Convexity at a point and Jensen inequality

Question

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 $c$.

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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).

Answer 1

For any real numbers $u,v,c$ such that $u\le c\le v$, let $\mu_{c;u,v}$ denote the unique probability distribution on the set $\{u,v\}$ with mean $c$.

Your generalization of Jensen's inequality follows immediately from the well-known fact that any probability distribution $\mu$ on $\mathbb R$ with a given mean $c\in\mathbb R$ is a mixture of probability distributions of the form $\mu_{c;u,v}$. See e.g. formula (2.13).

Details: Indeed, that formula implies $$Ef(X)=\int_{S_c} Ef(X_{u,v})\,\nu_X(du\times dv)$$ for some probability measure $\nu_X$ (depending on the distribution of $X$) on the set $S_c:=\{(u,v)\in\mathbb R^2\colon u\le c\le v\}$ and all functions $f\colon\mathbb R\to\mathbb R$ such the function $\mathbb R\ni x\mapsto f(x)-kx$ is bounded from below for some some real $k$.

Now, if $f$ is convex at $c$, then $Ef(X_{u,v})\ge f(EX_{u,v})=f(c)$ for all $(u,v)\in S_c$, and hence $Ef(X)\ge f(c)$.

Answer to your additional question concerning the strict convexity at $c$: Moreover, if $f$ is strictly convex at $c$, then $Ef(X_{u,v})> f(EX_{u,v})=f(c)$ for all $(u,v)$ in the set, say $S_c^\circ$, of all $(u,v)\in S_c$ such that $u<c<v$. Hence, $Ef(X)>f(c)$ unless $\nu(S_c^\circ)=0$. On the other hand, the condition $(u,v)\in S_c\setminus S_c^\circ$ implies that $Eg(X_{u,v})=g(c)$ for all functions $g\colon\mathbb R\to\mathbb R$. So, the condition $Ef(X)=f(c)$ implies $\nu_X(S_c^\circ)=0$, which in turn implies that $Eg(X)=\int_{S_c\setminus S_c^\circ} g(c)\,\nu_X(du\times dv)=g(c)$ for all (say) nonnegative $g\colon\mathbb R\to\mathbb R$, which means that $P(X=c)=1$.