# Problem understanding $\epsilon-\delta$ definition of continuity - a particular example.

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 $$M>0$$, $$M|g(x)-g(p)|. I don't understand how one could just assume that $$|g(x)-g(p)|<\epsilon/M$$.

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

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.

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"!

• How do you read "∃δ=δ(ϵ)>0:"? Dec 23, 2022 at 12:05
• @SageRenard that just means your $\delta$ might depend of $\epsilon.$ Most people just omit this.
– xyz
Dec 23, 2022 at 12:11

" 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 $$\alpha:=ε/M.$$

• 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. Dec 23, 2022 at 12:14