Fix $\displaystyle \alpha $ a positive real number, let $\displaystyle X=[0,1] \subseteq R $ and define $\displaystyle f:X \rightarrow R $ by

$\displaystyle f(x)= \alpha x ( 1- x) $ for $\displaystyle x \in X $

a. For what values of $\displaystyle \alpha $ does the mapping $\displaystyle f:X \rightarrow R $ have the property that $\displaystyle f(X) \subseteq X $?

b. For what values of $\displaystyle \alpha $ does the mapping $\displaystyle f:X \rightarrow R $ have the property that $\displaystyle f(X) \subseteq X $ and is a contraction?

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a. Since $\displaystyle f(x)= \alpha x ( 1- x) = \alpha x - \alpha x^2 $, then the derivative is $\displaystyle f'(x) = \alpha (1- 2x)$

Since $\displaystyle f(X) \subseteq X = [0,1] $, I need to find the maximum value of the function to be 1, since $\displaystyle f(x) \subseteq [0,1] $ to be a subset.

Therefore,

$\displaystyle f'(x) = \alpha (1- 2x) = 0 \implies x = \frac{1}{2}$

Plug x = 1/2 back into, $\displaystyle f(x) = \alpha x (1-x) \implies \alpha =4$

Thus, since we are given prior that alpha is a positive real number then $\displaystyle \boxed{0 < \alpha \leq 4}$ where $\displaystyle f(X) \subseteq X=[0,1]$

b. To be a contraction, then its Lipschitz constant C for the mapping is $\displaystyle 0 \leq c < 1$

Then $\displaystyle | f'(x) | \leq c$

So $\displaystyle | \alpha ( 1- 2x) | < c $

I am having troubles with this inequality. Since $\displaystyle x \in [0,1] $, then $\displaystyle 0 \leq x \leq 1 $ and $\displaystyle 0 \leq c < 1$, how would I solve for the value of $\displaystyle \alpha $ in $\displaystyle 0 \leq \alpha \leq 4$ ?

Thank you for your time.