Properties of Liminf and Limsup
Keefer Rowan
January 31, 2021
∞
Let {an }∞
1 , {bn }1 ⊆ R.
Lemma. inf j∈N aj + bj ≥ inf n∈N an + inf k∈N bk .
Proof. Let j ∈ N . Then aj ≥ inf n∈N an , bj ≥ inf k∈N bk , thus aj + bj ≥ inf n∈N an + inf k∈N bk .
Thus inf j∈N aj + bj ≥ inf n∈N an + inf k∈N bk .
Lemma. supj∈N aj + bj ≤ supn∈N an + supk∈N bk .
Proof. Similarly.
Lemma. lim inf(an + bn ) ≥ lim inf an + lim inf bn .
Proof. inf n≥j (an +bn ) ≥ inf n≥j an +inf n≥j bn , thus limj→∞ inf n≥j (an +bn ) ≥ limj→∞ (inf n≥j an +
inf n≥j bn ). Thus lim inf an + bn ≥ lim inf an + lim inf bn .
Lemma. lim sup(an + bn ) ≤ lim sup an + lim sup bn .
Proof. Similarly.
Lemma. Suppose an → a, then lim inf(an + bn ) = a + lim inf bn .
Proof. We have lim inf(an + bn ) ≥ a + lim inf bn by above. Then lim inf bn = lim inf((an +
bn ) − an ) ≥ lim inf(an + bn ) + lim inf(−an ) = lim inf(an + bn ) − a (as an → a, so −an → −a).
Thus a + lim inf bn = lim inf(an + bn ).
Lemma. Suppose an → a, then lim sup(an + bn ) = a + lim sup bn .
Proof. Similarly.
Lemma. inf −an = − sup an . Thus lim inf −an = − lim sup an .
Proof. Let x denote inf −an . Then by definition, x is the unique number s.t. ∀ n ∈ N : x ≤
−an and ∀ y ∈ R : (∀ n ∈ N : y ≤ −an ) =⇒ y ≤ x. Then we have ∀ n ∈ N : −x ≥ an and
∀ y ∈ R : (∀ n ∈ N : −y ≥ an ) =⇒ −y ≥ −x, or substituting z for −y, ∀ z ∈ R : (∀ n ∈
N : z ≥ an ) =⇒ z ≥ −x. Thus −x is the least upper bound for an , i.e. −x = sup an . Thus
inf −an = − sup an . The second result follows just by taking limits.
Lemma. sup −an = − inf an . Thus lim sup −an = − lim inf an .
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Proof. Similarly.
Lemma. If c ≥ 0, then sup can = c sup an and inf can = c inf an , thus lim sup can =
c lim sup an and lim inf can = c lim inf an .
Proof. We just show sup can = c sup an . The result is trivial for c = 0, so suppose c > 0. Let
n ∈ N. Then an ≤ sup aj , so can ≤ c sup aj . Thus c sup aj is an upper bound for caj , hence
sup caj ≤ c sup aj . Suppose c sup aj − was an upper bound for caj . Then for any n ∈ N,
can ≤ c sup aj − = c(sup aj − c−1 ), so for any n ∈ N, an ≤ sup aj − c−1 , so sup aj − c−1
is an upper bound for aj , a contradiction. Thus c sup aj = sup caj .
Lemma. If an , bn are sequences of nonnegative real numbers and an → a, then lim supn an bn =
a lim supn bn provided either a 6= 0 or lim supn bn < ∞.
Proof. If a = 0, then lim supn bn < ∞. Thus bn is eventually bounded, say by M , so that
for all n ≥ N , bn < M . Since an → 0, for any > 0, there is some M 0 , WLOG > M , s.t.
an < . Then for any n ≥ M 0 , we have that an bn < M . Hence lim sup an bn < M for any
> 0, Thus lim sup an bn = 0 = a lim sup bn .
Now suppose a 6= 0. Then for any 0 < < a, we can find M s.t. for all n ≥ M ,
a − < an < a + . Thus an bn < (a + )bn eventually, so lim sup an bn ≤ lim sup(a + )bn =
(a + ) lim sup bn . Similarly, (a − ) lim sup bn ≤ lim sup an bn . Taking → 0 gives the desired
result.
Lemma. Let an a sequence in R. Then
lim sup|an |= max(lim sup an , − lim inf an ).
Proof. There exists some subsequence |anj | s.t. |anj |→ lim sup|an |. Then infinitely many of
the anj must be positive (including 0) or infinitely many must be negative. Suppose the former, then taking a further subsequence, we get subsequence bn of an s.t. bn → lim supn |an |,
thus lim sup an ≥ limj bj = lim sup|an |. So lim sup|an |≤ max(lim sup an , − lim inf an ). Suppose the latter, i.e. infinitely many of anj are negative. Taking a further subsequence, we
have a subsequence cj s.t. cj < 0 and |cj |→ lim supn |an |. Thus −cj → lim supn |an |. Thus
cj → − lim supn |an | and then lim inf an ≤ limj cj = − lim supn |an |. Thus lim supn |an |≤
− lim inf an and lim sup|an |≤ max(lim sup an , − lim inf an ), which the holds in either case.
Then |an |≥ an so lim sup|an |≥ lim sup an . Also −|an |≤ an , so − lim sup|an |lim inf −|an |≤
lim inf an , so lim sup|an |≥ − lim inf an . Thus lim sup|an |≥ max(lim sup an , − lim inf an ), and
so we have equality.
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