The continuity of f

by macca1994
Tags: continuity
 P: 19 I have seen this theorem in a few books, but none of them give proofs, it says if f(x) is a continuous function then lf(x)l is a continuous function. What is the proof of this because i don't really understand why this holds, thanks
 Mentor P: 18,334 The function $|~|:\mathbb{R}\rightarrow \mathbb{R}$ is continuous. Composition of continuous functions is continuous.
 P: 3,002 It makes sense though. Consider y=-1/x : its continuous in the domain of x in (0,infinity) right? and its counterpart z=|-1/x| = 1/x is continuous in the same domain for x
 P: 19 The continuity of f Ahh, yeah i get that, but can you not prove it without using composition of two functions?
 P: 3,002 You could probably prove it via the Weierstrass definition: http://en.wikipedia.org/wiki/Continuous_function but Im not a mathematician so you'll have to hope HallsOfIvy checks this thread.
 P: 19 i might pm him, cheers
 Mentor P: 18,334 It is very easy to prove using the $\epsilon-\delta$ definition. Are you familiar with $\epsilon$and $\delta$ definitions?
 P: 19 yeah, that's about the limit of my analysis knowledge, how would you use the ε,δ definition?
P: 3,002
 Quote by macca1994 i might pm him, cheers
No need micro can handle it too. I think the epsilon / delta is the weierstrass definition.
Mentor
P: 18,334
 Quote by macca1994 yeah, that's about the limit of my analysis knowledge, how would you use the ε,δ definition?
Take an $a\in \mathbb{R}$. You will need to prove

$$\forall \varepsilon >0: \exists \delta >0: \forall x: |x-a|<\delta~\Rightarrow ||f(x)|-|f(a)||<\varepsilon$$

You are given that f is continuous in a, thus:

$$\forall \varepsilon >0: \exists \delta >0: \forall x: |x-a|<\delta~\Rightarrow |f(x)-f(a)|<\varepsilon$$

So, take $\varepsilon>0$ arbitrary. Take $\delta>0$ as in the previous definition: so it holds that

$$\forall x: |x-a|<\delta~\Rightarrow |f(x)-f(a)|<\varepsilon$$

Take an x arbitrary such that $|x-a|<\delta$. Then we know that

$$||f(x)|-|f(a)||\leq |f(x)-f(a)|<\varepsilon$$

So we have verified that $||f(x)|-|f(a)||<\varepsilon$ and thus we have verified that $\varepsilon-\delta$ definition of continuity. Thus $|f|$ is continuous.
 P: 19 ah that makes sense and is very obvious, do you need to show that limit of lf(x)l is in fact lf(a)l or is that just obvious?
Mentor
P: 18,334
 Quote by macca1994 ah that makes sense and is very obvious, do you need to show that limit of lf(x)l is in fact lf(a)l or is that just obvious?
I've just proven that here. It's because |f| is continuous in a.
 P: 19 oh okay, i get it, thanks for the help