Lipschitz function in Real Analysis

[mtayab1994]

Homework Statement

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)|&lt;k|x^{3}-y^{3}| where k is a positive real constant.

Homework 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&gt;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)?

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}|&lt;\eta\Longrightarrow|f(x)-f(y)|&lt;k|x^{3}-y^{3}|&lt;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}<br /> g(a)=f(a)-ka^{3}\geq0 &amp; ,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.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[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.

 

[mtayab1994]

And for number 4 do I have the right thought?

 

[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?

 

[mtayab1994]

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?

#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})|

 

[pasmith]

#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})|

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?

 

[mtayab1994]

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?

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?

 

[dirk_mec1]

No it should be 4b^2.

 

[mtayab1994]

No it should be 4b^2.

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.

 

[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.

 

[pasmith]

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.

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

 

[pasmith]

The definition of $u_n$ tells us that $x_n^3 = x_{n_1}/k$.

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

 

[brmath]

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

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

 

[mtayab1994]

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.

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?

 

[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.

 

[mtayab1994]

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.

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.

 

[pasmith]

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.

From the OP:

Let (Xn) be a sequence defined by X0=α given in ]a,b[

Thus a is one value which x_0 cannot take.

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.

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

 

[mtayab1994]

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 &lt; s or x_0 &gt; 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).

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?

 

[pasmith]

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?

Why does g being monotonic decreasing imply that if x_0 &gt; 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 &lt; s then (x_n) is increasing and bounded above by s.

 

[mtayab1994]

Why does g being monotonic decreasing imply that if x_0 &gt; 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 &lt; s then (x_n) is increasing and bounded above by s.

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?

 

[pasmith]

Actually, on closer inspection, it appears that (x_n) need not be monotonic. (If (x_n) were necessarily monotonic then parts 5 to 6 would be unnecessary, because as soon as it is monotonic it must converge, and the only point to which it can converge is s).

However, you can show directly that g is strictly decreasing implies that U_n is monotonic decreasing.

 

[brmath]

From the OP:

Thus a is one value which x_0 cannot take.

.

Hi Pasmith,

$x_0$ was defined to be a. So how is a the one value it cannot take?

 

[mtayab1994]

Actually, on closer inspection, it appears that (x_n) need not be monotonic. (If (x_n) were necessarily monotonic then parts 5 to 6 would be unnecessary, because as soon as it is monotonic it must converge, and the only point to which it can converge is s).

However, you can show directly that g is strictly decreasing implies that U_n is monotonic decreasing.

From part three we can see that $g$ is strictly decreasing so that means $(U_{n})$ is monotonic decreasing and bounded below by a>0 so therefore it converges to a number greater than 0.

 

[brmath]

From part three we can see that $g$ is strictly decreasing so that means $(U_{n})$ is monotonic decreasing and bounded below by a>0 so therefore it converges to a number greater than 0.

Do we know that $U_n$ > a for any n? I think you have to show it is, because otherwise your argument does not hold. And it was defined that $x_0$ = a.

I think it would be best to try to write out your entire argument for part 4, stating each step and the justification for it.

 

[mtayab1994]

Do we know that $U_n$ > a for any n? I think you have to show it is, because otherwise your argument does not hold. And it was defined that $x_0$ = a.

I think it would be best to try to write out your entire argument for part 4, stating each step and the justification for it.

X0 isn't a it's alpha α

 

[brmath]

X0 isn't a it's alpha α

That explains something. Imagine my misreading it.

Okay, why is $U_n$ bounded below by a? By zero yes.

 

[pasmith]

From part three we can see that $g$ is strictly decreasing so that means $(U_{n})$ is monotonic decreasing and bounded below by a>0 so therefore it converges to a number greater than 0.

No. In the case that x_0 = \alpha = s you have U_n = 0 for all n.

It is enough that U_n is bounded below by zero, and you need to show your working in support of the conclusion that U_n is monotonic decreasing.

 

[mtayab1994]

To show that $(U_n)$ is monotonic decreasing I thought of the following: $U_{n+1}-U_{n}=\frac{1}{k}(|f(x_{n})-f(s)|-|f(x_{n-1})-f(s)|)$, but I don't think I can remove the absolute value signs because we don't know if the difference is positive or not.

 

[brmath]

You want to use what you learned about g being monotonic in working through why U is.