Proving that if f' is bounded in the fixed point, then f' is bounded on some interval

Question

I have a question related to the Problem 10 in Exercises 1.7 (p. 35) in the book "Discrete chaos" , Saber Elaydi. The Problem is:

"Assume that $f$ is continuously differentiable at $x^*$. Show that if $|f'(x^*)| <1$, for a fixed point $x^*$ of $f$, then there exists an interval $I= (x^* -\delta, x^*+\delta)$ such that $|f'(x)| \leq M <1$ for all $x \in I$ and for some constant $M $. "

My attempt was: $f$ is continuously differentiable at $x^*$, so it is continuous at $x^*$, so for all $ \epsilon > 0$ $\exists \delta > 0 $ such that $ |x- x^* |< \delta \implies |f(x) - f(x^*) |< \epsilon $.

I wanted to prove that on the interval $(x^*-\delta, x^*+\delta) $ holds: $|f'(x)| \leq M <1$, for some $M$.

We can write $|f'(x)| $ as: $|f'(x)| = | f'(x) - f'(x^*) + f'(x^*) | \leq | f'(x) - f'(x^*)| + |f'(x^*) | \leq \epsilon + 1$, where $x \in (x^*-\delta, x^*+\delta) $

Now I don't know what to do, can someone help me? Thanks in advance.

Answer 1

You need to use the continuity of the derivative.

Since $|f'(x^*)|<1$ take a value $C$ s.t. $|f'(x^*)|\le C<1$.

Then, by continuity of $f'$ at $x^*$ , for every $\epsilon >0 $there is a $\delta >0$ s.t. if $x \in I$ then $|f'(x)-f'(x^*)| < \epsilon$

or $ |f'(x)| <|f'(x^*)|+\epsilon \le C+\epsilon $

Now just choose $\epsilon$ wisely.

Answer 2

If the conclusion is not true (i.e., for any M < 1 and for any positive delta, there is an x such that |x - x*| < delta and |f'(x)| > M), then there is a sequence (x_n) converging to x* such that |f'(x_n)| >= 1. As f' is continuous (|.| also), (|f'(x_n)|) converges to |f'(x*)| < 1 (a contradiction bcs (|f'(x_n)|) is a sequence of values greater than or equal to 1).

Answer 3

Remarks:

1) You need continuity of $|f'(x)|$ at $x_0$.

Use Triangle inequality:

$||f'(x)|-|f'(x_0)|| \le$

$ |f'(x)-f'(x_0)| <\epsilon$.

2) $\epsilon >0$ given there is a $\delta >0$ s.t.

$|x-x_0| <\delta$ implies $||f'(x)|-|f'(x_0)||<\epsilon$, or

$-\epsilon +|f(x_0)| <|f(x)| < |f(x_0)| +\epsilon$.

3)We have $|f'(x_0)|= M_0 <1$;

4) Choose $\epsilon=(1-M_0)/2$.

5) Then 2) reads:

$|f'(x)| <|f'(x_0)|+\epsilon$,

$|f'(x)|< M_0+(1-M_0)/2=$

$1/2+M_0/2=:M <1$, and we are done.