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Show that if $|g'(x)|\leq M|x-a|^n$ for $|x-a|<\delta$, then $|g(x) - g(a)|\leq M|x-a|^{n+1}/(n+1)$ for $|x-a|<\delta$.

My question is: are you allowed to integrate from $a$ to $x$ across the inequality $|g'(x)|\leq M|x-a|^n$ to obtain the result? If not why not? If so, what conditions enable us to?

Note: there is a duplicate question with alternate method.

zhw.
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helios321
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  • Is $g'$ continuous? – zhw. Oct 24 '17 at 09:55
  • g' is integrable by fundamental theorem of calculus, and https://math.stackexchange.com/questions/96950/if-f-is-a-riemann-integrable-prove-f-is-also-riemann-integrable , so $|(g(x)|$ is intergable – Belive Oct 24 '17 at 09:58
  • @Belive No the FTC does not imply that. You need to assume $g'$ is Riemann integrable to have the conclusion of the FTC. – zhw. Oct 24 '17 at 10:00

2 Answers2

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"are you allowed to integrate from $a$ to $x$ across the inequality $|g'(x)|\leq M|x-a|^n$ to obtain the result?" I'm not sure what that means.

Hint: Assuming $g'(x)$ is continuous on $(a-\delta, a + \delta),$ the FTC gives

$$g(x)-g(a) = \int_a^x g'(t)\,dt.$$

Slap absolute values on that, move them inside the integral, use the given estimate, etc.

zhw.
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  • The question is exactly stated as it is. Based on what you see can we say that g'(x) is continuous or not? If not, is it enough that g' is integrable? – helios321 Oct 24 '17 at 11:20
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Yes, you can.

We'll suppose $M\ge 0$.

Indeed, if $x\ge a$, the inequality with absolute values is equivalent to: $$-M(x-a)^n\le g'(x)\le M(x-a)^n $$ Integrating the right-hand side inequaity from $a$ to $x$ yields the result in this case.

If $x\le a$, it is equivalent to $$-M(a-x)^n\le g'(x)\le M(a-x)^n, $$ and integrating from $x$ to $a$, we obtain $$\int_x^a g'(t)\,\mathrm d t=g(a)-g(x)\le -M\frac{(a-t)^{n+1}}{n+1}\Biggl\vert_x^a=\frac{M(a-x)^{n+1}}{n+1}$$ and similarly $$g(a)-g(x)\ge M\frac{(a-t)^{n+1}}{n+1}\Biggl\vert_x^a=-\frac{M(a-x)^{n+1}}{n+1}.$$

Bernard
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