# Lipschitz function in Real Analysis

1. Dec 12, 2013

### mtayab1994

1. The problem statement, all variables and given/known data

Let f be a real function defined on the interval [a,b]/0<a<b:

$$\forall x,y\in[a,b],x\neq y/|f(x)-f(y)|<k|x^{3}-y^{3}|$$ where k is a positive real constant.

2. Relevant equations

1- Prove that f is uniformly continuous on [a,b]

2- We define a function g on [a,b] such that $$g(x)=f(x)-kx^{3}$$.
Prove that g is strictly monotone on [a,b] .

3- We suppose that for every x in [a,b] $$ka^{3}\leq f(x)\leq kb^{3}$$ . Deduct that there exists a unique number s such that g(s)=0.

4- Let (Xn) be a sequence defined by X0=α given in ]a,b[ . and $$X_{n}=(\frac{1}{k}f(X_{n-1}))^{\frac{1}{3}}$$.

- Prove that the function $$U_{n}=|x_{n}^{3}-s^{3}|,x>0$$ converges to a real number d≥0.

5- Deduct that there exists a sub-sequence (Xn) that converges to the real number $$c=\sqrt[3]{s^{3}+\epsilon d}$$ where (ε=1 or ε=-1)

6- Prove that f(c) takes either of the values $$k(x^{3}+d)$$ or $$k(x^{3}-d)$$

7- Deduct that d=0. What conclusion can we make about the sequence (Xn)?

3. The attempt at a solution

1- I let ε>0 and made η=(ε/k) so by definition of uniform continuity:

$$|x^{3}-y^{3}|<\eta\Longrightarrow|f(x)-f(y)|<k|x^{3}-y^{3}|<k\eta=\epsilon$$ So therefore f is uniformly continuous.

2- I derived g and I got that g'(x)=f'(x)-3kx^2. And since k>0 therefore for all x in [a,b] -3kx^2<0 but I don't know what to say about f'(x).

3- We have g(x)=f(x)-kx^3 and the fact that ka^3≤f(x)≤kb^3 so :$$\begin{cases} g(a)=f(a)-ka^{3}\geq0 & ,g(b)=f(b)-kb^{3}\leq0\end{cases}$$ .

which implies that g(a)*g(b)≤0 So by the intermediate value theorem there exists an s in [a,b] such that: g(s)=0.

4- I know that I have to use the lipschitz function, but I don't know how to start.

5-6-7 Nothing so far.

Any help would be appreciated. Thank you.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Dec 12, 2013
2. Dec 12, 2013

### brmath

Starting with #1, the definition of uniform continuity is that if $\epsilon$ is chosen there is a $\delta$ independent of x such that |$x_1 - x_2| < \delta \Rightarrow |f(x_1) - f(x_2)| < \epsilon$.

The definition doesn't say anything about |$x_1^3 - x_2^3| < \delta$ . You have to show there is a $\delta$ that works for the first powers. So you have to show the difference of the cubes being < $\delta$ implies that the linear difference gives you the right result.

3. Dec 13, 2013

### mtayab1994

And for number 4 do I have the right thought?

4. Dec 13, 2013

### brmath

Re # 4 you really need to work through 2 and 3 first, because they are giving you material for solving #4. Have you found a way to do #1 yet?

5. Dec 13, 2013

### mtayab1994

#3 is correct I believe and #2 I derived the function but I couldn't conclude anything. While for number one I couldn't really come up with anything besides of thinking of the fact that :

$|x^{3}-y^{3}|=|(x-y)(x^{2}+y^{2})|$

6. Dec 14, 2013

### pasmith

The correct expansion is $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$

If $x \in [a,b]$ and $y \in [a,b]$, what is an obvious upper bound for $x^2 + xy + y^2$?

7. Dec 14, 2013

### mtayab1994

Yes I did that part i got k|x^2+xy+y^2||x-y|<k4a^2|x-y| then i chose η=ε/(4ka^2) and the proof is done. Can i get a hint on how to start number 4?

8. Dec 14, 2013

### dirk_mec1

No it should be 4b^2.

9. Dec 14, 2013

### mtayab1994

I don't think so because (x^2+xy+y^2)k=3ka^2<4ka^2 that is when plugging in a for x and y.

10. Dec 14, 2013

### brmath

Hi,

Sorry I could not get back to you before -- I have been out sick. I can see you and pasmith are discussing how to get #1 correct, and really you should settle that down before proceeding.

To do #2, you do not want to take derivatives. While that is one way to show something is monotonic, it is not the only way. In this case we do not even know at this point that f has a derivative. The definition of f would allow us to conclude that f is differentiable almost everywhere, but a) you have not proved that and b) that doesn't help you here anyway.

To do #2, you should try a more primitive approach: if $x_1$ < $x_2$ is it true that $g(x_1) \le g(x_2)$? I say it is, and that you can show this by using the definition of g. Similarly, if $x_1$ > $x_2$.

Re part #3 you state correctly that g(a) > 0 and g(b) < 0. I don't understand why you referenced the product g(a)*g(b). You need to state that g is continuous, which is not something you have proved yet. If g is continuous then it must pass through zero on the way from f(a) to f(b) and the intermediate value theorem is indeed the reason.

To get started on part 4 note that s satisfies the equation g(s) = 0 or f(s) - k$s^3$ = 0. The definition of $u_n$ tells us that $x_n^3 = x_{n_1}/k$. You need to put together this information to get your result. Of course we know that g(s) = 0 is dependent on the assumptions in part 3.

11. Dec 14, 2013

### pasmith

But if $b > \frac{2a}{\sqrt{3}} > a$ then $4ka^2 < 3kb^2$, which is what you get by setting $x = y = b$.

12. Dec 14, 2013

### pasmith

I don't know how you got that; by definition
$$x_n^3 = \frac{f(x_{n-1})}{k}$$

13. Dec 14, 2013

### brmath

I got that by mistyping -- a common problem with me -- sorry.

14. Dec 14, 2013

### mtayab1994

For number 4 i was able to transform Un to the following:

$U_{n}=|X_{n}^{3}-s^{3}|=\frac{1}{k}|f(X_{n-1})-f(s)|$ . Now do I have to use the Lipschitz function that I had proved is continuous?

15. Dec 14, 2013

### brmath

Your conversion is correct : $U_n = (1/k)|f(x_{n-1}) -f(s)|$ The question was to show that $U_n$ converges to a real number d $\ge$ 0. Given the abs value signs, if it converges, it certainly will be to a non-negative number.

f(s) is a fixed number, so if $U_n$ is going to converge, it better be true that $x_n$ converges . Using whatever Lipsichitz conditions you have doesn't seem to me to bear directly on this problem. You know at least two things here that you have not considered using, and I believe you will need them to conclude that $U_n$ converges.

One is that g(x) is monotonic. The other is that $x_0$ = a.

I think you are hoping there is some fast way to do this one, but I don't see any cheap shots here. You are going to have to put together all the pieces from parts 1, 2, and 3. What this problem is doing is breaking a long and non-obvious proof down into smaller steps, so you can understand how to start with the hypotheses and arrive at the final conclusion.

16. Dec 15, 2013

### mtayab1994

This is what I've come up with: if $g(x)$ is strictly monotone increasing and $x_0$=a. Then Xn will converge because it's bounded above by b.

17. Dec 15, 2013

### pasmith

From the OP:

Thus $a$ is one value which $x_0$ cannot take.

Boundedness is not sufficient for convergence; you also need monotonicity of $(x_n)$, and the direction of the monotonicity depends on whether $x_0 < s$ or $x_0 > s$.

Also, the assumption made in part 3 - that $ka^3 \leq f(x) \leq kb^3$ - requires that $g$ is monotonic decreasing, since $g(a) = f(a) - ka^3 \geq 0$ and $g(b) = f(b) - kb^3 \leq 0$ (as you would have seen from using the IVT to deduce the existence of $s$).

18. Dec 15, 2013

### mtayab1994

So g is monotonic decreasing and $x_0>s$ so that means that it's decreasing and bounded below by s which means it's convergent?

19. Dec 15, 2013

### pasmith

Why does $g$ being monotonic decreasing imply that if $x_0 > s$ then $(x_n)$ is decreasing and bounded below by $s$? That's the bit you need to explain.

Also you need to explain why $g$ being monotonic decreasing implies that if $x_0 < s$ then $(x_n)$ is increasing and bounded above by $s$.

20. Dec 15, 2013

### mtayab1994

Well X0 is the first value Xn takes and if g is monotonic increasing then that means:

$g(x_{n})=f(x_{n})-kx_{n}^{3}$ is monotonic decreasing and since g is continuous that means:$g(x_{n})=0$ at a unique point which is s. So that means that $(x_{n})$ is bounded below by s hence it's convergent?