Show That a Function is Continuous at X0 Iff It is Finite Lsc and Usc There

3.7: Lower Semicontinuity and Upper Semicontinuity

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    • Portland State University via PDXOpen: Open Educational Resources

    The concept of semicontinuity is convenient for the study of maxima and minima of some discontinuous functions.

    Definition \(\PageIndex{1}\)

    Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x} \in D\). We say that \(f\) is lower semicontinuous (l.s.c.) at \(\bar{x}\) if for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that

    \[f(\bar{x})-\varepsilon<f(x) \text { for all } x \in B(\bar{x} ; \delta) \cap D .\]

    Similarly, we say that \(f\) is upper semicontinuous (e.s.c.) at \(\bar{x}\) if for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that

    \[f(x)<f(\bar{x})+\varepsilon \text { for all } x \in B(\bar{x} ; \delta) \cap D .\]

    It is clear that \(f\) is continuous at \(\bar{x}\) if and only if \(f\) is lower semicontinuous and upper semicontinuous at this point.

    Annotation 2020-08-30 212221.png

    Figure \(3.6\): Lower semicontinuity.

    Annotation 2020-08-30 212324.png

    Figure \(3.7\): Upper semicontinuity.

    Theorem \(\PageIndex{1}\)

    Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x} \in D\) be a limit point of \(D\). Then \(f\) is lower semicontinuous at \(\bar{x}\) if and only if

    \[\liminf _{x \rightarrow \bar{x}} f(x) \geq f(\bar{x}) .\]

    Similarly, \(f\) is upper semicontinuous at \(\bar{x}\) if and only if

    \[\limsup _{x \rightarrow \bar{x}} f(x) \leq f(\bar{x}) .\]

    Proof

    Suppose \(f\) is lower semiconitnuous at \(\bar{x}\). Let \(\varepsilon > 0\). Then there exists \(\delta_{0}>0\) such that

    \[f(\bar{x})-\varepsilon<f(x) \text { for all } x \in B\left(\bar{x} ; \delta_{0}\right) \cap D .\]

    This implies

    \[f(\bar{x})-\varepsilon \leq h\left(\delta_{0}\right),\]

    where

    \[h(\delta)=\inf _{x \in B_{0}(\bar{x} ; \delta) \cap D} f(x) .\]

    Thus,

    \[\liminf _{x \rightarrow \bar{x}} f(x)=\sup _{\delta>0} h(\delta) \geq h\left(\delta_{0}\right) \geq f(\bar{x})-\varepsilon .\]

    Since \(\varepsilon\) is arbitrary, we obtain \(\liminf _{x \rightarrow \bar{x}} f(x) \geq f(\bar{x})\).

    We now prove the converse. Suppose

    \[\liminf _{x \rightarrow \bar{x}} f(x)=\sup _{\delta>0} h(\delta) \geq f(\bar{x})\]

    and let \(\varepsilon > 0\). Since

    \[\sup _{\delta>0} h(\delta)>f(\bar{x})-\varepsilon ,\]

    there exists \(\delta > 0\) such that \(h(\delta)>f(\bar{x})-\varepsilon\). This implies

    \[f(x)>f(\bar{x})-\varepsilon \text { for all } x \in B_{0}(\bar{x} ; \delta) \cap D .\]

    Since this is also true for \(x = \bar{x}\), the function \(f\) is lower semicontinuous at \(\bar{x}\).

    The proof for the upper semicontinuous case is similar. \(\square\)

    Theorem \(\PageIndex{2}\)

    Let \(f: D \rightarrow \mathbb{R}\) and let \(\bar{x} \in D\). Then \(f\) is l.s.c. at \(\bar{x}\) if and only if for every sequence \(\left\{x_{k}\right\}\) in \(D\) that converges to \(\bar{x}\),

    \[\liminf _{k \rightarrow \infty} f\left(x_{k}\right) \geq f(\bar{x}) .\]

    Similarly, \(f\) is u.s.c. at \(\bar{x}\) if and only if for every sequence \(\left\{x_{k}\right\}\) in \(D\) that converges to \(\bar{x}\),

    \[\limsup _{k \rightarrow \infty} f\left(x_{k}\right) \leq f(\bar{x}) .\]

    Proof

    Suppose \(f\) is l.s.c. at \(\bar{x}\). Then for any \(\varepsilon > 0\), there exists \(\delta > 0\) such that (3.12) holds. Since \(\left\{x_{k}\right\}\) converges to \(\bar{x}\), we have \(x_{k} \in B(\bar{x} ; \delta)\) when \(k\) is sufficiently large. Thus,

    \[f(\bar{x})-\varepsilon<f\left(x_{k}\right)\]

    for such \(k\). It follows that \(f(\bar{x})-\varepsilon \leq \liminf _{k \rightarrow \infty} f\left(x_{k}\right)\). Since \(\varepsilon\) is arbitrary, it follows that \(f(\bar{x}) \leq \liminf _{k \rightarrow \infty} f\left(x_{k}\right)\).

    We now prove the converse. Suppose \(\liminf _{k \rightarrow \infty} f\left(x_{k}\right) \geq f(\bar{x})\) and assume, by way of contradiction, that \(f\) is not l.s.c. at \(\bar{x}\). Then there exists \(\bar{\varepsilon}>0\) such that for every \(\delta > 0\), there exists \(x_{\delta} \in B(\bar{x} ; \delta) \cap D\) with

    \[f(\bar{x})-\bar{\varepsilon} \geq f\left(x_{\delta}\right) .\]

    Applying this for \(\delta_{k}=\frac{1}{k}\), we obtain a sequence \(\left\{x_{k}\right\}\) in \(D\) that converges to \(\bar{x}\) with

    \[f(\bar{x})-\bar{\varepsilon} \geq f\left(x_{k}\right) \text { for every } \mathrm{k} .\]

    This implies

    \[f(\bar{x})-\bar{\varepsilon} \geq \liminf _{k \rightarrow \infty} f\left(x_{k}\right) .\]

    This is a contradiction. \(\square\)

    Definition \(\PageIndex{2}\)

    Let \(f: D \rightarrow \mathbb{R}\). We say that \(f\) is lower semicontinuous on \(D\) (or lower semicontinuous if no confusion occurs) if it is lower semicontinuous at every point of \(D\).

    Theorem \(\PageIndex{3}\)

    Suppose \(D\) is a compact set of \(\mathbb{R}\) and \(f: D \rightarrow \mathbb{R}\) is lower semicontinuous. Then \(f\) has an absolute minimum on \(D\). That means there exists \(\bar{x} \in D\) such that

    \[f(x) \geq f(\bar{x}) \text { for all } x \in D .\]

    Proof

    We first prove that \(f\) is bounded below. Suppose by contradiction that for every \(k \in \mathbb{N}\), there exists \(x_{k} \in D\) such that

    \[f\left(x_{k}\right)<-k .\]

    Since \(D\) is compact, there exists a subsequence \(\left\{x_{k_{\ell}}\right\}\) of \(\left\{x_{k}\right\}\) that converges to \(x_{0} \in D\). Since \(f\) is l.s.c., by Theorem 3.7.2

    \[\liminf _{\ell \rightarrow \infty} f\left(x_{k_{\ell}}\right) \geq f\left(x_{0}\right) .\]

    This is a contraction because \(\liminf _{\ell \rightarrow \infty} f\left(x_{k_{\ell}}\right)=-\infty\). This shows \(f\) is bounded below. Define

    \[\gamma=\inf \{f(x): x \in D\} .\]

    Since the set \(\{f(x): x \in D\}\) is nonempty and bounded below, \(\gamma \in \mathbb{R}\).

    Let \(\left\{u_{k}\right\}\) be a sequence in \(D\) such that \(\left\{f\left(u_{k}\right)\right\}\) converges to \(\gamma\). By the compactness of \(D\), the sequence \(\left\{u_{k}\right\}\) has a convergent subsequence \(\left\{u_{k_{\ell}}\right\}\) that converges to some \(\bar{x} \in D\). Then

    \[\gamma=\lim _{\ell \rightarrow \infty} f\left(u_{k_{\ell}}\right)=\liminf _{\ell \rightarrow \infty} f\left(u_{k_{\ell}}\right) \geq f(\bar{x}) \geq \gamma .\]

    This implies \(\gamma=f(\bar{x})\) and, hence,

    \[f(x) \geq f(\bar{x}) \text { for all } x \in D .\]

    The proof is now complete. \(\square\)

    The following theorem is proved similarly.

    Theorem \(\PageIndex{4}\)

    Suppose \(D\) is a compact subset of \(\mathbb{R}\) and \(f: D \rightarrow \mathbb{R}\) is upper semicontinuous. Then \(f\) has an absolute maximum on \(D\). That is, there exists \(\bar{x} \in D\) such that

    \[f(x) \leq f(\bar{x}) \text { for all } x \in D .\]

    For every \(a \in \mathbb{R}\), define

    \[\mathscr{L}_{a}(f)=\{x \in D: f(x) \leq a\}\]

    and

    \[\mathscr{L}_{a}(f)=\{x \in D: f(x) \leq a\} .\]

    Proof

    Add proof here and it will automatically be hidden

    Theorem \(\PageIndex{5}\)

    Let \(f: D \rightarrow \mathbb{R}\). Then \(f\) is lower semicontinuous if and only if \(\mathscr{L}_{a}(f)\) is closed in \(D\) for every \(a \in \mathbb{R}\). Similarly, \(f\) is upper semicontinuous if and only if \(\mathscr{U}_{a}(f)\) is closed in \(D\) for every \(a \in \mathbb{R}\).

    Proof

    Suppose \(f\) is lower semicontinuous. Using Corollary 2.6.10, we will prove that for every sequence \(\left\{x_{k}\right\}\) in \(\mathscr{L}_{a}(f)\) that converges to a point \(\bar{x} \in D\), we get \(\bar{x} \in \mathscr{L}_{a}(f)\). For every \(k\), since \(x_{k} \in \mathscr{L}_{a}(f)\), \(f\left(x_{k}\right) \leq a\).

    Since \(f\) is lower semicontinuous at \(\bar{x}\),

    \[f(\bar{x}) \leq \liminf _{k \rightarrow \infty} f\left(x_{k}\right) \leq a .\]

    Thus, \(\bar{x} \in \mathscr{L}_{a}(f)\). It follows that \(\mathscr{L}_{a}(f)\) is closed.

    We now prove the converse. Fix any \(\bar{x} \in D\) and \(\varepsilon > 0\). Then the set

    \[G=\{x \in D: f(x)>f(\bar{x})-\varepsilon\}=D \backslash \mathscr{L}_{f(x)-\varepsilon(f)}\]

    is open in \(D\) and \(\bar{x} \in G\). Thus, there exists \(\delta > 0\) such that

    \[B(\bar{x} ; \delta) \cap D \subset G .\]

    It follows that

    \[f(\bar{x})-\varepsilon<f(x) \text { for all } x \in B(\bar{x} ; \delta) \cap D .\]

    Therefore, \(f\) is lower semicontinuous. The proof for the upper semicontinuous case is similar. \(\square\)

    For every \(a \in \mathbb{R}\), we also define

    \[L_{a}(f)=\{x \in D: f(x)<a\}\]

    and

    \[U_{a}(f)=\{x \in D: f(x)>a\} .\]

    Corollary \(\PageIndex{6}\)

    Let \(f: D \rightarrow \mathbb{R}\). Then \(f\) is lower semicontinuous if and only if \(U_{a}(f)\) is open in \(D\) for every \(a \in \mathbb{R}\). Similarly, \(f\) is upper semicontinuous if and only if \(L_{a}(f)\) is open in \(D\) for every \(a \in \mathbb{R}\).

    Proof

    Add proof here and it will automatically be hidden

    Theorem \(\PageIndex{7}\)

    Let \(f: D \rightarrow \mathbb{R}\). Then \(f\) is continuous if and only if for every \(a,b \in \mathbb{R}\) with \(a < b\). the set

    \[O_{a, b}=\{x \in D: a<f(x)<b\}=f^{-1}((a, b))\]

    is an open in \(D\).

    Proof

    Suppose \(f\) is continuous. Then \(f\) is lower semicontinuous and upper semicontinuous. Fix \(a,b \in \mathbb{R}\) with \(a < b\). Then

    \[O_{a, b}=L_{b} \cap U_{a} .\]

    By Theorem 3.7.6, the set \(O_{a, b}\) is open since it is the intersection of two opens sets \(L_{a}\) and \(U_{b}\).

    Let us prove the converse. We will only show that \(f\) is lower semicontinuous since the proof of upper semicontinuity is simiilar. For every \(a \in \mathbb{R}\), we have

    \[U_{a}(f)=\{x \in D: f(x)>a\}=\cup_{n \in \mathbb{N}} f^{-1}((a, a+n))\]

    Thus, \(U_{a}(f)\) is open in \(D\) as it is a union of open sets in \(D\). Therefore, \(f\) is lower semicontinuous by Corollary 3.7.6. \(\square\)

    Exercise \(\PageIndex{1}\)

    Let \(f\) be the function given by

    \[f(x)=\left\{\begin{array}{ll}
    x^{2}, & \text { if } x \neq 0 \text {;} \\
    -1, & \text { if } x=0 \text {.}
    \end{array}\right .\]

    Prove that \(f\) is lower semicontinuous.

    Answer

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    Exercise \(\PageIndex{2}\)

    Let \(f\) be the function given by

    \[f(x)=\left\{\begin{array}{ll}
    x^{2}, & \text { if } x \neq 0 \text {;} \\
    1, & \text { if } x=0 \text {.}
    \end{array}\right .\]

    Prove that \(f\) is upper semicontinuous.

    Answer

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    Exercise \(\PageIndex{3}\)

    Let \(f, g: D \rightarrow \mathbb{R}\) be lower semicontinuous functions and let \(k > 0\) be a constant. Prove that \(f + g\) and \(kf\) are lower semicontinuous functions on \(D\).

    Answer

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    Exercise \(\PageIndex{4}\)

    Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a lower semicontinuous function such that

    \[\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow-\infty} f(x)=\infty .\]

    Prove that \(f\) has an absolute minimum at some \(x_{0} \in \mathbb{R}\).

    Answer

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    Source: https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/03%3A_Limits_and_Continuity/3.07%3A_Lower_Semicontinuity_and_Upper_Semicontinuity

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