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Setup

Suppose that $U$ is a subset of $L^{\infty}_{\mu}(\mathscr{F})\cap L^1_{\mu}(\mathscr{F})$ defined by $$ f\in U \Leftrightarrow g(f(x))\leq M \mbox{ and } f \in L^{\infty}_{\mu}(\mathscr{F})\cap L^1_{\mu}(\mathscr{F}) , $$ for a fixed convex and lower-semicontinuous function $g$ . Here $\mu$ is a (if acceptable $\sigma$-)finite measure on $\mathcal{X}$.

Question

Under what conditions can the interchanging of the $$ \inf_{f \in U}\int_{x \in \mathcal{X}}f^2(x)\mu(dx) = \int_{x \in \mathcal{X}}\inf_{f \in U}f^2(x)\mu(dx). $$ In other words....when can the infimum be taken pointwise?

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  • $\begingroup$ The usual approach would be to try to find a sequence $f_n$ converging a.e. to the infimum, and then try to use something like the dominated convergence theorem on the functions $f_n^2$. $\endgroup$
    – Nate Eldredge
    Commented Oct 13, 2018 at 16:01
  • $\begingroup$ That's interesting....I'll give that some though, at the moment I'm looking at this book: sites.math.washington.edu/~rtr/papers/… By Rockafellar and Wets. The last chapter may be able to help :) $\endgroup$
    – ABIM
    Commented Oct 13, 2018 at 16:30
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    $\begingroup$ If would be helpful if you could assure that the integrand on the right hand side, i.e. the function $g(x) = \inf_{f\in U} f^2(x)$, is in $U$, in other words, that $U$ is closed under taking infima. If not, I would suspect that the equality is not true in general. $\endgroup$
    – Dirk
    Commented Oct 16, 2018 at 12:17
  • $\begingroup$ Just noticed something strange: If $g$ in your question is convex and $f$ is real valued, then $g(f(x))\leq M$ just says that $f(x)$ is in some closed interval. Is it that what you mean? If yes, then the infimum is indeed attained pointwise and $f^2(x)$ should be equal to the smallest possible value everywhere. $\endgroup$
    – Dirk
    Commented Oct 16, 2018 at 12:22

1 Answer 1

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If $U$ is non-empty then the value of either side is $m \mu(\mathcal{X})$ where $m= \inf \{y^2 ~|~ g(y)<M\}$. If $\mu$ is non-finite then $m=0$.

To see this note first that for $f_1,f_2 \in U$ the function $f$ defined by $$ f(x)=\begin{cases} f_{1}(x) & f_{1}^{2}(x)\le f_{2}^{2}(x)\\ f_{2}(x) & f_{2}^{2}(x)<f_{1}^{2}(x). \end{cases} $$ is in $U$ as well. Also note $\int f^2 d\mu \le \int f_i^2d\mu$.

A second observation: if $U$ is non-empty then then either $f_{min} \equiv \sqrt{m}$ or $f_{min} \equiv -\sqrt{m}$ is in $U$ as $m \le f^2$ for all $f$ and either $g(\sqrt{m}) \le M$ or $g(-\sqrt{m}) \le M$ by definition of $m$. Note also $f_{min}^2(x) = \inf_{f\in U} f^2(x)$.

Combining those two facts one gets $$\inf_{f\in U} \int f^2 d\mu = \int \inf_{f\in U} f^2 d\mu = \int f_{min}^2 d\mu = m \mu(\mathcal{X}).$$

Side remark: This does not need any assumptions on $g$ except beeing somewhere finite-valued and $g(0) \le M$ for non-finite measures.

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