13.3 Sup transform[2CR]

Definition 349

[2CS]Suppose that either \(I=ℝ^+\) or \(I=ℝ\) in the following, for simplicity.

Let \(\varepsilon {\gt}0\); given a bounded function \(f:I→ℝ\) 1  , we define the ”sup transform” as the function \(g:I→ℝ\) given by

\begin{equation} \label{eq:trasf_ sup} g(x)=\sup _{y∈ (x,x+\varepsilon )} f(y)~ ~ . \end{equation}
350

We summarize this transformation with the notation \(g=F(\varepsilon ,f)\).

E350

[145]Prerequisites:349.Show that \(g\) is regulated.

E350

[146]Prerequisites:349.Show that \(g\) is lower semicontinuous.

E350

[147]Prerequisites:349.Show that \(f=g\) if and only if \(f\) is monotonic weakly decreasing and right continuous.

E350

[148]Prerequisites:349.Given

\[ g(x)= \begin{cases} -1 & x=4\\ 0 & x≠ 4 \end{cases} \]

find \(f\) such that \(g=F(1,f)\).

Hidden solution: [UNACCESSIBLE UUID ’149’]

E350

[14B]Prerequisites:349.Show that if \(f\) is continuous then \(g\) is continuous.

Hidden solution: [UNACCESSIBLE UUID ’14C’]

E350

[14D]Prerequisites:349,2.Let \(C_ b=C_ b(I)\) be the space of continuous bounded functions \(f:I:\to {\mathbb {R}})\). This is a Banach space (a complete normed space) with the norm \(\| f\| _∞=\sup _ x |f(x)|\).

Consider the map \(F:[0,∞)× C_ b→ C_ b\) transforming \(g=F(\varepsilon ,f)\), as defined in the eqn. ??.

Show that \(F\) is continuous.

E350

[14F]Prerequisites:349.How do previous exercises change if you define instead

\begin{equation} \label{eq:trasf_ sup_ aperto} g(x)=\sup _{y∈ [x,x+\varepsilon ]} f(y)~ ~ ? \end{equation}
353

Hidden solution: [UNACCESSIBLE UUID ’14G’]

[UNACCESSIBLE UUID ’14H’]

  1. The ”bounded” hypothesis is convenient, the following resulst are valid even without this hypothesis, with simple modifications.