# Uniform continuity

Discussion in 'Physics & Math' started by BLuEGoD, Aug 18, 2009.

1. ### BLuEGoDRegistered Senior Member

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9
Hello,

I have some troubles trying to solve whether some functions are uniformly continuous or not... Exactly I easily solve if x^2 is uniformly continuous in (0,1] as doing some math at $\left|f(x)-f(y)\right|=\left|x^2-y^2\right|$ then I get $\delta<\frac \epsilon2$

but if the function is not uniformly continuous the way as I could know is to fix an $\epsilon$, let's say $\epsilon=1$ and for $x^2$ in R. I have troubles finding which is the "x" or "y", in this example I saw $x=n+\frac \delta2$, y= n .. so $\left|f(x)-f(y)\right|=n\delta+\frac {\delta^2}4$ that is greater than 1....

So I get the point but in other functions I have troubles findin such things like ($x=n+\frac \delta2$).. Like 1/x (an easy one).. even if understanding the proof itself..
I mean, there must be different ways trying with other "x" but how to know one that works?

Thank you!

3. ### temurman of no wordsRegistered Senior Member

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1,330
Please show me an example of a function which is not uniformly continuous.

5. ### BLuEGoDRegistered Senior Member

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9
Already there.. I wrote $x^2$ in $\mathbb{R}$

7. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Thanks. I think a general principle would be to find a function g such that

$\left|f(x)-f(x+\delta)\right|=g(x,\delta)$

or more generally

$\left|f(x)-f(x+\delta)\right|\geq g(x,\delta)$

and then to prove that for any $\delta>0$ there is x such that

$g(x,\delta)\geq1$

Do you agree?

8. ### BLuEGoDRegistered Senior Member

Messages:
9
It's ok, but I saw other solutions like making the $\delta=0$
so I am a bit confused on the way of solving uniform continuity function exercises, just because the way is different, if it is supposed to be uniformly continuous then it seems that there's no need to fix the $\epsilon$..

Example:

$f(x)=\frac{x+1}{x^2}$
$a\in\mathbb{R}$
$0<a<1$

f uniformly continuous in [a,1) ?

$\left|f(x)-f(y)\right|=\frac{\left|x-y\right|\left|xy+x+y\right|}{x^2y^2}<\frac{3\left|x-y\right|}{a^4}$

(Also I can't understand why there's a 3 up and a^4 down.. I mean, it's ok that both are the bonds (1+1+1) and a^2a^2.. but If I have to do it by myself I am not sure I'll get that point and why)

so taking $\delta=\frac{a^4\epsilon}3$ it's ok.

But if it's in (0,1) the proof will be fixing the $\epsilon$..

But that's knowing it will not be u.c.!

9. ### AbsaneRocket SurgeonValued Senior Member

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8,989
It might help to think of uniform continuity this way...

With regular continuity, delta is dependent on the point you are checking and epsilon.

With uniform continuity, you are concerned with the metric distance between two points in X and epsilon.

I'm one of those people that cannot memorize theorems and definitions to save my life. It's amazing I graduated with a math degree when I can't remember anything... What saved me is what I did to compensate for memory: understand EVERYTHING.

If you understand what is going on with uniform continuity, then you might be better off answering the questions.

Draw a graph of e^x. Pick a random epsilon and place two points within it on the curve (x, f(x)) and (y, f(y)). Then, look and where x and y are on the horizontal axis. Pick a random delta to enclose them. Now, shift your epsilon towards the negative end of the horizontal axis. Again, pick two points within there. If you picked points far away from each other within epsilon, you'll notice that you need an even larger delta. So, if you continue this ad infinitum and place every delta you get into a set, you'll see that the set will fail to have a least upperbound (in other words, not the single delta we need).

To prove that a function is not uniformly continuous, assume to the contrary that it is and end the proof by showing that a set of deltas needed to satisfy EVERY f(x)-(y)<epsilon for a given epsilon is not bounded from above.

10. ### BLuEGoDRegistered Senior Member

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9
Thank you!, I am sure I have the same problem about memorizing, and exactly understanding all is what I do too.

Thank you for your reply, that's exactly what I have done, but anyway where I get

is looking at the way to prove that a function is not continuous

Exactly:

For $x^2$ in R.In this example I saw $x=n+\frac \delta2$, y= n .. so $\left|f(x)-f(y)\right|=n\delta+\frac {\delta^2}4$ that is greater than 1....

Could this be another good proof ? I was trying to think it by myself and put something else instead of $x=n+\frac \delta2$ more simple..

$y=2x$ so..

$\left|f(x)-f(y)\right|=\left|x+y\right|\left|x-y\right|=3x|x-y|<1$

$\delta<1/3x$ so taking x=1 $3\delta<1$ but $3/3<1$ <-wrong..

11. ### §outh§taris feeling causticRegistered Senior Member

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4,832
Well, you can ask yourself what uniformly continuous functions on a metric space do to Cauchy sequences