Assume that f is a differentiable function such that f(0)=f'(0)=0 and f''(0)0. Argue that there exists a positive constant a0 such that f(x)0 for all x in the interval (0,a). Can anything be concluded about f(x) for negative x's? [Hint: Start the proof by using the MVT on f'(x)]

Question

Assume that f is a differentiable function such that f(0)=f'(0)=0 and f''(0)>0. Argue that there exists a positive constant a>0 such that f(x)>0 for all x in the interval (0,a). Can anything be concluded about f(x) for negative x's? [Hint: Start the proof by using the MVT on f'(x)]

jamesj · Answer

If f''(0) > 0, then as f'' is continuous, there exists an interval [0,c), c > 0 where f''(x) is positive.

Choose any number c' < c.  Then for some y in (0,c'), by the MVT
$$ f''(y) = \frac{f'(c') - f(0)}{c' - 0} $$
which is positive.  Hence
$$ f'(c') - f(0) = c' f''(y) > 0 $$
i.e., f'(c') > 0.

Now as c' is an arbitrary number in (0,c), it follows that this relation is true for all
$$ x \ \in (0,c) $$

Therefore integrating f' over that interval, f(x) > 0 for all such x.