The real function $f$ is defined on $[0,\infty)$ and is increasing . The function $\phi$ is defined on $[0,\infty)$ by $$\phi (x)=\int_0^xf(t)dt$$
a) Prove that for all $x,y\ge 0$ $\phi (\frac{x+y}{2})\le \frac{1}{2}(\phi (x)+\phi (y))$.
b) Conclude that $\phi$ is convex .
i prove the 1st part using partition the interval ..but stuck in prove to 2nd part ..how prove it ??? Plz help ...
Let $M=\max(x,y)$ and note that $0\le f(t) \le M$ for all $t \in [0,M]$. In particular, $|\phi(a)-\phi(b)| \le M |a-b|$ for all $a,b \in [0,M]$ and so $\phi$ is continuous.
Fix $x,y$ and let $z_t = tx+(1-t)y$. Let $B= \{ t \in [0,1]| \phi(z_t) \le t \phi(x) + (1-t) \phi(y) \}$. Clearly $0,1 \in B$ and you have shown that ${1 \over 2} \in B$. Suppose $t_1,t_2 \in B$, then a) shows that $\phi(z_{{1 \over 2} (t_1+t_2) }) \le {1 \over 2} (\phi (z_{t_1})+\phi( z_{t_2}))$, from which it follows that ${1 \over 2} (t_1+t_2) \in B$. In particular, $B$ contains all points of the form ${k \over 2^n}$, $k=0,...,2^n$ for any $n$. Hence $B$ is dense in $[0,1]$ and since $\phi$ is continuous, it follows that $B$ is closed, and so $B=[0,1]$. Hence $\phi$ is convex.
