Are These Functions Uniformly Continuous on Their Given Intervals?

[mariama1]

determine if these functions are uniformly continuous ::


1- $$\ln x$$ on the interval (0,1)
2- $$\cos \ln x$$ on the interval (0,1)
3- $$x arctan x$$ on the interval (-infinty,infinty)
4- $$x^{2}\arctan x$$ on the interval (infinty,0

5- $$\frac{x}{x-1}-\frac{1}{\ln x}$$ on the interval (0,1)



Please help me .

 

[Char. Limit]

Start with the definitions. What does it mean for a function to be uniformly continuous?

 

[mariama1]

1- the first one if we find f`(x) = 1\x
and if we find the lim when x goes to 0 , then the limit does not exist
So , the function is not uniformly cont. on this interval
right ?
but how can i solve the next one ?

 

[Stardust*]

The property of uniform continuity is:

A function $$f :A \rightarrow R$$ is uniformly continuous is
$$\forall \varepsilon >0, \exists \delta >0: \forall x_1,x_2 \in A, |x_1-x_2|<\delta \Rightarrow |f(x_1)-f(x_2)|<\varepsilon$$

This means that for every interval of length $$2\varepsilon$$ in the image of f it is always possible to chose a good interval in A of length $$2\delta$$ so that the square resulting from the combination of those segments contains a "piece" of the function within the upper and lower sides of the rectangle. The choice of $$\varepsilon$$ is arbitrary, while $$\delta$$ depends on it.
I just wrote down these ideas to check out if they are correct.2)
$$cos(ln\,x)$$ in $$(0,1)$$
As $$x \rightarrow 0$$ the logarithm increases its slope and tends to $$-\infty$$. Meanwhile the cosine continuously changes between the maximum and minimum values, +1 and -1. Actually going to 1, $$ln(x)\rightarrow 0$$ and $$cos[ln(x)]\rightarrow 1$$.

A more formal explanation is:
$$|cos[ln(x_1)]-cos[ln(x_2)]|$$
$$=|-2sin\frac{ln(x_1)-ln(x_2)}{2}sin\frac{ln(x_1)+ln(x_2)}{2}|$$

$$\leq 2|sin{\frac{ln(x_1)-ln(x_2)}{2}}|$$

$$\leq |ln(x_1)-ln(x_2)|$$Now the problem is reduced to show that it is impossible to have such a difference $$0<|ln(x_1)-ln(x_2)|<\varepsilon$$ can always correspond to an interval $$\delta$$ in A made small or big anyway, since ln(x) diverges as it approaches 0+. So the function is not uniformly continuous in (0,1).
Is this reasoning correct?

 

[Stardust*]

3)
$$f(x)=x \,arctg(x)$$
Well, I think the work is more or less the same...
Let's take $$x_1,x_2 \in R$$ so that
$$|x_1\,arctg(x_1)-x_2\, arctg(x_2)|\leq \pi|x_1-x_2|\leq \varepsilon$$
So we can take whatever x1,x2 , have the inequality above for $$\varepsilon \geq \pi |x_1-x_2|$$ and take our $$|x_1-x_2|\leq \delta$$
with [tex]\delta \leq \varepsilon[\tex].
It seems the function is uniformly continuous.

 

[mariama1]

3)
$$f(x)=x \,arctg(x)$$
Well, I think the work is more or less the same...
Let's take $$x_1,x_2 \in R$$ so that
$$|x_1\,arctg(x_1)-x_2\, arctg(x_2)|\leq \pi|x_1-x_2|\leq \varepsilon$$
So we can take whatever x1,x2 , have the inequality above for $$\varepsilon \geq \pi |x_1-x_2|$$ and take our $$|x_1-x_2|\leq \delta$$
with [tex]\delta \leq \varepsilon[\tex].
It seems the function is uniformly continuous.

Thanks
Yes , i think you proof is right
because x arctan x seems to be uniformly cont. because it is bounded between 90, -90

but about cos ln x , if we take the limit of f`(X) = - sin 1\x
the limit when x goes to 0 does not exist .
i think it is enoygh to prove that this function is not uniformly cont.
Right ?

and what about Q4 ??

 

[Stardust*]

Now a proposal of solution for 4:
Let's keep $$\varepsilon\geq 0$$ fixed and choose $$x_1,x_2 \in R$$ so that:
$$0<x_1<x_2=x_1+\delta /2$$

The variation in the 'height' of the function between the two points is:
$$|f(x_1)-f(x_2)|=|x_1^2arctg(x_1)-x_2^2arctg(x_2)|\leq |\pi(x_1^2-x_2^2)|=\pi|x_1\delta+\delta ^2 /4|$$
but this last quantity is always:
$$\pi|x_1\delta+\delta ^2 /4|\geq|x_1\delta|$$.
So we have three conditions
a)$$|x_1-x_2|<\delta$$
b)$$|x_1^2arctg(x_1)-x_2^2arctg(x_2)|<\varepsilon$$
c)$$|x_1^2arctg(x_1)-x_2^2arctg(x_2)|\geq x_1 \delta$$

(a, b derive from the definition of uniform continuity, while c is the result of the previous passages) How can we get to a contraddiction? I believe this is possible by showing that
$$|x_1^2arctg(x_1)-x_2^2arctg(x_2)$$ can be both bigger and smaller than the chosen $$\varepsilon$$ at the same time, for the same x1. This is possible if we use $$\varepsilon >x_1 \delta$$. The only way out from this situation is to admit f cannot be uniformly continuous.
I hope this last passage is correct, but I still have some doubt. Any idea to improve it?


Anyway, for exercise 2):
the fact the derivative doesn not exists [many physicists I know would better say "it explodes" :-) ] only tells us that the function is not Lipschitzian. f being Lipschtizian means f being uniformly continuous, but I don't think the inverse is so sure. Indeed, at least a fractal function I saw this week looked very uniformly continuous, but not lipschitzian...

 

[Stardust*]

Now, I think there's something wrong in what I said for 4).
I'll try again.

So we have:
$$f(x)=x^2 arctg(x)=x^2 \cdot h(x)$$
Let's consider two points:
$$x_1,x_2 \in [0,\infty]\, , \,x_2<x_1\, , \, x_1=x_2+\omega$$, where $$\omega>0$$.

We get to:
$$\left|(x_2+\omega)^2 h(x)-x_2h(x)\right|=\left|x_2^2[h(x_2+\omega)-h(x_2)]+2x_2\omega h(x_2+\omega)+\omega^2 h(x_2+\omega)\right|$$(*)
Looking at $$y=x^2$$, it is clear it is strictly increasing in the interval $$[0, \infty]$$;
$$y=arctg(x)=h(x)$$ is strictly increasing too. So the quantity in (*) is always positive since all the pieces of the sum are positive. This means we can always take $$x_2$$ big enough to make it greater than an arbitrary value $$M>\varepsilon \in R$$
[tex][/tex]
[tex][/tex]
[tex][/tex]

 

[Stardust*]

The last part of my previous post is:
Looking at $$y=x^2$$, it is clear it is strictly increasing in the interval $$[0, \infty]$$;
$$y=arctg(x)=h(x)$$ is strictly increasing too. So the quantity in (*) is always positive since all the pieces of the sum are positive. This means we can always take $$x_2$$ big enough to make it greater than an arbitrary value $$M>\varepsilon \in R$$, so this difference (*) is not as small as necessary to say $$f(x)$$ is uniformly continuous.

 

[mariama1]

The last part of my previous post is:
Looking at $$y=x^2$$, it is clear it is strictly increasing in the interval $$[0, \infty]$$;
$$y=arctg(x)=h(x)$$ is strictly increasing too. So the quantity in (*) is always positive since all the pieces of the sum are positive. This means we can always take $$x_2$$ big enough to make it greater than an arbitrary value $$M>\varepsilon \in R$$, so this difference (*) is not as small as necessary to say $$f(x)$$ is uniformly continuous.

Thanks for your notes , but the f(x) seems to be not uniformly continuous , but I am not sure

 

[micromass]

Stardust*, your solutions are very good. But it's actually against PF policy to give complete solutions to homework problems. Since the OP will only learn this things if he figures it out himself...

 

[Stardust*]

But it's actually against PF policy to give complete solutions to homework problems. Since the OP will only learn this things if he figures it out himself...

I'm sorry, I did not want to annoy anyone, I was just trying to see if I could give a hand and if I could solve such exercises in a more or less correct way (these are the first ones of this kind I do, so this is good practice for me, too).
I'll be more careful in future...

Thanks for your notes , but the f(x) seems to be not uniformly continuous , but I am not sure

Yes, I guess so. Probably I expressed myself in a not clear way, this is what I mean by saying you can make the difference (*) big as you want, while you need it to be small.

For 5) , I still have no clues. Any idea?