Help with Epsilon - Delta proofs

[MelchyBeau]

Hi,

I am taking Real Analysis at the moment. (I have a B.S. in physics but need Real Analysis and Abstract Algebra) for my Masters in Math. Unfortunately my book(Strichartz) is crappy, and my teacher is very confusing.

Anyway does anyone have an example of how to work out an epsilon delta proof to show a function is continous (not uniformly, I can do those fairly easily). Also how to show a function is not uniformly continous, and a function is not continous. If you know of a website that has these examples, that would work too.

I am not liking real analysis, maybe it is because of the teacher. I love combinatorics though, that crap is fun.

Thanks in advance,
Melch

[Trantor]

[ QUOTE ]
Hi,

I am taking Real Analysis at the moment. (I have a B.S. in physics but need Real Analysis and Abstract Algebra) for my Masters in Math. Unfortunately my book(Strichartz) is crappy, and my teacher is very confusing.

Anyway does anyone have an example of how to work out an epsilon delta proof to show a function is continous (not uniformly, I can do those fairly easily). Also how to show a function is not uniformly continous, and a function is not continous. If you know of a website that has these examples, that would work too.

I am not liking real analysis, maybe it is because of the teacher. I love combinatorics though, that crap is fun.

Thanks in advance,
Melch

[/ QUOTE ]

Is this the sort of thing you are after?

http://www.karlscalculus.org/pr3_0a.html

[GreekHouse]

I'm not sure exactly what you're asking. These problems can be tricky sometimes or they can be really easy. If you are talking about a really complicated one, it can be tricky just to keep your conditionals in order.

The key is just to assume you're given some epsilon and then to find a delta such that for any y close enough to x (i.e. |x-y|<delta) you will have f(x) within epsilon of f(y) (i.e. |f(x)-f(y)|<epsilon). How you go about doing this relies heavily on the context of the problem.

[MathEconomist]

I'm also not sure what you're asking. Do you not understand the logic/form behind these proofs, or do you just have trouble actually working them out? If it's the latter, nothing but practice will help.

For the former, I take the following intuitive approach. Think of continuity proofs as a challenge and response. I'm giving you an epsilon and challenging you to find a delta s.t. if d(x,y)<delta, d(f(x),f(y))<epsilon. To make the proof formal, you specify a function that chooses delta given x,epsilon so that whatever challenge I present to you, you have a rule for how to respond (so delta=f(x,epsilon)). For UC proofs, we need to exhibit a function that chooses delta as a function of epsilon only (so that where we're at doesn't influence how close we need to be.)

To prove not continuous is usually easy. Just find a point and an epsilon s.t. no delta will work (in other words, wherever the function "jumps" at, pick an epsilon smaller than the size of the jump). This is usually done just by looking at the function and figuring out where it's discontinuous at. To prove not UC is a bit harder, but basically you need to show that to the function is changing at a greater and greater rate somewhere, so that around some point you have to be arbitarily close to get the fucntion to change by only epsilon for a fixed epsilon. Proving not UC are the hardest proofs usually, and they take more practice, but the point is to show that the delta you pick HAS to depeond on the x value in addition to epsilon.

I hope this helps. Again, if you're just having trouble working them out only practice will help. Just keep plugging away and working out as many of these as you can and it will become easier. Some of these proofs can be quite difficult and involve a lot of solving, but I find the above to be the simplest conceptual explanation of what you're trying to do.

[jason_t]

Practice is the only way.

[MelchyBeau]

I understand I need to practice them. The issue for me is I haven't seen one written out. If I could see a proof that a function (this function should not be uniformly continous) is continuous, I believe I could figure out the rest. A big issue for me is the formalism. I believe I can get that from an example

Melch

[MathEconomist]

Well, I really think it's better if you figure out the exact proofs on your own, but here's a more detailed explanation of a proof for a non-UC continuous function. We'll be given x and epsilon. We need to specify how to choose delta. It might be a hassle to solve this explicity, but the general form is that we know that we need abs(f(y)-f(x))<e. This means -e<f(y)-f(x)<e. So -e+f(x)<f(y)<e+f(x). Now, this is where this might get to be a hassle. Basically we'll need to take f inverse of everything, and depeding on what f is this may involve flipping the inequalities. At any rate, this will give us upper and lower bounds on y, and these implicitly define what our delta is (and by the way, if these don't depend on x, the function is going to be UC). If we need to define it more explicitly, we can subtract off x from everything, and then choose delta to be the min (in absolute value) of those two bounds. So that's pretty much the exact way that you do it. Start with epsilon and x, assume that our condition holds and solve for what that implies about delta.

I hope this helps. I think you will learn a lot more by doing these yourself and seeing how this general idea applies to specific cases than by just seeing one of these done for you and learning to parrot the results.

[jason1990]

The function f(x)=x^2 is continuous.

Proof: Let x in R and ep>0 be arbitrary. Define del to be the minimum of 1 and ep/(2|x|+1).

Let y in R be arbitrary such that |y-x|<del. Then

|f(y) - f(x)| = |y^2 - x^2|
= |y - x||y + x|
< del(|y| + |x|).

Note that |y|<=|y-x|+|x|<del+|x|<=1+|x|. Thus,

|f(y) - f(x)| < del(2|x| + 1) <= ep. QED

[BruceZ]

Think of it as too guys arguing about whether a function f(x) is continuous at a x = c. The first guy names a small positive number epsilon and says "give me a delta so that the function f(x) is within epsilon of a particular number whenever x is within delta of c". If the second guy can always provide a delta for every epsilon, no matter how small, then the function is continuous at x = c. If there is some epsilon for which it is impossible to find a corresponding delta, then the function is not continuous at x = c. You have to prove one of these cases, but how you do this depends on the problem. If it is continuous, then the fixed number that f(x) is within epsilon of is the limit of f(x) as x -> c.

[MelchyBeau]

I understand what is going on, I just needed to see an example of a proof just so I knew how to actually write it out.

thank you though. I got a good example of what it should look like now.

Melch