Let $(X,A,\mu)$ be a set, a $\sigma$-algebra and a measure. Suppose that $\mu(X) = 1$. Let $u : X \rightarrow {\mathbb{R}}$ and $f : \mathbb{R} \rightarrow \mathbb{R} \, \cup \, \{+\infty \} $ be an integrable function and a convex, lower semi-continous function respectively. Then $$\int_{X} f \circ u \, d\mu \geq f \left( \int_{X} u \, d\mu \right)$$ This is the version of Jensen's Inequality I'm working with (note that convex functions with values in $\mathbb{R} \cup {+\infty}$ are not automatically continous and that LSC is required in order to have the lower affine approximants required to prove the theorem). I'd like to find a counterexample to this inequality when $f$ is convex but not necessarily lower semi-continous. I expected it to be quite easy to find, but the fact that $E = \{ x \in \mathbb{R} | f(x) < +\infty \}$ has to be an interval and that $u(x)$ has to be in $E$ for almost every $x \in X$ for the left-hand side of the equation not to be $+\infty$ is kinda annoying!
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Do you have reference for the version of Jensen's inequality in your OP. Lower semicontinuity seems to be irrelevant in general. Here is one version where lac does not hold. Suppose $a,b$ are real numbers $a<b$ and $\phi:(a.b)\rightarrow\mathbb{R}$ convex. Extend $\phi$ to $\mathbb{R}$ as $\infty$ outside $(a,b)$. If $X\in L_1(P)$ and $a<X<b$, then $\phi(\int X ,dP)\leq \int\phi\circ X,dP$. $\phi$ is still convex bu not lsc. – Mittens Jan 15 '22 at 22:01
2 Answers
It seems to me the hypothesis of lower semi-continuity is not needed.
The left side of your inequality is well defined, though perhaps equal to $+\infty$. In this case the inequality holds trivially. If $\int_X f\circ u \,d\mu<\infty$, then $\mu(\{x\in X: u(x)\in\{f=+\infty\}\})=0$. Let $G$ denote $\{x\in X: u(x)\in\{f<+\infty\}\}$. Let $g$ be $f$ redefined (if necessary) at the finite endpoints of $\{f<+\infty\}$ so as to be lsc; that is, if $b$ is such an endpoint then $g(b):=\liminf_{t\to b} f(t)$. Then $$ \int_X f\circ u \, d\mu=\int_G f\circ u \, d\mu=\int_G g\circ u \, d\mu\ge g\left(\int_G u \, d\mu\right)=g\left(\int_X u \, d\mu\right)=f\left(\int_X u \, d\mu\right). $$ Here the inequality follows from the lsc form of Jensen that you noted, and the final equality is true because $\int_X u \, d\mu\in \{f<+\infty\}$.
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Sorry, I am not sure this approach is quite valid, because the set of values such that ${f<+\infty}$ can include the endpoints, and therefore include the set of points where $g\neq f$ (so the final equality is not necessarily true). Concretely, suppose $f:\mathbb{R}\to\mathbb{R}\cup{\infty}$ is defined by $f(x)=0$ on the open interval $(a,b)$, $f(a)=f(b)=1$, and $f(x)=\infty$ otherwise. In that case if we just let $\mu$ be some measure such that $\int_X u d\mu = b$, then $g\left(\int_X u d\mu\right) \neq f\left(\int_X u d\mu\right)$. – helloworld Jun 24 '23 at 21:03
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(And also we can pick that measure in that example while ensuring the left side of the inequality is less than $+\infty$, e.g. using just a point measure on $b$. Though on an informal level, I think that in some sense this suggests it suffices to just do a special-case analysis for the scenarios where $\int_X u , d\mu$ is a boundary point but $\int_X f \circ u , d\mu$ is finite.) – helloworld Jun 24 '23 at 21:11
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Agreed. The case in which $\int_X u,d\mu$ an endpoint of ${f<+\infty}$ can be dealt with separately. In any event, the lower semi-continuity hypothesis is unnecessary. – John Dawkins Jun 26 '23 at 15:14
The discussion in the comments of John Dawkins' answer suggests that there won't be a counterexample when the convex function $f$ is defined on $\mathbb{R}$, and I think it would be possible to use an inductive argument to show there are also no counterexamples when it is defined on any finite-dimensional space. But the following seems to give a counterexample when $f$ is defined on the sequence space $\ell^2$, building off the construction in this answer: define $f:\ell^2\to\mathbb{R}\cup\{+\infty\}$ via $f(x)=0$ if $x$ only has finitely many non-zero terms, and $f(x)=+\infty$ otherwise. I believe is indeed a valid convex function (for any $x=tx_1 + (1-t)x_2$, do a case analysis on whether $x_1,x_2$ have finitely many non-zero terms to see that $f(x)\leq tf(x_1) + (1-t)f(x_2)$), though it is not lsc. Now if you consider the point $\sum_k \frac{1}{2^k} e_k$ where $e_k$ are the canonical basis vectors (note that this means we only need to use a discrete measure for this counterexample), we have
$$f\left(\sum_k \frac{1}{2^k} e_k\right) = +\infty \quad \text{since $\sum_k \frac{1}{2^k} e_k$ has infinitely many non-zero terms},$$ $$\sum_k \frac{1}{2^k} f\left(e_k\right) = \sum_k \frac{1}{2^k} (0) = 0,$$
and hence Jensen's inequality does not hold for this point. (Basically this counterexample arises from having the "infinite convex combination" converge to a point in the closure of $\operatorname{dom}(f)$ that is not in $\operatorname{dom}(f)$ itself.)
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