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## Homework Statement

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

[tex]\forall x,y\in[a,b],x\neq y/|f(x)-f(y)|<k|x^{3}-y^{3}|[/tex] 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 [tex]g(x)=f(x)-kx^{3}[/tex].

Prove that g is strictly monotone on [a,b] .

3- We suppose that for every x in [a,b] [tex]ka^{3}\leq f(x)\leq kb^{3}[/tex] . 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 [tex]X_{n}=(\frac{1}{k}f(X_{n-1}))^{\frac{1}{3}}[/tex].

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

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

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

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:

[tex]|x^{3}-y^{3}|<\eta\Longrightarrow|f(x)-f(y)|<k|x^{3}-y^{3}|<k\eta=\epsilon[/tex] 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 :[tex]\begin{cases}

g(a)=f(a)-ka^{3}\geq0 & ,g(b)=f(b)-kb^{3}\leq0\end{cases}[/tex] .

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

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