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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2

Statement

Clause 1 is about an arbitrary norm; clauses 2 to 4 are about Rn with n≥1.

  1. Finite and reverse triangle inequalities. Let V be a vector space over R and N a norm on it (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). For every p∈N and every list u:p→V (Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S), N(∑j<puj)  ≤  ∑j<pN(uj), and for all u,w∈V, ∣N(u)−N(w)∣  ≤  N(u−w).

Now let n∈N with n≥1, let Rn carry the norms of The p-norms ∥x∥p for rational p≥1, and ∥x∥∞ and write ι for the canonical natural (The canonical natural ι(n)=n⋅1F of a field).

  1. Every norm is dominated by the 1-norm. Let N be a norm on Rn and put C:=max⁡{ N(ek):k<n }, a maximum over a nonempty finite set of reals (The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0, Every nonempty finite set of reals has a maximum and a minimum). Then C≥0 and N(x)  ≤  C ∥x∥1for every x∈Rn.
  2. The comparison chain. For every x∈Rn, ∥x∥∞  ≤  ∥x∥2  ≤  ∥x∥1  ≤  ι(n) ∥x∥∞,∥x∥1  ≤  ι(n)  ∥x∥2. In particular ∥⋅∥1, ∥⋅∥2 and ∥⋅∥∞ are pairwise equivalent norms on Rn, with the constants displayed (Equivalent norms, and the dictionary with equivalent metrics).
  3. Every norm is Lipschitz for the Euclidean metric. With N and C as in clause 2, N:(Rn,d2)→(R,dR) is Lipschitz with constant Cι(n) (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, The absolute value makes R a metric space: d(x,y)=∣x−y∣ is a metric, its open balls are the intervals (x−r,x+r), and it is unbounded), hence uniformly continuous and continuous (Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent, Continuity of a map between metric spaces, at a point and globally, in the ε-δ form).

Where n≥1 enters. Clauses 2 and 4 need the maximum defining C to exist, and clause 3 mentions ∥⋅∥∞; at n=0 each is a maximum over the empty index set and does not exist, exactly as in Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it and The p-norms ∥x∥p for rational p≥1, and ∥x∥∞. Clause 1 carries no hypothesis on the dimension and no hypothesis on the space.

Facts & Assumptions

Given: A vector space V over R with a norm N (Vector space over a field, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms); and, for clauses 2 to 4, a natural n≥1, the space Rn, a norm N on it, and vectors x,y∈Rn.

[L1]

The norm axioms: N(v)=0 exactly when v=0V; N(λv)=∣λ∣N(v); N(u+w)≤N(u)+N(w); and N(v)≥0 (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L3]

The induction principle (The principle of mathematical induction).

[L4]

Laws of finite sums of reals (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, monotonicity, ∑k<nλ=ι(n)λ, a sum of nonnegative terms is nonnegative, and every single term is at most such a sum.

[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, which belongs to the set and bounds it above.

[L7]

The three norms (The p-norms ∥x∥p for rational p≥1, and ∥x∥∞, Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page): ∥x∥1=∑k<n∣xk∣, ∥x∥2=∑k<nxk2, ∥x∥∞=max⁡{∣xk∣:k<n}, and each induces the correspondingly named published metric.

[L8]

Cauchy-Schwarz in root form (The Cauchy-Schwarz inequality for finite sums): ∣∑k<nakbk∣≤∑k<nak2∑k<nbk2.

[L9]

Square roots and squaring (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Squaring is monotone on the nonnegatives): every c≥0 has a unique c≥0 with (c)2=c; for a,b≥0, a≤b exactly when a2≤b2.

[L10]

Absolute value (Basic properties of the absolute value): ∣t∣≥0, ∣t∣2=t2, ∣st∣=∣s∣∣t∣, ∣−t∣=∣t∣, and ∣t∣ equals t or −t.

Proof

technique · direct
1.1

The finite triangle inequality holds by induction on p: at p=0 both sides are 0, since ∑j<0uj=0V and N(0V)=0 and the empty real sum is 0; and if N(∑j<puj)≤∑j<pN(uj), then N(∑j<p+1uj)=N(∑j<puj+up)≤N(∑j<puj)+N(up)≤∑j<pN(uj)+N(up)=∑j<p+1N(uj).

L1L2L3L4
1.2

For u,w∈V: N(u)=N((u−w)+w)≤N(u−w)+N(w), so N(u)−N(w)≤N(u−w); and N(w−u)=N((−1)(u−w))=∣−1∣N(u−w)=N(u−w), so the same argument with u and w exchanged gives N(w)−N(u)≤N(u−w). Since ∣N(u)−N(w)∣ is one of N(u)−N(w) and N(w)−N(u), the reverse triangle inequality follows, completing clause 1.

L1L2L10
1.3

For every j<n: xj2≤∑k<nxk2, since every single term of a sum of nonnegative terms is at most the sum; taking nonnegative square roots and using ∣xj∣2=xj2 gives ∣xj∣≤∥x∥2.

L4L7L9L10
1.4

For every j<n: ∣xj∣≤∑k<n∣xk∣=∥x∥1, again because a single term is at most the sum.

L4L7L10
1.5

∑k<n∣xk∣≤∑k<n∥x∥∞=ι(n)∥x∥∞, since ∣xk∣≤∥x∥∞ for every k<n and a constant list sums to ι(n) times its value; so ∥x∥1≤ι(n)∥x∥∞.

L4L6L7L11
1.6

Instantiating [L8] at ak:=∣xk∣ and bk:=1 gives ∥x∥1=∣∑k<n∣xk∣⋅1∣≤∑k<n∣xk∣2 ∑k<n1=∥x∥2ι(n).

L4L7L8L10
1.7

The set {N(ek):k<n} is a nonempty finite set of reals because n≥1, so C=max⁡{N(ek):k<n} exists, belongs to the set, satisfies N(ek)≤C for every k<n, and is ≥0 since every value of N is.

L1L5L6
1.8

x=∑i<nxiei, the coordinate list of x with respect to the ordered basis e being i↦x(i)=xi.

L5
2.1

∥x∥∞ is one of the numbers ∣xj∣ with j<n, so step 1.3 gives ∥x∥∞≤∥x∥2.

step 1.3L6L7
2.2

∑k<nxk2=∑k<n∣xk∣ ∣xk∣≤∑k<n∣xk∣ ∥x∥1=∥x∥1∑k<n∣xk∣=∥x∥12, using step 1.4 termwise, monotonicity and scaling; taking nonnegative square roots gives ∥x∥2≤∥x∥1.

step 1.4L4L7L9L10
2.3

Applying step 1.1 to the list i↦xiei and then (N2): N(x)=N(∑i<nxiei)≤∑i<nN(xiei)=∑i<n∣xi∣ N(ei)≤∑i<n∣xi∣ C=C∥x∥1, the last inequality by monotonicity from step 1.7. This is clause 2.

step 1.1step 1.7step 1.8L1L4L7
3.1

Steps 2.1, 2.2, 1.5 and 1.6 are the four inequalities of clause 3; since ι(n)>0 and ι(n)>0, they exhibit positive constants in both directions for each of the three pairs, so the three norms are pairwise equivalent.

step 1.5step 1.6step 2.1step 2.2L11L9
3.2

By step 1.2 applied on Rn, then step 2.3, then step 1.6: ∣N(x)−N(y)∣≤N(x−y)≤C∥x−y∥1≤Cι(n)  ∥x−y∥2.

step 1.2step 1.6step 2.3L4
4.1

Since ∥x−y∥2=d2(x,y) and ∣N(x)−N(y)∣=dR(N(x),N(y)), step 3.2 says exactly that N is Lipschitz with the nonnegative constant Cι(n), hence uniformly continuous and continuous; this is clause 4, and with steps 1.2, 2.3 and 3.1 all four clauses are proved.

step 1.2step 2.3step 3.1step 3.2L7L12∎

Remarks

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