Suppose \(f:\mathbb R\to\mathbb R\) is differentiable and satisfies \(f'(x)>f(x)\) for all real \(x\). Show that if \(f(0)=0\), then \(f(x)>0\) for all \(x>0\).
Solution
Since \(f ^ { \prime } ( 0 ) > 0\) we have \(f ( x ) = x \cdot f ^ { \prime } ( 0 ) + o ( | x | )\) in a neighborhood of zero, so there is a \(t > 0\) such that f is positive on \(( 0 , t )\) . Assume for contradiction that \(f ( x ) \leq 0\) for some \(x > 0\) and let \(x _ { 0 }\) be the first such x. Then \(f ( x ) > 0\) on \(( 0 , x _ { 0 } )\) , which means that \(f ^ { \prime } ( x ) > 0\) on \(( 0 , x _ { 0 } )\) , so \(f ( x _ { 0 } ) > f ( 0 ) = 0\) , a contradiction