Question

Let f be any function defined on R and let it satisfy the condition :
\(|f(x)-f(y)|\le|(x-y)^2|,\ \forall(x,y)\in R\)
If f(0) = 1, then :

• A. f(x) can take any value in R
• B. \(f(x)\lt0,\ \forall x\in R\)
• C. \(f(x)\gt0,\ \forall x\in R\)
• D. \(f(x)=0,\ \forall x\in R\)

Hint:

Divide by \(|x-y|\) and let \(y\to x\) to show \(f'(x)=0\), so f is constant.

Answer 1

The Correct Option is C

For \(x\ne y\): \(\left|\frac{f(x)-f(y)}{x-y}\right|\le|x-y|\). Letting \(y\to x\), \(|f'(x)|\le0\), so \(f'(x)=0\) for all x.
Hence f is constant, and since \(f(0)=1\), \(f(x)=1\) for all x. Thus \(f(x)\gt0\) for all \(x\in R\).
Answer: (3) \(f(x)\gt0,\ \forall x\in R\)