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Let $f$ and $g$ be defined in $\mathbb R$ and suppose that exists $M>0$ such that

$$|f(x)-f(p)|\leqslant M|g(x)-g(p)|, \quad \text{ for every } x.$$

Prove that if $g$ is continuous at $p$, then $f$ is also continuous at $p$.

I understand the overall idea of what should be done in this problem, but I don't understand the step below.

$$ |x-p|<\delta \implies |g(x)-g(p)|< \epsilon/M $$

In my thought process, if $g$ is continuous at $p$, then $|g(x)-g(p)|<\epsilon$ and, since $M>0$, $M|g(x)-g(p)|<M\epsilon$. I don't understand how one could just assume that $|g(x)-g(p)|<\epsilon/M$.

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    $\begingroup$ Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. $\endgroup$ Dec 23, 2022 at 11:36

3 Answers 3

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Since $g$ is supposed to be continuous at $p$, $$\forall \epsilon > 0, \exists \delta > 0, \ \forall x \in \mathbb R, |x - p| < \delta \implies |g(x) - g(p)| < \epsilon.$$ I think you got that. It means that you can take whatever $\epsilon$ you want and get a $\delta$ for that $\epsilon$. So it is perfectly correct to take $\epsilon/M$ (replace $\epsilon$ by $\eta$ in the definition of continuity and set $\eta := \epsilon/M$.

I guess the proof your looking at took $\epsilon/M$ so that it matches the definition of continuity at $p$ that we want $f$ to verify. So that the constants cancel well. With more practice, you'll see that exactly matching the constants doesn't really matter.

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This is a very typical doubt when starting to learn $\epsilon-\delta$ proofs. I will try to be as detailed as possible.

Since $g$ is continuous, you know, by definition that,

$$ \forall \epsilon >0, \exists \delta = \delta(\epsilon) > 0\colon |x-p| < \delta \implies |g(x)-g(p)| < \epsilon,$$ here, $x$ is arbitrary and $p$ is some value in the domain of the function $g$.

$\color{red}{NOTE:}$ In the definition above, $\epsilon$ is ANY positive value! For example, you could replace $\epsilon$ with $0.001, 0.1,$ with $1,10,20,$ and so on. The idea of the assumption $|g(x)-g(p)| < \frac{\epsilon}{M}$ is based of this.

To be more specific, you fix some arbitrary $\epsilon > 0$ and you know, by hypothesis, that $M>0.$ Then, the value $$ \frac{\epsilon}{M}$$ is also strictly positive. Thus, according to the NOTE. you can replace $\epsilon$ by $\frac{\epsilon}{M}!$ Doing so, you get the follwing:

$$ \exists \delta = \delta\left(\frac{\epsilon}{M}\right) > 0\colon |x-p| < \delta \implies |g(x)-g(p)| < \frac{\epsilon}M.$$ I think this clarifies your main doubt. If you have doubts in the rest of the proof make yourself "noisy"!

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  • $\begingroup$ How do you read "∃δ=δ(ϵ)>0:"? $\endgroup$ Dec 23, 2022 at 12:05
  • $\begingroup$ @SageRenard that just means your $\delta$ might depend of $\epsilon.$ Most people just omit this. $\endgroup$
    – xyz
    Dec 23, 2022 at 12:11
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" if g is continuous in p, then |g(x)-g(p)|<ε": no! If g is continuous in p, then for every $\alpha>0$, there exists $\beta>0$ s.t. $|x-p|<\beta\Rightarrow |g(x)-g(p)|<\alpha.$ Now, given some $ε>0,$ apply this to $\alpha:=ε/M.$

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    $\begingroup$ Welcome to club 10k @Anne! Use your freshly earned privileges wisely. FWIW as a general observation I very much approve of your approach to the site. $\endgroup$ Dec 23, 2022 at 12:14

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