Question

Continuity Party (tutorial about delta epsilon proofs of continuity of real valued functions)

Answer 1

So After all the fun we had last night with \(\epsilon-delta\) proofs, I thought I would do something similar for continuous functions. Lets start with the definition.


Definition: A functions is said to be continuous on \(D\) provided that \[\forall c \in \in D \ \forall \ \epsilon > 0 \ \exists \delta > 0 \ \forall x\in D \ (|x-c|<\delta \implies |f(x)-f(c)|<\epsilon)\]



This reads "For all \(x\) in the domain of our function and for all epsilon greater that zero there is a positive number delta such that when the distance from \(x\) to \(x\) is less than delta we have that the distance between the image of \(f(x)\) and \(f(c)\) is less than epsilon. \\



(for fun you might start to think about what happens if we switch around the quantifiers) \\

Let us look at the example \[f:(0,\infty) \underset{x\mapsto \frac{1}{x}}{\longrightarrow} \mathbb{R}\]

The first thing I do is consider what the result would look like.

\[|\frac{1}{x}-\frac{1}{c}|=\frac{|x-c|}{xc}\]

The real problem here is the \(x\) on the bottom. Since \(c,\epsilon\) come before \(\delta\) in our definition, it is ok do define \(\delta\) dependent on them, but since \(x\) comes after \(\delta\) we can't have \(\delta\) depending on \(x\). (\(\delta\) has been chosen before \(x\) even entered in the picture). So it would be nice if we could find a number \(y\) s.t. we have \(\frac{1}{x}<y\) because then we could say \(\frac{|x-c|}{xc}<y\frac{|x-c|}{c}\). Then we could replace \(|x-c|\) with \(\delta\), then shove it all under \(\epsilon\) and solve for \(\delta\)(remember we are trying to hunt for a delta). So we would have \(y\frac{|x-c|}{c}<\epsilon\) then \(y\frac{\delta}{c}<\epsilon\) and we solve for \(\delta\) and get \(\delta < c\frac{\epsilon}{y}\) so we would have our delta, and we would be done. So now we find the \(y\). \\

One thing I always do is start with \(|x-c|<\delta\) and see what we can do. Well this implies \(c-\delta<x<c+\delta\) and if \(c-\delta>0\) we can do this \(\frac{1}{x}<\frac{1}{c-\delta}\), So we found our \(y\). To make sure \(c-\delta > 0\) we replace \(\delta\) with \(\frac{c}{2}\). So if \(\delta < \frac{c}{2}\) we have \(|x-c|<\delta \implies |x-c|<\frac{c}{2}\implies \frac{1}{x}< \frac{1}{c-\frac{c}{2}} = \frac{2}{c}\). \\

So now we have \(\frac{|x-c|}{xc}<\frac{2}{c}\frac{|x-c|}{c}=\frac{2|x-c|}{c^2}\) and now I replace \(|x-c|\) with \(\delta\) and shove it all under \(\epsilon\) and then solve for \(\delta\). \(\frac{2\delta}{c^2}<\epsilon \iff \delta < \frac{c^2\epsilon }{2}\). So now we tie it all together. \\ \\

\begin{proof}{(\(f:\mathbb{R}^+ \underset{x\mapsto \frac{1}{x}}{\longrightarrow} \mathbb{R}\) is continuous.)}
Let \(c\in (0,\infty)\). Let \(\epsilon > 0\) be given. Set \(\delta < \min\{\frac{c}{2}, \frac{c^2\epsilon}{2}\}\). Then \(\forall x\in (0, \infty)\) we have, \[|x-c|<\delta \implies | \frac{1}{x}-\frac{1}{c} | = \frac{|x-c|}{xc} < \frac{2|x-c|}{c^2} < \frac{2\delta}{c^2}<\epsilon\] as required. \\


\end{proof}


Lets look at another example.


Consider \[f:\mathbb{R}\underset{x\mapsto 3x^2}{\longrightarrow} \mathbb{R}\]

What would it look like at the end of the proof ?\[|3x^2-3c^2| = 3|x-c||x+c|<\epsilon\]

Well, we can easily deal with the \(|x-c|\) factor, but the \(|x+c|\) is sort of a problem. To take care of this we use a ``trick''. Note that \(|x+c| = |x-c+2c|\le|x-c|+2|c|\). Sweet! we (still) know that we can control \(|x-c|\), and \(2|c|\) is also controllable. (I don't think control is the correct math term to use here, but it serves its purpose. I hope). \\

So now we have \(|3x^2-3c^2| = 3|x-c||x+c| \le 3|x-c|(|x-c|+2|c|)\). My first thought here was to distribute the \(|x-c|\) here, but that quickly gets ugly. Remember we can always have something like \(\delta < \min\{a,b,c,...\}\). We can take advantage of this (often this happens) and ``get rid of'' one of the \(|x-c|\)'s. So we can say \(\delta < |c|\) and from this we get \(|x-c| < \delta \implies |x-c|< |c|\implies 3|x-c|(|x-c|+2|c|) < 3|x-c|(|c|+2|c|)\\=3|x-c|3|c|= 9|x-c||c|\), now we replace \(|x-c|\) with \(\delta\) and shove it all under \(\epsilon\) and then solve for \(\delta\). \(9\delta|c|<\epsilon \implies \delta < \frac{\epsilon}{9|c|}\). Now we have all the tools to actually prove what we wanted. \\

Proof:


Let \(c\in \mathbb{R}\). Let \(\epsilon > 0\) be given. Set \(\delta < \min\{|c|, \frac{\epsilon}{9|c|}\}\). Then \(\forall x\in \mathbb{R}\) we have


\(|x-c|< \delta \implies |3x^2-3c^2| = 3|x-c||x+c|
\le 3|x-c|(|x-c|+2|c|)
\\< 3|x-c|(|c|+2|c|)
= 9|x-c||c|
< \epsilon \)



I WIN AGAIN!



I hope this helps.

Answer 2

If \(f\) is continuous on \(A\) and \(B\), and \(D=A\cup B\), can you prove/disprove the \(f\) is continuous on \(D\)?

Answer 3

Its a weird way to think about it. Because the function is defined by the domain

Answer 4

But I know what you are saying....the rule on different domains...

Answer 5

I am trying to think of a counter example

Answer 6

You said continuous on \(D\), not that it was the domain.

Answer 7

nope

think of absolute value, it is just x on [0,infinity] which is continuous

and -x on (-infinity, 0) which is continuous

but |x| is not

Answer 8

on R

Answer 9

It's possible that \(f(t) = u(3-t)\) is a counter example, or maybe it doesn't meet the conditions set forth.

Answer 10

u is some constant?

Answer 11

Wait, are you saying \(|x|\) isn't continuous? I was pretty sure it was.

Answer 12

if so then this is just an affine function and all affine\linear functions are continuous on any subset of R

Answer 13

its not at 0

Answer 14

\[
u(t-c) = \frac{|x-c|}{2(x-c)} + \frac 12
\]

Answer 15

I find it strange that \(|x|\bigg|^{x=0} = 0\) and \(\lim_{x\to 0}|x| = 0\) and yet it isn't continuous.

Answer 16

\[
|x-0|=|x|<\delta \implies ||x|-|0|| = ||x|| = |x|<\epsilon
\]

Answer 17

I can't pick \(\delta = \epsilon\)?

Answer 18

omg im an idiot I am thinking differentiability, I am writing a tut for differentiability on R

Answer 19

on how to prove things are not differentiable....

Answer 20

So I cant think of a counter example, lets prove it

Answer 21

I just don't know how we are going to do the notation...

Answer 22

Okay how about \(f(x) = \frac{|x|}x\)

Answer 23

because we cant call a function with different domains the same thing

Answer 24

I would do \(\epsilon_A\) and \(\epsilon_B\) if you try to prove it.

Answer 25

OK we say

Answer 26

\(f(x)\) I gave is counter example if it meets the conditions of continuity, but I'm not sure if it does.

Answer 27

Because I'm not very clear on what happens at boundaries.

Answer 28

It looks as though one sides limits are sufficient for boundaries?

Answer 29

\(f:\mathbb{R}\rightarrow \mathbb{R}\) and talk about \(f|_A\) and \(f|_B\)

the restriction of \(f\) to \(A\) and \(B\). And we assume those are both continuous, and show either \(f|_{A\cup B}\) is or is not continuous

that seem right?

Answer 30

Yeah I think that works but we should prove it

Answer 31

well \(0\) is not in the domain of \(\frac{1}{x}\) and is for \(|x|\), the function is not defined \(\frac{|x|}{x}\)on the union

Answer 32

and I think that would be continuous outside of 0

Answer 33

Okay, sure, I guess technically \(|x|/x\) isn't defined at \(0\).
So how about: \[
g(x) = \lim_{y\to x}\frac{|y|}{y}
\]?

Answer 34

that thing looks crazy

Answer 35

I am not sure because it would be a crazy proof with the double epsilons because of limit and continuity

Answer 36

looks continuous to me

Answer 37

looks?

Answer 38

I think we would have to say something along the lines of \(x\) is in \(A\) or \(x\) is in \(B\).

Answer 39

http://www.wolframalpha.com/input/?i=g%28x%29+%3D+%5Clim_%7By%5Cto+x%7D%5Cfrac%7B%7Cy%7C%7D%7By%7D

Answer 40

yeah for sure we would, but then you have to use the definition of the limit inside the definition of continuity. It could be done, but its early and I don't want to do it lol

Answer 41

You don't have to put a limit into a limit. It's just a piece-wise function.

Answer 42

I am saying to prove it is continuous you need an epsilon, and we are talking about a function that is defined as a limit, so to talk about the limit we will need another epsilon. Unless we are going to use limit properties which I figured we would not.