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This question arises from a paper I read.

Let $f$ be a measurable function on $\mathbb{R}^n$. The essential supremum of $f$ is defined by,

$\operatorname{ess}\sup _{x \in \mathbb{R}^n} f(x) = \inf \{t : f(x) \leq t \text{ for almost all } x \in \mathbb{R}^n\}$.

Now, let $0 < \lambda <1$ and let $f,g \in L^1 (\mathbb{R})$ be non-negative. Let,

$s(x) = \operatorname{ess}\sup f\left(\dfrac{x-y}{1- \lambda}\right)^{1- \lambda} g\left(\dfrac{y}{\lambda}\right)^ \lambda$ (over all $y \in \mathbb{R}^n$) .

Then, $s$ can be written as,

$$s(x) = \sup_{\phi \in D} \int_{\mathbb{R^n}} f\left(\dfrac{x-y}{1- \lambda}\right)^{1- \lambda} g\left(\dfrac{y}{\lambda}\right)^ \lambda \phi(y) dy$$,

Where $D$ is a countable dense subset of the unit ball of $L^1 (\mathbb{R}^n)$.

Can someone explain how $s$ is written in terms of an integral. In the paper, the author has trivialized this but I don't get the idea behind it. I assume this is some sort of a standard representation of the essential supremum of a measurable function on $\mathbb{R}^n$.

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1 Answer 1

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If you take $\phi$ in the unit ball of $L^1(\mathbb{R}^n)$ in the supremum, then this equality is a consequence of the isometric isomorphism between the dual of $L^1$ and $L^\infty$. That you can pass to a dense subset is a standard argument: The supremum of a continuous function on a dense subset is equal to the supremum on the entire space.

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