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

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.

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.

• Ok, I can choose $\epsilon = 1- C$ and then the proof follows. Thank you. Feb 23 '20 at 9:27
• @user121 exactly, nice. Feb 23 '20 at 9:37

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.

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).