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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 $x<u<y$ by putting $p(z):=sup_{\{x \in I: x<z\}} \frac{f(z)-f(x)}{z-x}$.

I obtain for $x<z$ : $\frac{f(z)-f(x)}{z-x} \leq p$ and for $z<x$: $p\leq \frac{f(x)-f(z)}{x-z}$. But I need a strong inequalities.

How to improve its or to give another algebraic proof that for strictly convex $f$ the following holds

$$f(x) > f(z)+p (x-z) \textrm{ for } z\neq x, z,x \in I?$$

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up vote 3 down vote accepted

You are close. If $y=tx+(1-t)z$ for $x<z, t\in(0,1)$, then (by replacing $y$ and using the definition of strict convexity, after a short calculation (note $x-z<0$)) $$\frac{f(x)-f(z)}{x-z} < \frac{f(y)-f(z)}{y-z}$$ that is, if you fix $z$, $$\phi:x\mapsto \frac{f(x)-f(z)}{x-z}$$ is strictly increasing. Hence your $p(z)$ actually equals $\lim_{x \rightarrow z, x < z} \phi(x)$ and the inequality you are asking for for those $x$ to the left of $z$ also follows by strict monotonicity of $\phi$. The case $y>z$ I leave to you.

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If $f(x)$ is differentiable, the solution is even simpler:

$f(x)$ is strictly convex $\Leftrightarrow$ $f''(x)>0$


Hence $f(z)+p(x-z)=f(z)+f'(z)(x-z)<f(z)+f'(z)(x-z)+f''(z)(x-z)^2/2=f(x)$

The last equality holds due to Taylor expansion.

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