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LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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Each p\lVert\cdot\rVert_p is a norm on Rn\mathbb{R}^n, and the induced metrics are exactly d1d_1, d2d_2 and dd_\infty of the published metric-spaces page

Statement

Let nNn \in \mathbb{N} and let pQp \in \mathbb{Q} with p1p \ge 1, with the norms of The pp-norms xp\lVert x\rVert_p for rational p1p \ge 1, and x\lVert x\rVert_\infty. Then:

  1. p\lVert\cdot\rVert_p is a norm on Rn\mathbb{R}^{n} (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).
  2. For n1n \ge 1, \lVert\cdot\rVert_\infty is a norm on Rn\mathbb{R}^{n}.
  3. The dictionary. For n1n \ge 1 and all x,yRnx, y \in \mathbb{R}^{n}, xy1=d1(x,y),xy2=d2(x,y),xy=d(x,y),\lVert x-y\rVert_1 = d_1(x,y), \qquad \lVert x-y\rVert_2 = d_2(x,y), \qquad \lVert x-y\rVert_\infty = d_\infty(x,y), where d1d_1, d2d_2, dd_\infty are the metrics of the published Rn\mathbb{R}^n as the set of functions nRn \to \mathbb{R}, and d1d_1, d2d_2, dd_\infty are metrics on it. So the metric induced by each of these three norms (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) is the correspondingly named published metric, not merely one equivalent to it.

Consequence, used repeatedly below and stated once here. By clause 3 at p=2p = 2, the metric space (Rn,d2)(\mathbb{R}^{n}, d_2) of the published metric-spaces page and the metric space underlying the normed space (Rn,2)(\mathbb{R}^{n}, \lVert\cdot\rVert_2) of this page are the same object. Hence completeness (R\mathbb{R} and Rn\mathbb{R}^n for n1n \ge 1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R\mathbb{R} clause 2), Heine-Borel (Heine-Borel in Rn\mathbb{R}^n: with the Euclidean metric a subset of Rn\mathbb{R}^n is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line clause 2) and the compactness equivalences (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice) are statements about this page's normed space, with their hypothesis n1n \ge 1 inherited unchanged and not weakened. Nothing below cites any of those three theorems for n=0n = 0.

Why this lemma exists. Without it the library would hold a norm-induced metric on Rn\mathbb{R}^{n} and a separately published metric on the same set with no recorded relation, and every later citation would have to guess which was meant. The proof of clause 3 is a comparison of two written expressions; the value is that the comparison is made and recorded.

Facts & Assumptions

Given: A natural number nn, a rational p1p \ge 1, vectors x,yRnx, y \in \mathbb{R}^{n} and a real λ\lambda; write S(x):=k<nxkpS(x) := \sum_{k<n}|x_k|^{p}, so that xp=S(x)1/p\lVert x\rVert_p = S(x)^{1/p} (The pp-norms xp\lVert x\rVert_p for rational p1p \ge 1, and x\lVert x\rVert_\infty, Finite sums and finite products, by recursion).

[L1]

Rational powers (Rational powers ara^r of a positive base, Laws of rational exponents): for a,b0a, b \ge 0 and rationals r,s>0r, s > 0 one has ar0a^{r} \ge 0, (ab)r=arbr(ab)^{r} = a^{r}b^{r}, 0r=00^{r} = 0, and ar>0a^{r} > 0 when a>0a > 0; and for a>0a > 0, (ar)s=ars(a^{r})^{s} = a^{rs} and a1=aa^{1} = a.

[L2]

Monotonicity in the base (Monotonicity of rarr \mapsto a^{r} and of aara \mapsto a^{r} clause 2): for a rational r>0r > 0 and reals 0a<b0 \le a < b one has ar<bra^{r} < b^{r}; hence aba \le b implies arbra^{r} \le b^{r}, the case a=ba = b being trivial, and ar=0a^{r} = 0 only for a=0a = 0.

[L3]

Laws of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, monotonicity; a sum of nonnegative terms is nonnegative, each single term is at most such a sum, and a sum of nonnegative terms that vanishes has every term 00.

[L4]

Minkowski's inequality for finite sums at rational p1p \ge 1 (Minkowski's inequality for finite sums (rational exponent)): (k<nak+bkp)1/p(k<nakp)1/p+(k<nbkp)1/p\bigl(\sum_{k<n}|a_k+b_k|^{p}\bigr)^{1/p} \le \bigl(\sum_{k<n}|a_k|^{p}\bigr)^{1/p} + \bigl(\sum_{k<n}|b_k|^{p}\bigr)^{1/p}.

[L5]

Absolute value (Basic properties of the absolute value, Absolute value in an ordered field, The triangle inequality): t0|t| \ge 0; t=0|t| = 0 exactly when t=0t = 0; st=st|st| = |s|\,|t|; s+ts+t|s+t| \le |s|+|t|; and t2=t2|t|^{2} = t^{2}.

[L6]

Maxima (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set): a nonempty finite set of reals has a maximum, the maximum belongs to the set and bounds it above, and a set with an upper bound belonging to it has that element as its maximum.

[L7]

Order arithmetic: multiplying an inequality by a nonnegative real preserves it (Sign rules for products and monotonicity of multiplication in its strict form, together with the case of equality settled by totality), and \le is transitive (Ordered field).

[L9]

The published metrics on Rn\mathbb{R}^{n} for n1n \ge 1 are d1(x,y)=k<nxkykd_1(x,y) = \sum_{k<n}|x_k-y_k|, d2(x,y)=k<n(xkyk)2d_2(x,y) = \sqrt{\sum_{k<n}(x_k-y_k)^{2}} and d(x,y)=max{xkyk:k<n}d_\infty(x,y) = \max\{|x_k-y_k| : k<n\}, and each is a metric (Rn\mathbb{R}^n as the set of functions nRn \to \mathbb{R}, and d1d_1, d2d_2, dd_\infty are metrics on it, Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

technique · direct
1.1

Every term xkp|x_k|^{p} is nonnegative, so S(x)0S(x) \ge 0 and xp=S(x)1/p\lVert x\rVert_p = S(x)^{1/p} is defined and nonnegative.

L1L3
1.2

S(x)=0S(x) = 0 holds exactly when xkp=0|x_k|^{p} = 0 for every k<nk<n, a vanishing sum of nonnegative terms having every term 00; and xkp=0|x_k|^{p} = 0 exactly when xk=0|x_k| = 0, that is exactly when xk=0x_k = 0.

L1L2L3L5
1.3

For every k<nk<n, (λx)kp=(λxk)p=λpxkp|(\lambda x)_k|^{p} = \bigl(|\lambda|\,|x_k|\bigr)^{p} = |\lambda|^{p}|x_k|^{p}, so S(λx)=λpS(x)S(\lambda x) = |\lambda|^{p}S(x) by scaling of finite sums.

L1L3L5
1.4

Instantiating [L4] at ak:=xka_k := x_k and bk:=ykb_k := y_k, and using (x+y)k=xk+yk(x+y)_k = x_k+y_k, gives x+ypxp+yp\lVert x+y\rVert_p \le \lVert x\rVert_p + \lVert y\rVert_p, which is axiom (N3) for p\lVert\cdot\rVert_p.

L4L8
1.5

Under [A1] the set {xk:k<n}\{|x_k| : k<n\} is nonempty and finite, so x\lVert x\rVert_\infty exists, is one of the xk|x_k|, and satisfies xkx|x_k| \le \lVert x\rVert_\infty for every k<nk<n; in particular x0\lVert x\rVert_\infty \ge 0.

A1L5L6
1.6

Under [A1], xy1=k<nxkyk\lVert x-y\rVert_1 = \sum_{k<n}|x_k-y_k| by the case p=1p=1 of the definition, and that is the written expression for d1(x,y)d_1(x,y).

L1L9
1.7

Under [A1], xy2=(k<nxkyk2)1/2=k<n(xkyk)2\lVert x-y\rVert_2 = \bigl(\sum_{k<n}|x_k-y_k|^{2}\bigr)^{1/2} = \sqrt{\sum_{k<n}(x_k-y_k)^{2}}, using t2=t2|t|^{2} = t^{2} and the identification of the exponent 1/21/2 with the nonnegative square root, and that is the written expression for d2(x,y)d_2(x,y).

L5L8L9
1.8

Under [A1], xy=max{xkyk:k<n}\lVert x-y\rVert_\infty = \max\{|x_k-y_k| : k<n\} by definition, and that is the written expression for d(x,y)d_\infty(x,y).

L9
2.1

xp=0\lVert x\rVert_p = 0 holds exactly when S(x)=0S(x) = 0, since S(x)>0S(x) > 0 would give S(x)1/p>0S(x)^{1/p} > 0 and 01/p=00^{1/p} = 0.

step 1.1L1L2
2.2

Under [A1]: x=0\lVert x\rVert_\infty = 0 forces xk0|x_k| \le 0 and xk0|x_k| \ge 0 for every k<nk<n, hence x=0x = 0; and 0=0\lVert 0\rVert_\infty = 0. This is (N1) for \lVert\cdot\rVert_\infty.

step 1.5L5L8
2.3

Under [A1]: for every k<nk<n, (λx)k=λxkλx|(\lambda x)_k| = |\lambda|\,|x_k| \le |\lambda|\,\lVert x\rVert_\infty, and choosing j<nj<n with xj=x|x_j| = \lVert x\rVert_\infty gives (λx)j=λx|(\lambda x)_j| = |\lambda|\,\lVert x\rVert_\infty; so λx|\lambda|\lVert x\rVert_\infty belongs to the set and bounds it above, whence λx=λx\lVert \lambda x\rVert_\infty = |\lambda|\lVert x\rVert_\infty. This is (N2) for \lVert\cdot\rVert_\infty.

step 1.5L5L6L7
2.4

Under [A1]: for every k<nk<n, (x+y)k=xk+ykxk+ykx+y|(x+y)_k| = |x_k+y_k| \le |x_k| + |y_k| \le \lVert x\rVert_\infty + \lVert y\rVert_\infty; choosing j<nj<n with (x+y)j=x+y|(x+y)_j| = \lVert x+y\rVert_\infty gives x+yx+y\lVert x+y\rVert_\infty \le \lVert x\rVert_\infty + \lVert y\rVert_\infty, which is (N3) for \lVert\cdot\rVert_\infty.

step 1.5L5L6L7
3.1

By steps 2.1 and 1.2, xp=0\lVert x\rVert_p = 0 exactly when xk=0x_k = 0 for every k<nk<n, that is exactly when x=0x = 0; this is axiom (N1) for p\lVert\cdot\rVert_p.

step 2.1step 1.2L8
3.2

Steps 2.2, 2.3 and 2.4 are (N1), (N2) and (N3) for \lVert\cdot\rVert_\infty under [A1], so clause 2 holds.

step 2.2step 2.3step 2.4A1L8
4.1

If λ=0\lambda = 0 then λx=0\lambda x = 0 and both sides of (N2) are 00 by step 3.1; if λ0\lambda \ne 0 then λ>0|\lambda| > 0, and step 1.3 with the power laws gives λxp=(λpS(x))1/p=(λp)1/pS(x)1/p=λp(1/p)xp=λxp\lVert \lambda x\rVert_p = \bigl(|\lambda|^{p}S(x)\bigr)^{1/p} = \bigl(|\lambda|^{p}\bigr)^{1/p}S(x)^{1/p} = |\lambda|^{p\cdot(1/p)}\lVert x\rVert_p = |\lambda|\,\lVert x\rVert_p; this is axiom (N2).

step 1.3step 3.1L1L5L8
5.1

Steps 3.1, 4.1 and 1.4 are (N1), (N2) and (N3) for p\lVert\cdot\rVert_p, so clause 1 holds.

step 1.4step 3.1step 4.1L8
6.1

Steps 1.6, 1.7 and 1.8 give clause 3, and with steps 5.1 and 3.2 all three clauses are proved; in particular the metric induced by 2\lVert\cdot\rVert_2 on Rn\mathbb{R}^{n} for n1n \ge 1 is the published d2d_2, which is the consequence recorded in the Statement.

step 5.1step 3.2step 1.6step 1.7step 1.8L9

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