Question

(Still hoping for more ideas :( )Suppose f, g: E -> R are uniformly continuous functions where E is a subset of R. If f and g are both bounded, then f*g is uniformly continuous. Prove that if one of f or g are not bounded (the other must be bounded) then f*g is not necessarily uniformly continuous on E.

Answer 1

The common counterexample to show this is something like f = x and g - sinx. However we haven't defined trigonometric functions in our analysis class, so I'm not allowed to use anything trigonometric. So I either need a different counterexample somehow or a way to prove this is true in the general case. Either way, I haven't quite thought of a way to do this.

Answer 2

g = sinx, not g - sinx

Answer 3

question f*g or f of g?

Answer 4

f*g.

Answer 5

If f and g are both bounded and both uniformly continuous, then f*g are uniformly continuous. We have to show that this is not necessarily true if one of the functions is unbounded. So either an example that shows this or a general case proof, either or. But I haven't been able to get help outside of trigonometric functions, which I unfortunately cannot use for this problem :(

Answer 6

how about both f and g =x, ? f*g= x^2 which is not uniformly continous?

Answer 7

But that would make both f and g unbounded. I need one of the functions to actually be bounded.

Answer 8

:( I hate real analysis.!!

Answer 9

Aww. I think its really fun, just difficult, lol.

Answer 10

It takes me down since I get stuck everywhere, hehehe.. I am current student. At first, I think it is easy, but it is not. :(

Answer 11

Well, its easy to understand, not to show. I dont know if its better to mess with the general case or to just keep trying two functions to see if I can get that example I need, lol.

Answer 12

Well, sorry it took a while, I'm new to this subject.
I want to share you with my thoughts.

Well, from what I understood, the fact that the functions are uniformly continuous means that for any range \(\epsilon\) you allow me for the function to change around each point, I can find a \(\delta\) to describe a neighborhood around the point that the function could not change that much.
That means, that in order for the function to be uniformly continuous, its rate of change has to be limited, so I can tell the maximum change the function could change in a neighborhood.
The rate of change is described by the derivative of the function, and limited rate of change means that the derivative function has to be a bounded function.

We can now say the opposite as well. In order for a function not to be uniformly continuous its rate of change must not be limited, which means I cannot tell the maximum change in a given neighborhood, because somewhere along the way the function changes even more in such a neighborhood.
This means that for non-uniformly continuous function the derivative must be unbound.

Now, our function is \(f(x) \cdot g(x)\) and therefore the derivative:
$$
\Big( f(x) \cdot g(x) \Big)' = f'(x)\cdot g(x) + f(x) \cdot g'(x)
$$
Now, let's say \(f(x)\) is our unbounded function and \(g(x)\) the bounded function.
Let's even go further and say \(f(x) = x\) and we get
$$
\Big(x \cdot g(x) \Big)' = g(x) + x \cdot g'(x)
$$ Now, since \(g(x)\) is a bounded function and therefore its value is limited, it is clearly not what's going to make this derivative unbounded.
We have to relay on the other part \(x \cdot g'(x)\).
In order for that to do the job \(g'(x)\) must keep away from zero. That means \(g(x)\) must be constantly changing, and yet be bounded. Which implies it has to be periodic.
Ofc, the ideas that come first to mind are \(sin(x)\) and \(cos(x)\), but there are other options: http://en.wikipedia.org/wiki/Sawtooth_wave#mediaviewer/File:Waveforms.svg
Though if you can't use sin(x) then I don't see why are you allowed to use any of those just like that.

Any ideas?

Answer 13

Hmm....could we maybe restrict the interval then? If we can say that it's not uniformly continuous on a subset, then that should suffice I would think. Keep away from zero makes sense. but of course sinx and cosx have critical points where the derivative is 0. So does it have to be periodic?

Answer 14

The critical points are ok. keeping away from zero means that it won't cause the function to be bounded.
For that to happen it has to 'over-take' the unbounded nature of \(x\).
A bad example would be if \(g'(x) = \frac{1}{x}\) and then \(x \cdot \frac{1}{x} = 1\).
This is a bad example because that means \(g(x) = ln(x)\) which is not a bounded function, but it just shows the idea.

if the function is periodic, then there will always be points that it 'keeps away from zero' which allow \(x \cdot g'(x)\) to get to 'new heights' and therefore make the function unbound.

Look at the graph of \(x \cdot sin(x)\) for an example:
http://www.wolframalpha.com/input/?i=%283x%29%27

Yes, the graph is not always 'away from zero', but it is not limited (it's unbound), that is what I meant.

Perhaps you can just describe this 'excistance' of g(x) that satisfies those condition as a theoretical case which is a good enough answer idk.

Answer 15

Well, we haven't quite gotten to derivatives in our class, but knowing that idea can help me try to think of a function. Do you think there is a rational function or polynomial that could satisfy this?

Answer 16

I am jealous!! you guys are soooooooo good at the field.

Answer 17

@Loser66 Thanks =)

Well.. I couldn't think of any polynomial that satisfies that. I don't think there is a 'periodic polynomial'.
Even taylor series are accurate for functions like \(sin(x)\) in a limited neighborhood and then get off the track.

There was something I was thinking about.. lemme see if I can define this

Answer 18

Sure, that would be appreciated. Thanks for your input so far ^_^

Answer 19

Here, maybe something of that kind:
http://www.wolframalpha.com/input/?i=mod%28IntegerPart%28x%29%2C+2%29+-+0.5

Answer 20

ugly one isn't it.. but you gotta admit the idea is simple =)
It should mean \(g(x)\) would be a triangular saw function.

In the picture I showed before:
http://en.wikipedia.org/wiki/Sawtooth_wave#mediaviewer/File:Waveforms.svg
Then g(x) would be something of the kind of the third (triangular) and g'(x) something like the second (square)

Answer 21

It's so disappointing how limited we are in ideas. But because of the restrictions I have, it seems like the best bet might be some sort of general case proof. Basically, the way the course goes is anything that hasnt been officially defined in our notes or lecture is off limits, unless its thought to be necessary knowledge in order to use whats been defined. But yeah, maybe a piecewise function? Is the interval restriction a bad idea?

Answer 22

Well, uniformly continuous functions are not defined locally, but globally. so we can't just look on an interval, it is meaningless.
If you want to defined g(x) with an interval and then repeat it or so, then ye maybe you can if you make sure it comes out continuous.
But that wouldn't be any prettier than the other ideas =\

Well, unless I miss something we need some kind of periodic function. there are plenty you could make. Just make sure they are continuous.

Answer 23

`I am jealous!! you guys are soooooooo good at the field. ` @Loser66
Well not really.. I never faced this field.
What I'm saying I am able to say because I've spent a couple of hours thinking about it. It not unlikely that I've missed something. =)

Answer 24

Well, for example, we had a problem before where we needed to show that 1/x^2 was not uniformly continuous on \((0,\infty)\). The way we ended up doing it was by showing it was not uniformly continuous on \((0,1]\). Thats where my idea came from.

Answer 25

Yes but that is because you can show the function is unbounded because of the asymptote at x=0 which is covered by your interval.
Notice that you proved it is not uniformly continuous showing a point that doesn't match the condition.
Showing it is uniformly continuous would mean showing there aren't such points. So you cannot just limit it to the interval.

However, you are asked to prove that a condition such that multiplication of bound uniformly continuous function with an unbound uniformly continuous function might end up with non-uniformly continuous function.
A general proof seems fine to me.. but if you didn't learn derivatives in class and can't use it, then maybe there is another way showing the same idea...
Though.. how could it be that no derivatives are taught and yet you work with such limits?

Answer 26

Limits of functions were done a couple sections before this. It came at the end of last semester, actually. Derivatives are coming in the next chapter. But Im not sure, it seemed like our professor as well as our GA had no objections to the question. But our GA said for us to try and come up with a function that might work first. So I don't know.

Answer 27

So you're familiar with derivatives, but since not formally learnt in class you cannot use it??
Well limits of functions is one thing.. this is not just simple limits to evaluate.. This seems like a level above derivatives.. but oh well =)

Answer 28

Basically, yes. If so, the sinx idea would have made this problem so much easier, lol. But that's how it has worked, if it hasnt been introduced, then we cannot use it :/But I see how much derivatives come into play with this, so I dont know. I did find something, though. Maybe you can gather something from it?

By definition of uniform continuity on E, for \(x,y \in E\) and any \(\epsilon\ > 0, \ \exists \delta > 0\) such that \(|x - y| < \delta\ \implies\ | f(x) - f(y) | < \epsilon\)
Given that, the inequality to be used for f*g would be \(|f(x)g(x) - f(y)g(y)| < \epsilon\). So this is what I came across:
\(|f(x)g(x) - f(y)g(y)| \ge |g(y)||f(x) - f(y)| - |f(y)||g(x)-g(y)| - |f(x)-f(y)||g(x) - g(y)|\)

I guess the last g(y) gets cut off at the end, but yeah. Supposedly this inequality is supposed to say something about f(x) and g(x). This inequality and idea assumes g(x) is the unbounded function.

Answer 29

Yeah, at the end, it should be \(-|f(x) - f(y)||g(x) - g(y)|\)

Answer 30

hmm... \(\epsilon\) is an arbitrary value.. saying \(|f(x) - f(y)| < \epsilon\) means nothing on its own..
That's where the \(\delta\) comes to play. You have to show that for some value of \(\delta\) (which is the neighborhood) the function wouldn't change more than \(\epsilon\), which could be set to anything, at any point.

Just to make sure we're talking on the same thing, I watched this nice video before answering to get a grasp of what uniform continuous function even means:
https://www.youtube.com/watch?v=hXkQqCBLRp8
Maybe it won't say anything new to you, but still, see how he visualizes the concept of the neighborhood \(\delta\) getting smaller for the permitted 'error' \(\epsilon\)

That's why I wanted to use the derivative to see if there could exist a neighborhood where the function couldn't pass the permitted error, no matter where.

Answer 31

Here, for better persepective, this is what I was referring to

http://math.stackexchange.com/questions/673612/if-f-g-are-uniform-continuous-with-f-bounded-then-fg-is-uniform-continuou

I think I understand the idea and why the derivatives say so much about it, but Ill take a look at that, too :)

Answer 32

Do you mean the first answer or the second?
The first one just gives a solution you cannot use, without any theory behind it.
This solution works ofc because it matches the ideas we discussed here..
I'll read the second monstrous answer as well =)

Answer 33

Haha, well, its towards the bottom where the long inequality is introduce and it uses the idea that |a + b + c | >= |a| - |b| - |c|

Answer 34

Well.. sometimes I feel algebra is just not my language..
Damn, I just can't understand what he's up to.. he didn't even finish the proof..
Sorry =\

Answer 35

Yeah, I know. He kind of left it at that and seemed to think the rest was implied.

Answer 36

Well, it's getting late (actually it gets early now..) in here, so I think I'll go to sleep
But if I'll come up with a way showing it without using derivatives (and who knows what else I can't use...) then I'll let you know =)

Answer 37

ALright, thank you, I appreciate your time and effort ^_^

Answer 38

Oh well, guess not going to get much more out of this question. Thanks @pitamar

Answer 39

Sure np =)
Sorry I couldn't be more help, but I couldn't think of anything new.