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A corollary of the mean value theorem states if if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and if $|f'(x)| \le M $ for all $x \in [a,b]$, then $\displaystyle \frac{|f(x_{2})-f(x_{1})|}{|x_{2}-x_{1}|} \le M$ for all $x_{1},x_{2} \in [a,b]$.

But is it also true that $\displaystyle \frac{|f(x_{2}) - f(a)|}{b-a} \le M$?

That seems to be what was used in another thread to argue that $ \displaystyle\sum_{n=1}^{\infty} \frac{\sin (\sqrt{n})}{n}$ converges.

Convergence of $\sum _{k=1}^\infty \sin \left(\sqrt{k}\right)/k$

Community
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user26647
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1 Answers1

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We have

$$\frac{\lvert f(x_2) - f(a)\rvert}{b-a} = \frac{\lvert x_2-a\rvert}{b-a}\frac{\lvert f(x_2) - f(a)\rvert}{\lvert x_2-a\rvert} \leqslant \frac{\lvert f(x_2) - f(a)\rvert}{\lvert x_2-a\rvert}$$

for all $a < x_2 \leqslant b$, since $\lvert x_2 - a\rvert \leqslant b-a$ then.

Daniel Fischer
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