A function $f$ defined on an interval $A \subseteq \mathbb R$ is said to be convex if for all $a,b \in A$ and all $\lambda \in [0,1]$ it holds that \begin{equation} f ( (1- \lambda) a + \lambda b ) \leq ( 1- \lambda) f(a) + \lambda f(b). \end{equation} Similarly, if the latter inequality is a strict inequality for all $a,b \in A$ with $a\neq b$ and all $\lambda \in ]0,1[$, then $f$ is said to be strictly convex.
A well-known characterization for convex functions defined on an open interval is the following: If $A$ is open, then we have that $f$ is convex if and only if for each $c \in A$ there exists an $m \in \mathbb R$ such that $f(x) \geq m(x-c) + f(c) $ for all $x \in A$.
Now, I wonder if the following analogous result for strictly convex functions (also defined on an open) is true as well:
$f$ is strictly convex if and only if for each $c \in A$ there exists an $m \in \mathbb R$ such that $f(x) > m(x-c) + f(c) $ for all $x \in A$ with $x\neq c$.
In fact, I think I have a proof of this result, but it's rather sketchy (as it involves some suprema etc.), so it would be really great and helpful if somebody could confirm that the latter is indeed true. It is not necessary to provide help with a proof or so.
Many thanks in advance!