Assume that $I$ is an interval in $\mathbb{R}$ and $f: I\rightarrow \mathbb{R}$.
$f$ is called strictly convex if $f(tx+(1-t)y) < t f(x)+(1-t)f(y)$ for $x\neq y$, $x,y \in I$, $t\in (0,1)$. How to show that for $z \in int I$ there exists a $p\in \mathbb{R}$ such that $f(x) > f(z)+p (x-z) \textrm{ for } z\neq x, z,x \in I?$
I try using inequality $ \frac{f(u)-f(x)}{u-x} < \frac{f(y)-f(u)}{y-u}$ for $x by putting $p(z):=sup_{\{x \in I: x
I obtain for $x
How to improve its or to give another algebraic proof that for strictly convex $f$ the following holds
$f(x) > f(z)+p (x-z) \textrm{ for } z\neq x, z,x \in I?$