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You know that for some $a$, $f(a) = g(a)$.

  1. If it is also true that $\frac{\mathrm df(x)}{\mathrm dx} = \frac{\mathrm dg(x)}{\mathrm dx}$, does it follow that $f(x) = g(x)$ for all $x$?
  2. If so, is there a name for this form of proof? I have seen something like it in proving the uniqueness of functions.
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    $\begingroup$ This is certainly true if $f$ and $g$ are $C^1$ functions (derivatives are continuous), which can be proved by integrating. $\endgroup$
    – Kenta S
    Aug 15, 2020 at 9:24
  • $\begingroup$ Thanks. The second question was whether there was a specific name for this form of proof. $\endgroup$
    – buckner
    Aug 15, 2020 at 12:07

2 Answers 2

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If $f,g$ are differentiable, then by assumption so is $h = f-g$, and $h'(x) = 0$ for all $x$. If $x,y$ are points, then $h$ is differentiable on $(x,y)$ and continuous on $[x,y]$ by assumption. So the mean value theorem implies that there is $\xi$ such that

$$ 0 = h'(\xi) = \frac{h(x) - h(y)}{x-y} \implies h(x) = h(y) .$$

It follows that $h$ is constant. The assumption that $f(a) = g(a)$ then implies that $f = g$

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  • $\begingroup$ Thanks. The second question was whether there was a specific name for this form of proof. $\endgroup$
    – buckner
    Aug 15, 2020 at 12:07
  • $\begingroup$ I don't think so. You could talk about uniqueness of solutions, but that is generally when there is only one "unknown" function, instead of two. This result tells us that functions that have the same derivative differ by a constant, so I guess you could interpret it as a result about the kernel of the "derivative operator". $\endgroup$
    – Wizact
    Aug 15, 2020 at 12:15
  • $\begingroup$ Many thanks Wizact :) $\endgroup$
    – buckner
    Aug 15, 2020 at 13:03
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    $\begingroup$ This assumes that the domain of $f$ (and $g$) is all of $\Bbb R$ or at least is an interval. The original claim is false with non-connected domains. $\endgroup$ Aug 15, 2020 at 21:05
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If f'(x) and g'(x) are integrable, then:

\begin{align} \frac{df(x)}{dx} &= \frac{dg(x)}{dx}\\ \therefore \int\frac{df(x)}{dx}dx&=\int\frac{dg(x)}{dx}dx\\ \therefore f(x)+C_1&=g(x)+C_2\\ \therefore f(a)+C_1&=g(a)+C_2 \textit{ where }f(a)=g(a)\\ \therefore C_1&=C_2\\ \therefore f(x)&=g(x) \end{align}

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    $\begingroup$ This assumes that $f'$ and $g'$ are integrable, which is not necessarily true under the given hypotheses. $\endgroup$
    – user169852
    Aug 15, 2020 at 9:37
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    $\begingroup$ @Bungo This is a good point and I will be deleting my answer very shortly (once you see this comment). Wizacts already answers the question anyway. $\endgroup$
    – user400188
    Aug 15, 2020 at 9:38
  • $\begingroup$ @Bungo On second thought, I might leave the answer up, as I am having trouble understanding Wizacts line on the mean value theorem. OP if you understand Wizacts answer, please accept it over mine because it arrived first and was correct while mine was not. $\endgroup$
    – user400188
    Aug 15, 2020 at 9:42

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