I am looking for a reference for the following claim:
Let $\phi:\mathbb (a,b) \to \mathbb R$ be a continuous function, and let $c \in (a,b)$ be fixed.
Suppose that "$\phi$ is convex at $c$". i.e. for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =c$, we have $$ \phi(c)=\phi\left(\alpha x_1 + (1- \alpha)x_2 \right) \leq \alpha \phi(x_1) + (1-\alpha)\phi(x_2) . $$
Then $\phi$ satisfies Jensen ineqaulity "at $c$".
Finite form:
Given $\lambda_i \in [0,1],x_i\in(0,\infty),i=1,\dots,k$ such that $\sum_{i=1}^k \lambda_i=1,\sum_{i=1}^k \lambda_ix_i=c$, we have $$\phi(\sum_{i=1}^k \lambda_ix_i) \le \sum_{i=1}^k \lambda_i \phi(x_i).$$
A more general probabilistic (measure-theoretic) form:
Given a random variable $X \in (a,b)$ with expectation $E(X)=c$, we have $$ \phi(c)=\phi(E(X)) \le E(\phi(X)). $$
In addition, if $\phi$ is strictly convex at $c$, then equality holds if and only if $X$ is constant a.e..
Both of these forms of Jensen inequality follow from the existence of a supporting line to the graph of $\phi$ at $c$.
The proof of the latter fact is not hard, but I couldn't find a source in the literature that presents this "localized" form of Jensen inequality, under the sole assumption of "convexity at a point". (In fact, I couldn't even find the term "convex at a point" anywhere...).
I find it impossible to believe that this doesn't show up in existing literature. Any help would be welcomed.
Comment:
Convexity at $c$ does not imply that the one-sided derivatives exist, so the standard proof for the existence of a supporting line (subgradient) does not apply here. (When the function is convex on an interval, every number between the two-sided derivatives form a subgradient).