rama Murthy Thanks, i just fixed the typo and rewrote my proof. $\endgroup$ – Manx Jun 9 at 3:11
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This is false. (Perhaps you have copied $|f'(x)| \leq M$ as $|f(x)| \leq M$).

Counterexample: if $f(x)=\sqrt x$ on $[0,1]$ then the hypothesis holds with $M=1$ but the right hand inequality in the conclusion fails for $a$ and $b$ close to $0$.

Answer to the edited question: just apply MVT: $f(b)-f(a)=(b-a)f'(x)$ for some $x$ and $-M \leq f'(x) \leq M$. Can you complete the proof using this?

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  • $\begingroup$ Thanks for you reponding, I had prove there exists some x in [a,b] s.t. statement hold with MVT, how do I show for any x in [a,b] statement still hold? $\endgroup$ – Manx Jun 9 at 3:28
  • $\begingroup$ You are asked to prove some inequalities involving $f(a)$ and $f(b)$. You are not asked to prove anything that involves an arbitrary point $x \in [a,b]$. My $x$ is same as your $c$. $\endgroup$ – Kavi Rama Murthy Jun 9 at 4:40

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