# Convexity at a point and Jensen inequality

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).

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 $$u. Hence, $$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$$.

• Thank you for this answer. Can you please elaborate on how proposition 3.18 implies the required form of Jensen inequality? I am having a bit of trouble deducing it. BTW, this approach seems to be a bit of an overkill, isn't it? (it seems less elementary than just proving the existence of a supporting line. I hoped there would be a more elementary reference, say some book on convex analysis, rather than a paper from 2009. I am quite sure this version of Jensen was known long before that...). – Asaf Shachar Jul 9 at 5:24
• Anyway, I am interested in understanding better this more general approach, so if you could elaborate more on how it implies Jensen, it would be great. – Asaf Shachar Jul 9 at 5:24
• Oh, and last question: Does the approach you have described imply that if $\phi$ is strictly convex at $c$, and there is an equality in Jensen, then the random variable is constant a.e.? – Asaf Shachar Jul 9 at 8:39
• @AsafShachar : I have added the details you requested, and even an (affirmative answer) to your additional question concerning the strict convexity at $c$. – Iosif Pinelis Jul 9 at 14:27
• @WillieWong : This is so. Indeed, consider first the probability measures with a compact support. For any compact $K\subset\mathbb R^n$, let $P_{K,c}$ be the set of all probability measures with support $\subseteq K$ and mean $c$. Then $P_{K,c}$ is convex and compact and hence, by the Choquet–Bishop–de Leeuw theorem, any point in $P_{K,c}$ is a mixture of extreme points of $P_{K,c}$, which latter are probability measures with support of cardinality $\le n+1$. It remains to note that any probability measure on $\mathbb R^n$ is a mixture of probability measures with a compact support. – Iosif Pinelis Jul 9 at 21:12