Tangents of a Strictly Convex Fuction

Question

Let $f:\mathbb R \rightarrow \mathbb R$ be differentiable and strictly convex. Is it true that $x,y \in \mathbb R$ and $x \neq y$ imply \begin{equation} f'(x)(y-x) <f(y) - f(x) \end{equation} If so, how can we prove it?

Answer 1

Pick an arbitrary $x,y \in \mathbb R$ such that $x \neq y$.

Since $f$ is convex (strict convexity not required yet), we can conclude that \begin{equation} f'(x)(z-x) \leq f(z) - f(x), \ \ \forall z \in \mathbb R. \end{equation} In particular, the inequality holds for $z = tx + (1-t)y$, where $t \in (0,1)$: \begin{equation} f'(x)(tx + (1-t)y-x) \leq f(tx + (1-t)y) - f(x). \end{equation} We will now analyze both sides of this inequality. For the LHS, we can simply rearrange terms to obtain \begin{equation} f'(x)(tx + (1-t)y-x) = (1-t)f'(x)(y-x). \end{equation} For the RHS, we can apply the strict convexity of $f$ and rearrange terms to obtain \begin{equation} f(tx + (1-t)y) - f(x)< t(f(x) + (1-t)f(y) - f(x) = (1-t)(f(y) - f(x)). \end{equation} Bringing it all together, we have \begin{equation} (1-t)f'(x)(y-x) < (1-t)(f(y) - f(x)). \end{equation} Since $t \in (0,1)$, this is equivalent to the desired result: \begin{equation} f'(x)(y-x) < f(y) - f(x). \end{equation}

Answer 2

Yes this is true. An equivalent condition for convexity is that $f(x)$ lies above its tangent at every point. This is precisely the statement that $f'(x) < \frac{f(y) - f(x)}{y-x}$, i.e. the derivative should be smaller than the slope of the secant line.