# Proving that the absolute value of a function is continuous if the function itself is continuous

I would like to prove that if the function $$f : D \rightarrow \mathbb{R}$$ is continuous, and the function $$|f| : D \rightarrow \mathbb{R}$$ is defined by $$|f|(x) = |f(x)|$$ for all $$x$$, then the function $$|f| : D \rightarrow \mathbb{R}$$ is also continuous.

I think that the correct way to do to this is to use the fact that for two functions $$f : D \rightarrow \mathbb{R}$$ and $$g : U \rightarrow \mathbb{R}$$ such that $$f(D)$$ is contained in $$U$$, the composition of functions, $$g \circ f: D \rightarrow \mathbb{R}$$ is continuous if $$f$$ and $$g$$ are continuous.

So, does that mean to prove my original assertion, I should introduce a new function $$f : D \rightarrow \mathbb{R}$$ defined and $$g : D \rightarrow \mathbb{R}$$ defined by $$g(x) = |x|$$. Then, I should prove that $$f$$ and $$g$$ are continuous, and by the continuity of composition of functions, I can conclude that $$g \circ f$$ is continuous, which completes my proof?

I don't know if $$f$$ and $$g$$ should be defined on $$D$$, though. I believe that now I just need to show $$g$$ is continuous to complete my proof.

Could someone please help me check this method, and help me with this exercise? This isn't a homework problem, I'd just like to learn some analysis on my own.

My attempt at the proof: First of all, here is the definition of continuity that I am following:

Definition: A function $$f : D \rightarrow \mathbb{R}$$ is said to be continuous at the point $$x_{0}$$ in $$D$$ provided that whenever $$\{x_{n}\}$$ is a sequence in $$D$$ that converges to $$x_{0}$$, the image sequence $$\{f(x_{n}\}$$ converges to $$f(x_{0})$$. The function is said to be continuous provided that it is continuous everywhere in $$D$$.

Lemma 1: The function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ defined by $$f(x) = |x|$$ is continuous.

Proof of Lemma 1: Select a point $$x_{0}$$ in $$\mathbb{R}$$ and let $$\{x_{n}\}$$ be a sequence that converges to $$x_{0}$$. Then,

$$\lim_{n\to\infty}f(x_{n}) = \lim_{n\to\infty}|x_{n}| = |x_{0}| = f(x_{0}).$$

So, $$f$$ is continuous at $$x_{0}$$.

Proof of Assertion: Let $$f : D \rightarrow \mathbb{R}$$ and $$g : D \rightarrow \mathbb{R}$$ be functions, where $$g(x) = |x|$$. By our assumption, $$f$$ is continuous. Also, by Lemma $$1$$, $$g$$ is continuous. Therefore, by the continuity of composite functions, $$(g \circ f)(x)$$ is also continuous. This completes our proof.

• $f$ should be $f$, not $f(x)=x$. – Randall Oct 14 '18 at 3:20
• sorry, I am confused by what you mean by this – user400359 Oct 14 '18 at 3:22
• Hint: Reverse triangle inequality – Jacky Chong Oct 14 '18 at 3:23
• Your idea is the right one. However, with YOUR choices of $f$ and $g$, the composition $g \circ f$ is just $(g \circ f)(x) = |x|$. You want to get $|f(x)|$ as an output instead. – Randall Oct 14 '18 at 3:23
• Oh, so if I let $f = f$ and $g = |x|$, then $g \circ f$ becomes $|f(x)|$. I understand that part now, thanks. – user400359 Oct 14 '18 at 3:24

To show that $$|f(x)|$$ is continuous at a point $$x=a$$ we need to show that given an $$\epsilon>0$$, there exist a $$\delta$$ such that $$|x-a| <\delta \implies ||f(x)|-|f(a)||<\epsilon$$
Since $$f(x)$$ is continuous at $$x=a$$ for the given $$\epsilon$$ we have a $$\delta$$ such that $$|x-a| <\delta \implies |f(x)-f(a)|<\epsilon$$
Note that if $$|x-a| <\delta$$ then $$||f(x)|-|f(a)||\le |f(x)-f(a)|<\epsilon$$