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Chapter 3: Problem 59

Prove that if \(f\) is differentiable on \((-\infty, \infty)\) and \(f^{\prime}(x)<1\) for all real numbers, then \(f\) has at most one fixed point. A fixed point of a function \(f\) is a real number \(c\) such that \(f(c)=c\).

Short Answer

Expert verified
If a function f has derivative less than 1 at every point and is differentiable over all real numbers, then it has at most one fixed point.

Step by step solution

01

Assume there are two fixed points

Assume, for the sake of contradiction, that there are two different fixed points of f, say c and d (where c<d). By definition, these satisfy \(f(c) = c\) and \(f(d) = d\).
02

Apply the Mean Value Theorem

Since f is differentiable on \((-\infty, \infty)\), we can apply the Mean Value Theorem to f on the interval [c,d]. Hence, there exists a point e in the interval (c,d) such that the derivative at e, \(f'(e)\), is equal to the average rate of change over [c,d], which is \((f(d)-f(c))/(d-c)\) by definition.
03

Substitute the equalities from step 1

Substitute \(f(c) = c\) and \(f(d) = d\) into the mean value equation from step 2. This leads to \(f'(e) = (d - c) / (d - c) = 1\).
04

Contradiction

But given in the problem \(f'(x) < 1\) for all x, so certainly \(f'(e) < 1\). This contradicts the result from step 3 where we found \(f'(e) = 1\).
05

Conclusion

Therefore, the assumption that there are two different fixed points of f must be wrong. Hence f can have at most one fixed point.

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