3. LIMITS AND CONTINUITY
3.7 LOWER SEMICONTINUITY AND UPPER SEMICONTINUITY
The concept of semicontinuity is convenient for the study of maxima and minima of some discontinuous functions.
Definition 3.7.1 Let f:D→Rand let ¯x∈D. We say that f islower semicontinuous(l.s.c.) at ¯xif for everyε>0, there existsδ >0 such that
f(x¯)−ε< f(x)for allx∈B(x¯;δ)∩D. (3.12)
Similarly, we say that f isupper semicontinuous(u.s.c.) at ¯xif for everyε>0, there existsδ>0
such that
f(x)< f(x¯) +ε for allx∈B(x¯;δ)∩D.
It is clear that f is continuous at ¯xif and only if f is lower semicontinuous and upper semicontin- uous at this point.
Figure 3.6: Lower semicontinuity.
Figure 3.7: Upper semicontinuity.
Theorem 3.7.1 Let f:D→Rand let ¯x∈Dbe a limit point ofD. Then f is lower semicontinuous at ¯xif and only if
lim inf
x→x¯ f(x)≥f(x¯).
Similarly, f is upper semicontinuous at ¯xif and only if lim sup
x→x¯
f(x)≤ f(x¯).
Proof:Suppose f is lower semicontinuous at ¯x. Letε >0. Then there existsδ0>0 such that
f(x¯)−ε< f(x)for allx∈B(x¯;δ0)∩D. This implies f(x¯)−ε≤h(δ0), where h(δ) = inf x∈B0(¯x;δ)∩Df(x). Thus, lim inf x→x¯ f(x) =sup δ>0 h(δ)≥h(δ0)≥ f(x¯)−ε.
96 3.7 LOWER SEMICONTINUITY AND UPPER SEMICONTINUITY
Sinceεis arbitrary, we obtain lim infx→x¯f(x)≥ f(x¯).
We now prove the converse. Suppose lim inf
x→x¯ f(x) =sup
δ>0
h(δ)≥ f(x¯)
and letε>0. Since sup
δ>0
h(δ)> f(x¯)−ε,
there existsδ >0 such thath(δ)>f(x¯)−ε. This implies
f(x)> f(x¯)−ε for allx∈B0(x¯;δ)∩D.
Since this is also true forx=x¯, the function f is lower semicontinuous at ¯x.
The proof for the upper semicontinuous case is similar.
Theorem 3.7.2 Let f:D→Rand let ¯x∈D. Then f is l.s.c. at ¯xif and only if for every sequence
{xk}inDthat converges to ¯x, lim inf
k→∞
f(xk)≥ f(x¯).
Similarly, f is u.s.c. at ¯xif and only if for every sequence{xk}inDthat converges to ¯x, lim sup
k→∞
f(xk)≤ f(x¯).
Proof: Suppose f is l.s.c. at ¯x. Then for anyε >0, there existsδ >0 such that (3.12) holds. Since
{xk}converges to ¯x, we havexk∈B(x¯;δ)whenkis sufficiently large. Thus,
f(x¯)−ε< f(xk)
for suchk. It follows that f(x¯)−ε≤lim infk→∞f(xk). Sinceε is arbitrary, it follows that f(x¯)≤ lim infk→∞f(xk).
We now prove the converse. Suppose lim infk→∞f(xk)≥ f(x¯)and assume, by way of contra-
diction, that f is not l.s.c. at ¯x. Then there exists ¯ε>0 such that for everyδ >0, there exists
xδ ∈B(x¯;δ)∩Dwith
f(x¯)−ε¯ ≥ f(xδ).
Applying this forδk=
1
k, we obtain a sequence{xk}inDthat converges to ¯xwith f(x¯)−ε¯ ≥ f(xk)for every k. This implies f(x¯)−ε¯ ≥lim inf k→∞ f(xk). This is a contradiction.
Definition 3.7.2 Let f:D→R. We say that f islower semicontinuous on D(or lower semicontinu-
ous if no confusion occurs) if it is lower semicontinuous at every point ofD.
Theorem 3.7.3 SupposeDis a compact set ofRand f:D→Ris lower semicontinuous. Then f
has an absolute minimum onD. That means there exists ¯x∈Dsuch that
f(x)≥ f(x¯)for allx∈D.
Proof:We first prove that f is bounded below. Suppose by contradiction that for everyk∈N, there
existsxk∈Dsuch that
f(xk)<−k.
SinceDis compact, there exists a subsequence{xk`}of{xk}that converges tox0∈D. Since f is
l.s.c., by Theorem3.7.2
lim inf
`→∞
f(xk`)≥ f(x0).
This is a contraction because lim inf`→∞f(xk`) =−∞. This shows f is bounded below. Define
γ=inf{f(x):x∈D}.
Since the set{f(x):x∈D}is nonempty and bounded below,γ∈R.
Let{uk}be a sequence inDsuch that{f(uk)}converges toγ. By the compactness ofD, the
sequence{uk}has a convergent subsequence{uk`}that converges to some ¯x∈D. Then
γ=lim
`→∞
f(uk`) =lim inf
`→∞
f(uk`)≥ f(x¯)≥γ. This impliesγ= f(x¯)and, hence,
f(x)≥ f(x¯)for allx∈D.
The proof is now complete.
The following theorem is proved similarly.
Theorem 3.7.4 SupposeDis a compact subset ofRand f:D→Ris upper semicontinuous. Then
f has an absolute maximum onD. That is, there exists ¯x∈Dsuch that
f(x)≤ f(x¯)for allx∈D.
For everya∈R, define
La(f) ={x∈D: f(x)≤a}
and
Ua(f) ={x∈D: f(x)≥a}.
Theorem 3.7.5 Let f:D→R. Then f is lower semicontinuous if and only ifLa(f)is closed inD
for everya∈R. Similarly, f is upper semicontinuous if and only ifUa(f)is closed inDfor every
98 3.7 LOWER SEMICONTINUITY AND UPPER SEMICONTINUITY
Proof: Suppose f is lower semicontinuous. Using Corollary2.6.10, we will prove that for every
sequence {xk} inLa(f) that converges to a point ¯x∈D, we get ¯x∈La(f). For everyk, since
xk∈La(f), f(xk)≤a.
Since f is lower semicontinuous at ¯x,
f(x¯)≤lim inf k→∞
f(xk)≤a.
Thus, ¯x∈La(f). It follows thatLa(f)is closed.
We now prove the converse. Fix any ¯x∈Dandε>0. Then the set
G={x∈D: f(x)> f(x¯)−ε}=D\Lf(¯x)−ε(f)
is open inDand ¯x∈G. Thus, there existsδ>0 such that
B(x¯;δ)∩D⊂G.
It follows that
f(x¯)−ε< f(x)for allx∈B(x¯;δ)∩D.
Therefore, f is lower semicontinuous. The proof for the upper semicontinuous case is similar.
For everya∈R, we also define
La(f) ={x∈D: f(x)<a} and
Ua(f) ={x∈D: f(x)>a}.
Corollary 3.7.6 Let f:D→R. Then f is lower semicontinuous if and only ifUa(f)is open inD
for everya∈R. Similarly, f is upper semicontinuous if and only ifLa(f)is open inDfor every
a∈R.
Theorem 3.7.7 Let f: D→R. Then f is continuous if and only if for everya,b∈Rwitha<b, the set
Oa,b={x∈D:a< f(x)<b}=f−1((a,b)) is an open set inD.
Proof: Suppose f is continuous. Then f is lower semicontinuous and upper semicontinuos. Fix
a,b∈Rwitha<b. Then
Oa,b=Lb∩Ua.
By Theorem3.7.6, the setOa,bis open since it is the intersection of two open setsLaandUb.
Let us prove the converse. We will only show that f is lower semicontinuous since the proof of
upper semicontinuity is similar. For everya∈R, we have
Ua(f) ={x∈D: f(x)>a}=∪n∈Nf−1((a,a+n))
Thus,Ua(f)is open inDas it is a union of open sets inD. Therefore, f is lower semicontinuous by
Exercises
3.7.1 Let f be the function given by
f(x) =
(
x2, if x6=0;
−1, if x=0.
Prove that f is lower semicontinuous.
3.7.2 Let f be the function given by
f(x) =
(
x2, if x6=0;
1, if x=0.
Prove that f is upper semicontinuous.
3.7.3 Let f,g:D→Rbe lower semicontinuous functions and letk>0 be a constant. Prove that
f+gandk f are lower semicontinous functions onD.
3.7.4 ILet f: R→Rbe a lower semicontinuous function such that
lim
x→∞f(x) =x→−lim∞f(x) =∞.
THE MEAN VALUE THEOREM
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TAYLOR’S THEOREM
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