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Let $f\in C^1([a,b],\mathbb{R})$ such as $\dfrac{f(b)-f(a)}{b-a}=\sup\left\{f^{'}(x) : x\in[a,b]\right\}$. Prove that $f$ is linear

Proving that $f$ is linear means to prove that $\forall x\in [a,b],\, f^{'}(x) = \dfrac{f(b)-f(a)}{b-a}$. Because $f\in C^1$ then $\exists M\in[a,b],\,\dfrac{f(b)-f(a)}{b-a}=f^{'}(M)$. So $\forall x\in[a,b],\,f^{'}(x)\leq f{'}(M).$

From here I can see that I have to prove that $f{'}(x)\geq f{'}(M),$ But I can't quite see the continuation of this proof. Although I can graphically see by plotting some graphs that if $f$ was not linear then eventually there would be a point where $\dfrac{f(b)-f(a)}{b-a}$ is not a supremum.

Marco
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Sewshley
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    $f(b) - f(a) = \int_a^b f'(t) dt\le (b-a)\sup f'(x)$, the equal sign occurs if all $f'$ equals to $ \sup f' $ by continuity of $f'$. – Yimin Jan 31 '24 at 23:50
  • @Yimin I didn't quite understand your idea. Can you explain it more. Thanks in advance. – Sewshley Feb 01 '24 at 13:13

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Let $F(x)=f(x)-\frac{f(b)-f(a)}{b-a}x$, then $F'\leq 0$ so $F$ is decreasing. But $F(a)=F(b)$, thus $F$ has to be constant, and $f$ must be linear.

Eric Ley
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