上一篇的例 8 過關，靠的是一句「誤差被 1/n 壓住，而這個界不含 x」。這句話其實在說一件事：把 fₙ 與 f 之間所有點的誤差裡最大的那個挑出來，得到一個數；這個數趨於零。既然是一個數，就可以用前三節的全部工具對付它——均勻收斂於是化約成一條普通數列的收斂。
「最大的誤差」要存在，得先確定它不會是無窮大。所以先立一個詞：函數 f : D → ℝ^q 稱為 bounded（有界），意思是存在 M > 0 使 ‖f(x)‖ ≤ M 對每個 x ∈ D 成立。以下一律假設 D 非空——空的定義域上沒有函數值可取，下面那個 sup 會落在空集合上而不存在，這個退化情形不予討論。D 非空且 f 有界時，值域的長度所成的集合非空且上有界，§6-2 的完備性公理（6.4）保證下面這個數存在：
For D ⊆ ℝᵖ, write B_{pq}(D) (or simply B(D)) for the collection of all bounded functions from D into ℝ^q. Addition and scalar multiplication in B_{pq}(D) are defined pointwise: (f + g)(x) = f(x) + g(x) and (cf)(x) = c f(x) for all x ∈ D. The zero function0 : D → ℝ^q is given by 0(x) = 0 for all x ∈ D.
B_{pq}(D) 是 D 上所有有界函數（值在 ℝ^q）所成的集合。加法與純量倍逐點定義；零元是處處取零的那個函數。
正例：f(x) = sin x 屬於 B_{11}(ℝ)——因為 |sin x| ≤ 1，取 M = 1 即可，而 ‖f‖_ℝ = 1。反例：f(x) = x不屬於 B_{11}(ℝ)——它無界，sup {|x| : x ∈ ℝ} 在 ℝ 中不存在，‖f‖_ℝ 根本寫不出來。同一個 f 若把定義域縮成 [0, 1] 就進得來了，且 ‖f‖_{[0,1]} = 1——有界與否取決於定義域。
17.8 LEMMA
(a) Under the operations of Definition 17.7 the set B_{pq}(D) is a vector space. (b) The assignment f ↦ ‖f‖_D = sup {‖f(x)‖ : x ∈ D} is a norm on B_{pq}(D).
Proof. (i) Each ‖f(x)‖ is non-negative, so their supremum ‖f‖_D is non-negative. (ii) The zero function clearly has ‖0‖_D = 0; conversely if ‖f‖_D = 0, then 0 ≤ ‖f(x)‖ ≤ ‖f‖_D = 0 for each x, so ‖f(x)‖ = 0 and hence f(x) = 0 for every x — that is, f is the zero function.
(iii) For c ∈ ℝ, homogeneity of the norm in ℝ^q gives ‖(cf)(x)‖ = |c| ‖f(x)‖ for each x, and taking suprema yields ‖cf‖_D = |c| ‖f‖_D. (iv) For every x ∈ D, ‖(f + g)(x)‖ = ‖f(x) + g(x)‖ ≤ ‖f(x)‖ + ‖g(x)‖ ≤ ‖f‖_D + ‖g‖_D. Hence ‖f‖_D + ‖g‖_D is an upper bound for the set {‖(f + g)(x)‖ : x ∈ D}, and being at least the least such bound, ‖f + g‖_D ≤ ‖f‖_D + ‖g‖_D.
Proof. Suppose the convergence is uniform, and let ε > 0. Definition 17.4 supplies K(ε) such that ‖fₙ(x) − f(x)‖ < ε for all n ≥ K(ε) and all x ∈ D. For such n the number ε is an upper bound of {‖(fₙ − f)(x)‖ : x ∈ D}, so ‖fₙ − f‖_D ≤ ε. As ε > 0 was arbitrary, ‖fₙ − f‖_D → 0.
Conversely, suppose ‖fₙ − f‖_D → 0 and let ε > 0. There is K(ε) with ‖fₙ − f‖_D < ε for n ≥ K(ε). Since the supremum dominates every member of the set it bounds, each x ∈ D satisfies ‖fₙ(x) − f(x)‖ ≤ ‖fₙ − f‖_D < ε. That single K(ε) therefore serves every x at once, which is exactly Definition 17.4.
反方向才是這條定理真正有用的一半，而且它把「均勻」這個字的來歷說清楚了。‖fₙ − f‖_D 是一個數，它的收斂只牽涉 n，交出來的 K(ε) 當然不含 x；而這個數又罩得住每一點的誤差——於是同一個 K 自動對全體 x 通用。從此判定均勻收斂不必再回到 17.4 逐點檢查：算出 ‖fₙ − f‖_D 這個 sup，再問這條實數列趨不趨於零就好。