Question

Assume that \(|f(x)| \leq 1\) and \(\left|f^{\prime \prime}(x)\right| \leq 1\) for all \(x\) on an interval of length at least \(2 .\) Show that \(\left|f^{\prime}(x)\right| \leq 2\) on the interval.

Short answer

The inequality \(\left|f^{\prime}(x)\right| \leq 2\) holds for all \(x\) in the interval.

Step-by-step solution

Assumptions

Assume that \(f(0)=f(1)=0\). Any other interval of length 1 can be shifted and scaled to match this. Then the maximum value of \(f'(x)\) would be the same.

Apply the Mean Value Theorem

By the Mean Value Theorem, there exists a \(c\) in the interval \((0, 1)\) such that\[f^{\prime}(c)=f(1)-f(0).\]From the assumption we know that \(|f(x)| \leq 1\), so\[|f^{\prime}(c)|=|f(1)-f(0)| \leq |f(1)|+|f(0)| \leq 2.\]

Interval Halving Method

Now consider any \(x\) in the interval \((0, 1)\). Split this interval into two halves, \((0, x)\) and \((x, 1)\). Apply the Mean Value Theorem to each interval, and a similar argument as above shows that \(|f^{\prime}(c)| \leq 2\) for some \(c\) in each interval. So on any interval of length at least 2, the maximum of \(|f^{\prime}(x)|\) is at most 2.