Question

Let \(f\) be a differentiable function from \(\mathbf{R}\) to \(\mathbf{R}\) such that \(|f(x)-f(y)|\le 2|x-y|^{3/2}\), for all \(x,y\in\mathbf{R}\). If \(f(0)=1\) then \(\int_0^1 f^2(x)\,dx\) is equal to :

• A. \(1\)
• B. \(2\)
• C. \(\frac{1}{2}\)
• D. \(0\)

Hint:

Divide by \(|x-y|\) and let \(y\to x\): the derivative is zero, so \(f\) is constant.

Answer 1

The Correct Option is A

For \(x\ne y\): \(\left|\frac{f(x)-f(y)}{x-y}\right|\le 2|x-y|^{1/2}\). Letting \(y\to x\), \(|f\,\prime(x)|\le0\), so \(f\,\prime(x)=0\) for all \(x\).
Thus \(f\) is constant; \(f(0)=1\Rightarrow f(x)=1\).
\(\int_0^1 f^2(x)\,dx=\int_0^1 1\,dx=1\).
Answer: (1) \(1\)