Proof. For each i let rᵢ = (cᵢ₁, ⋯, c_{ip}) ∈ ℝᵖ be the i-th row of the matrix. By 21.2 the i-th coordinate of f(x) is yᵢ = rᵢ · x. Hence 8.8 gives |yᵢ| ≤ ‖rᵢ‖ ‖x‖, and squaring, yᵢ² ≤ (Σⱼ cᵢⱼ²) ‖x‖².
Summing these q inequalities and using ‖f(x)‖² = Σᵢ yᵢ², ‖f(x)‖² ≤ (ΣᵢΣⱼ cᵢⱼ²) ‖x‖² = A²‖x‖². Both sides are non-negative, so taking square roots gives ‖f(x)‖ ≤ A‖x‖.
這張圖在說證明的第一步在辨認什麼：表的一整條橫列本身就是 ℝᵖ 裡的一個點，而輸出的第 i 個座標恰好是它與輸入的內積。認出內積，Schwarz 不等式就直接套得上。逐列估完之後平方相加，各列的 ‖x‖² 是共同因子，提出來就得到整張表的平方和。
上面的估計是對「一個點」講的。連續談的是「兩個點的距離」，而線性恰好把兩者接了起來。
21.3 THEOREM
If f is a linear function with domain ℝᵖ and range in ℝ^q, then there is a constant A ≥ 0 such that ‖f(u) − f(v)‖ ≤ A‖u − v‖ for all u, v ∈ ℝᵖ. Consequently f is continuous at every point of ℝᵖ.
線性函數不必逐點驗連續，一句話就全部到位。而且 A > 0 時 δ(ε) = ε/A對每一點都通用——不像平方函數或倒數函數那樣要跟著位置調整。
Proof. Let A be as in the Lemma and let u, v ∈ ℝᵖ. Applying linearity with a = 1 and b = −1 gives f(u − v) = f(u) − f(v). Hence, putting x = u − v in the Lemma, ‖f(u) − f(v)‖ = ‖f(x)‖ ≤ A‖x‖ = A‖u − v‖.
If A > 0, then at every point a ∈ ℝᵖ the hypothesis of the local linear estimate is met with M = A and with r as large as we please, so f is continuous at a and δ(ε) = ε/A may be used. If A = 0, then every cᵢⱼ vanishes, so f is the zero function, which is continuous everywhere. In either case f is continuous on ℝᵖ.
這一步把估計翻成連續。引理要求 M > 0，所以 A = 0 得單獨處理：A 是一堆平方的和開根號，它等於零只可能是每一項都等於零，於是整張表全是零、f 把每個點都送到零向量，而常數函數的連續在前一節第一篇就驗過。A > 0 時有效範圍 r 可以取任意大，inf {r, ε/A} 裡只剩第二項。