陶哲轩实分析（上）9.9及习题-Analysis I 9.9

这一节讲一致连续，是全局性质或者区间性质而非点的性质。一致连续初看不太好理解，但在后续章节中就会发现一致连续是比逐点连续更强的性质。

Exercise 9.9.1

If $(a_n )_{n=1}^∞$ and $(b_n )_{n=1}^∞$ are equivalent, then $∀ϵ>0,∃N>1$, s.t. $(a_n )_{n=N}^∞$ and $(b_n )_{n=N}^∞$ are $ϵ$-close. Thus
$|a_n-b_n |<ϵ,\quad ∀n>N$
This means $\lim_{n→∞}(a_n-b_n)=0$, and all steps above can be reversed.

Exercise 9.9.2

( a ) implies ( b ):
$∀ϵ>0,∃δ>0$, s.t. $|f(x)-f(y)|<ϵ$ whenever $|x-y|<δ$. As $(x_n )_{n=1}^∞$ and $(y_n )_{n=1}^∞$ are equivalent, we can find $N$ such that $|x_n-y_n |<δ,∀n>N$, so we shall get
$|f(x_n )-f(y_n)|<ϵ,\quad ∀n>N$
This means the sequence $(f(x_n ))_{n=1}^∞$ and $(f(y_n))_{n=1}^∞$ are equivalent.
( b ) implies ( a ):
Assume $f$ is not uniformly continuous on $X$, then we shall be able to find a $ϵ_0>0$ s.t. for every $n∈\mathbf N,∃x_n,y_n∈X,|x_n-y_n |<1/n$, but $|f(x_n )-f(y_n)|≥ϵ_0$. By Archimedean principle we know the sequence $(x_n )_{n=1}^∞$ and $(y_n )_{n=1}^∞$ are equivalent, but
$\lim_{n→∞}(f(x_n )-f(y_n))≥ϵ_0>0$
Which means the sequence $(f(x_n ))_{n=1}^∞$ and $(f(y_n))_{n=1}^∞$ are not equivalent, a contradiction.

Exercise 9.9.3

$f$ is uniformly continuous on $X$ means $∀ϵ>0, ∃δ>0$, s.t. $|f(x)-f(y)|<ϵ$ whenever $|x-y|<δ$. As $(x_n )_{n=0}^∞∈X$ is a Cauchy sequence, we can find $N>0$ s.t.
$|x_n-x_m |<δ,\quad ∀n,m>N$
This means
$|f(x_m )-f(x_n)|<ϵ,\quad ∀n,m>N$
Which shows $(f(x_n ))_{n=0}^∞$ is a Cauchy sequence.

Exercise 9.9.4

Since $x_0$ is an adherent point of $X$, there’s a sequence $(x_n )_{n=1}^∞∈X$ which converges to $x_0$, in particular, this sequence is a Cauchy sequence by Theorem 6.4.18, thus by Proposition 9.9.12, the sequence $(f(x_n ))_{n=1}^∞$ is a Cauchy sequence, by Theorem 6.4.18, $(f(x_n ))_{n=1}^∞$ is convergent, let the number being convergent be $L∈\mathbf R$.
To finish the proof, let $(x_n' )_{n=1}^∞∈X$ be any sequence which converges to $x_0$, then $(x_n' )_{n=1}^∞$ and $(x_n )_{n=1}^∞$ are equivalent sequence, thus $(f(x_n' ))_{n=1}^∞$ and $(f(x_n ))_{n=1}^∞$ are also equivalent sequence, thus $(f(x_n' ))_{n=1}^∞$ converges to $L$. By Proposition 9.3.9, $f$ converges to $L$ at $x_0$.
An alternate demonstration of Example 9.9.10: if $f$ is uniformly continuous, then since $0$ is an adherent point of the sequence $(0,2)$, the limit $\lim_{x→0;x∈(0,2)}1/x$ should exist, but in fact $\lim_{x→0;x∈(0,2)}1/x=+∞$.

Exercise 9.9.5

Assume $f(E)$ is unbounded, then $∀n∈\mathbf N^+$, we can find $x_n∈E$, s.t. $|f(x_n )|>n$. Since $E$ is bounded, we know the sequence $(x_n )_{n=1}^∞$ is bounded, thus has a convergent subsequence $(x_{n_j} )_{j=1}^∞$, we denote $\lim_{j→∞}x_{n_j }=x_0$, by Corollary 9.9.14, $\lim_{j→∞}f(x_{n_j } )$ exists, thus is bounded, but $|f(x_{n_j} )|>n_j>j$, which is a contradiction.

Exercise 9.9.6

Let $∀ϵ>0$, then since $g$ is uniformly continuous, $∃δ_1>0$, s.t.
$|g(y)-g(y')|<ϵ,\quad ∀y,y'∈Y,|y-y' |<δ_1$
For this $δ_1$, since $f$ is uniformly continuous, $∃δ_2>0$, s.t.
$|f(x)-f(x')|<δ_1,\quad ∀x,x'∈X,|x-x' |<δ_2$
So we have
$|(g∘f)(x)-(g∘f)(x')|=|g(f(x))-g(f(x' ))|<ϵ,\quad ∀x,x'∈X,|x-x' |<δ_2$
Thus $g∘f$ is uniformly continuous on $X$.