13.1 Semi continuity[2CV]

Let \((X,𝜏)\) be a topological space.

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[139] Let \(f : ℝ → ℝ\) be defined as \(f (x) = 1\) if \(x ∈ ℝ ⧵ ℚ\), \(f(0)=0\), and \(f (x) = 1/q\) if \(|x| = p/q\) with \(p, q\) coprime integers, \(q\ge 1\). Show that f is continuous on \(ℝ ⧵ ℚ\) and discontinuous in every \(t ∈ ℚ\).

Show that the described function is u.s.c. Hidden solution: [UNACCESSIBLE UUID ’13B’]

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[13C]Prerequisites:1.

Construct a monotonic function with the same property as the one seen in the exercise 1.

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[13D]Let \(f:X→ ℝ\); the following assertions are equivalent.

1. \(f\) is lower semicontinuous.

2. For every \(t\), we have that the sublevel

   \[ S_ t = \{ x∈ X, f(x)≤ t \} \]

   is closed.

3. The epigraph

   \[ E = \{ (x,t)∈ X×ℝ, f(x)≤ t \} \]

   is closed in \(X× ℝ\).

Note that the second condition means that \(f\) is continuous from \((X,𝜏)\) to \(ℝ,𝜏_+\) where \(𝜏_+=\{ (a,∞):a∈ℝ\} ∪\{ ∅,ℝ\} \) is the set of half-lines, which is a topology (easy verification).

Then formulate the equivalent theorem for functions upper semicontinuous.

Hidden solution: [UNACCESSIBLE UUID ’13F’]

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[13G]If \(f,g:X→ ℝ\) are lower semicontinuous, then \(f+g\) is l.s.c. Hidden solution: [UNACCESSIBLE UUID ’13H’]

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[13J]Let \(I\) be a family of indices. Supposte that, for \(n∈ I\), \(f_ n:X→ℝ\) are l.s.c. functions. We define \(f{\stackrel{.}{=}}\sup _{n∈ I}f_ n\), then \(f\) is l.s.c. (defined as \(f:X→ℝ∪\{ +∞\} \)). ¹ . Hidden solution: [UNACCESSIBLE UUID ’13K’]

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[13M]Vice versa, given \(f:ℝ→ℝ∪\{ +∞\} \) l.s.c., there exists an increasing sequence of continuous functions \(f_ n:ℝ→ℝ\) such that \(f_ n(x)→_ n f(x)\). Hidden solution: [UNACCESSIBLE UUID ’13N’]

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[13P]Topics:inf-convolution. Difficulty:*. When \((X,d)\) is a metric space, and \(f:X→ℝ∪\{ +∞\} \) is l.s.c. and bounded from below, let

\[ f_ n(x) {\stackrel{.}{=}}\inf _{y∈ X} \{ f(y) + n d(x,y) \} \]

be the inf-convolution. Show that the sequence \(f_ n\) is an increasing sequence of Lipschitz functions with \(f_ n(x)→_ n f(x)\). Hidden solution: [UNACCESSIBLE UUID ’13Q’]

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[13R]Given \(f:X→ℝ\), define

\[ f^{*}(x)=f(x)∨ \limsup _{y→ x} f(y) \quad ; \]

show that \(f^{*}(x)\) is the smallest upper semicontinuous function that is greater than or equal to \(f\) at each point.

Similarly, define

\[ f_{*}(x)=f(x)∧ \liminf _{y→ x} f(y) \]

then \(-(f^{*})=(- f)_{*}\), and therefore \(f_{*}(x)\) is the greatest lower semicontinuous function that is less than or equal to \(f\) at each point.

Finally, note that \(f^*≥ f_*\).

Hidden solution: [UNACCESSIBLE UUID ’13S’]

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[13T]Topics:oscillation.

Given any \(f:X→ℝ\), we define oscillation function \({\operatorname {osc}}(f)\)

\[ {\operatorname {osc}}(f) (x) {\stackrel{.}{=}}f^{*}(x)-f_{*}(x) \]

1. Note that \({\operatorname {osc}}(f)≥ 0\), and that \(f\) is continuous in \(x\) if and only if \({\operatorname {osc}}(f)(x)=0\).

2. Show that \({\operatorname {osc}}(f)\) is upper semicontinuous.

3. If \((X,d)\) is a metric space, note that

   \[ {\operatorname {osc}}(f) (x) {\stackrel{.}{=}}\lim _{\varepsilon → 0+} \sup \{ |f(y) - f(z)| ~ ,~ d(x,y){\lt}\varepsilon ,d(x,z){\lt}\varepsilon \} \quad . \]

Hidden solution: [UNACCESSIBLE UUID ’13V’]

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[13W]Let \((X,𝜏)\) be a topological space and \(f:X→ℝ\) a function. Let \(\overline x∈ X\) be an accumulation point. Let eventually \(U_ n\) be a family of open neighbourhoods of \(\overline x\) with \(U_ n⊇ U_{n+1}\). Then there exists a sequence \((x_ n)⊂ X\) with \(x_ n∈ U_ n\) and \(x_ n≠ \overline x\) and such that

\[ \lim _{n→∞}f(x_ n)=\liminf _{x→ \overline x}f(x)~ ~ . \]

(Note that in general we do not claim neither expect that \(x_ n→\overline x\)). Hidden solution: [UNACCESSIBLE UUID ’13X’]

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[13Y] Let \((X,𝜏)\) be a topological space and \(f:X→ℝ\) a function; let \(\overline x∈ X\) be an accumulation point; let \(A\) be the set of all the limits \(\lim _ n f(x_ n)\) (when they exist) for all sequences \((x_ n)⊂ X\) such that \(x_ n→ \overline x\); then

\[ \liminf _{x→ \overline x}f(x)≤ \inf A~ ~ ; \]

moreover, if \((X,𝜏)\) satisfies the first axiom of countability, then equality holds and \(\inf A=\min A\).

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[13Z]Let \(f_ 1:[0,∞]→ [0,∞]\) monotonic function (weakly increasing) and right continuous. Let then \(f_ 2:[0,∞)→[0,∞]\) be given by

\[ f_ 2 (s) = \sup \{ t≥ 0 : f_ 1 (t) {\gt} s\} \]

(with the convention that \(\sup ∅ =0\)) and then again \(f_ 3:[0,∞)→[0,∞]\) defined by

\[ f_ 3 (s) = \sup \{ t≥ 0 : f_ 2 (t) {\gt} s\} \quad : \]

then \(f_ 1≡ f_ 3\).

Hidden solution: [UNACCESSIBLE UUID ’140’]