Indian Statistical Institute B.Math & B.Stat : Real Analysis

Suppose $f$ is a differentiable and increasing function on $[0, 1]$ such that $f(0) < 0 < f(1)$. Let $F(t) = \int_{0}^{t} f(x) dx$. Then show that $F$ has a unique minimum in the open interval $(0, 1)$. Since $f$ is differentiable, $f$ is continuous. Therefore $F(t)$ is differentiable and $F'(t) = f(t) , t \in [0,1]$ ( students interested in the proof are advised to read INTRODUCTION TO REAL ANALYSIS, Author : ROBERT G. BARTLE, DONALD R. SHERBERT ). Given that $f$ is a differentiable and increasing function on $[0, 1]$ such that $f(0) < 0 < f(1)$ it follows that $F'(t)$ satisfies the same conditions of $f$ in $[0,1]$ since $F'(t) = f(t)$. Now see that $F'(0)F'(1) = f(0)f(1) < 0$, continuity of $F'$ implies $\exists$ $c \quad \in (0,1)$ such that $F'(c)=0$. Again since $F'$ is increasing, the points at which $F'$ vanishes forms a sub-interval $(c,d)$ of $[0,1]$ (i.e, the points at which the function $F'$ vanishes forms a connected set !) where $0 < c < d < 1$. Thus $F'(t) < 0$ for $t \in [0,c)$ and $F'(t) > 0$ for $t \in (d,1]$ which implies $F$ is strictly decreasing in $[0,c)$, strictly increasing in $(d,1]$ and $F$ is constant on $[c,d]$ which in turn implies $F$ has a unique minimum on $[c,d]$. Note that if it was given $f$ to be strictly increasing then $c=d$ and the unique minimum will be attained at a unique point.