Problem 59

Question

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 \(f(c)=c\).

Step-by-Step Solution

Verified

Answer

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.

1Step 1: 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

2Step 2: 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.

3Step 3: 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 \(f'(e) = (d - c) / (d - c) = 1\).

4Step 4: 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 \(f'(e) = 1\).

5Step 5: 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.

Problem 59