陶哲轩实分析（上）9.3及习题-Analysis I 9.3

9.3的题目叫limiting values of functions，在Analysis II里有一节同样题目的章节。这一节限制在R上讨论，函数的极限值是后续讨论continuity的基础，函数极限可以和数列极限等价起来，函数极限是唯一的，满足极限算律，极限也是局部的性质（即只和某一点附近的情况有关）

Exercise 9.3.1

(a) implies (b):
If $f$ converges to $L$ at $x_0$ in $E$, then $∀ϵ>0,∃δ>0, s.t. |x-x_0 |<δ ⇒|f(x)-L|<ϵ$. Let $(a_n )_{n=0}^∞∈E$ such that $a_n→x_0$, then $∃N∈\mathbf N, s.t. |a_n-x_0 |<δ,∀n>N$, so if $n>N$, we have $|f(a_n )-L|<ϵ$, so $(f(a_n ))_{n=0}^∞→L$.
(b) implies (a):
Assume $f$ doesn’t converge to $L$ at $x_0$ in $E$, then due to axiom of choice, there’s a $ϵ_0>0$, such that for $∀N∈\mathbf N,∃x_n∈E, s.t. (|x_n-x_0 |<δ)∧(|f(x_n )-L|≥ϵ)$, the sequence $(x_n )_{n=0}^∞∈E$, but $(f(x_n ))_{n=0}^∞↛L$, a contradiction.

Exercise 9.3.2

Since $x_0$ is an adherence point of $E$, we can find $(a_n )_{n=0}^∞∈E$ such that $a_n→x_0$, then we have $(f(a_n ))_{n=0}^∞→L$ and $(g(a_n ))_{n=0}^∞→M$, by Proposition 9.3.9. Thus:
$((f-g)(a_n ))_{n=0}^∞→L-M ⇒ (f(a_n ))_{n=0}^∞-(g(a_n ))_{n=0}^∞→L-M$ by Theorem 6.1.19 (d)
$((\max⁡(f,g) )(a_n ))_{n=0}^∞→\max(L,M) ⇒\max⁡((f(a_n ))_{n=0}^∞,(g(a_n ))_{n=0}^∞ )→\max(L,M)$ by Theorem 6.1.19 (g)
$((\min⁡(f,g) )(a_n ))_{n=0}^∞→\min(L,M) ⇒\min⁡((f(a_n ))_{n=0}^∞,(g(a_n ))_{n=0}^∞ )→\min⁡(L,M)$ by Theorem 6.1.19 (h)
$((fg)(a_n ))_{n=0}^∞→LM ⇒ (f(a_n ))_{n=0}^∞ (g(a_n ))_{n=0}^∞→LM$ by Theorem 6.1.19 (b)
$((f/g)(a_n ))_{n=0}^∞→L/M ⇒ (f(a_n ))_{n=0}^∞/(g(a_n ))_{n=0}^∞→L/M$ by Theorem 6.1.19 (f)
Then all functions have asserted limits at $x_0$ in $E$ since the above conclusions are true for arbitrary $(a_n )_{n=0}^∞∈E$ such that $a_n→x_0$, and thus by Proposition 9.3.9.

Exercise 9.3.3

$\lim_{x→x_0;x∈E}⁡f(x)=L$ iff
$∀ϵ>0,∃δ'>0,s.t.(x∈E)∧(|x-x_0 |<δ' )⇒|f(x)-L|<ϵ$
iff
$∀ϵ>0,∃δ'>0,s.t.(x∈E)∧(|x-x_0 |<\min⁡(δ,δ' )≤δ' )⇒|f(x)-L|<ϵ$
iff
$∀ϵ>0,∃δ'>0,s.t.(x∈E∩(x_0-δ,x_0+δ))∧(|x-x_0 |<δ' )⇒|f(x)-L|<ϵ$
iff
$\lim_{x→x_0;x∈E}⁡f(x)=L$

Exercise 9.3.4

A definition for limit superior: we say $\limsup_{x→x_0;x∈E}⁡f(x)=M$ iff for $∀ϵ>0,∃δ>0$, s.t.
$\Big((|x-x_0 |<δ)⇒(f(x)<M+ϵ)\Big)∧\Big(∃x'∈(x_0-δ,x_0+δ),f(x' )>M-ϵ\Big)$
A definition for limit inferior: we say $\liminf_{x→x_0;x∈E}f(x)=m$ iff for $∀ϵ>0,∃δ>0$, s.t.
$\Big((|x-x_0 |<δ)⇒(f(x)>m-ϵ)\Big)∧\Big(∃x'∈(x_0-δ,x_0+δ),f(x' )<m+ϵ\Big)$

An analogue of Proposition 9.3.9 (limsup case):
Let $E⊆X⊆R, f:X→R, x_0∈\overline E$, then the following two statements are logically equivalent:
(a) $\limsup_{x→x_0;x∈E}f(x)=M$
(b) For any sequence $(a_n )_{n=0}^∞∈E,a_n→x_0$, we have $\limsup_{n→∞}f(a_n)≤M$, and there is a sequence $(b_n )_{n=0}^∞∈E, b_n→x_0$, such that $\lim_{n→∞}f(b_n)=M$.

An analogue of Proposition 9.3.9 (liminf case):
Let $E⊆X⊆R, f:X→R, x_0∈\overline E$, then the following two statements are logically equivalent:
(a) $\liminf_{x→x_0;x∈E}f(x)=m$
(b) For any sequence $(a_n )_{n=0}^∞∈E,a_n→x_0$, we have $\lim⁡inf_{n→∞}f(a_n)≥m$, and there is a sequence $(b_n )_{n=0}^∞∈E, b_n→x_0$, such that $\lim_{n→∞}f(b_n)=m$.

Proof: I only prove the case for limit superior.
(a) implies (b):
If $\limsup_{x→x_0;x∈E}f(x)=M$, then on one hand, $∀ϵ>0$, the set
$A_n=\{x∈E: |x-x_0 |<1/n\text{ and }M-ϵ<f(x)<M+ϵ\}$
is non-empty, use axiom of choice to find a $b_n∈A_n$ and form a sequence $(a_n )_{n=0}^∞∈E$, it’s easy to see that $b_n→x_0$ and $\lim_{n→∞}f(b_n)=M$.
On the other hand, assume $∃(a_n )_{n=0}^∞∈E, a_n→x_0$, and $\limsup_{n→∞}f(a_n)=M'>M$, then let
$ϵ=(M'-M)/2>0$
We can first have a $δ>0$, s.t. $(|x-x_0 |<δ)⇒(f(x)<M+ϵ)$, and there’s a $N∈\mathbf N$, s.t.
$|a_n-x_0 |<δ,\quad ∀n>N$
Which means
$f(a_n )<M+ϵ,\quad ∀n>N$
But since $\limsup_{n→∞}f(a_n)=M'$, we can find a $K>N$, s.t. KaTeX parse error: Expected group after '^' at position 10: f(a_K )>M^̲'-ϵ, now we have
$f(a_K )<M+ϵ=M'-ϵ<f(a_K )$
This is a contradiction.
(b) implies (a):
Assume the first condition is wrong, then we shall find a $ϵ_0>0$, s.t. $∀N∈\mathbf N$, the set
$A_n=\{x∈E:|x-x_0 |<1/n\text{ and }f(x)≥M+ϵ_0\}$
is non-empty, choose $a_n∈A_n$ by axiom of choice, then $a_n→x_0$ as $n→∞$, and
$M<M+ϵ_0≤\inf⁡(a_n )≤\limsup_{n→∞}⁡f(a_n )≤M$
This is a contradiction.
Assume the second condition is wrong, then $∀x∈(x_0-δ,x_0+δ),f(x)≤M-ϵ_0$ for some $ϵ_0>0$ and any $δ>0$, then any sequence $(a_n )_{n=0}^∞∈E$ such that $a_n→x_0$ and $f(a_n )$ converges would have $\lim_{n→∞}f(a_n)≤M-ϵ_0<M$, this contradicts the existence of $(b_n )_{n=0}^∞$.

Exercise 9.3.5

Let $(a_n )_{n=0}^∞∈E$ such that $a_n→x_0$, by Proposition 9.3.9, we know that both $(f(a_n ))_{n=0}^∞$ and $(h(a_n ))_{n=0}^∞$ converge to $L$, thus for $∀ϵ>0$, there’s $N_1,N_2∈\mathbf N$, such that
$|f(a_n )-L|<ϵ,\quad∀n>N_1\\ |h(a_n )-L|<ϵ,\quad∀n>N_2$
Let $N=\max(N_1,N_2)$, then we shall have
$L-ϵ<f(a_n )≤g(a_n )≤h(a_n )<L+ϵ,\quad∀n>N$
This means $(g(a_n ))_{n=0}^∞$ converges to $L$, this is true for arbitrary sequence $(a_n )_{n=0}^∞∈E$ which satisfies $a_n→x_0$, so we can conclude $\lim_{x→x_0;x∈E}⁡g(x)=L$.