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The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ and is Lipschitz, hence continuous, for $d_2$

Statement

Clause 1 is about an arbitrary norm; clauses 2 to 4 are about $\mathbb{R}^{n}$ with $n \ge 1$ .

Finite and reverse triangle inequalities. Let $V$ be a vector space over $\mathbb{R}$ 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 $p \in \mathbb{N}$ and every list $u : p \to V$ (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ), $N\Bigl(\sum_{j<p} u_j\Bigr) \;\le\; \sum_{j<p} N(u_j),$ and for all $u, w \in V$ , $\bigl|N(u) - N(w)\bigr| \;\le\; N(u - w).$

Now let $n \in \mathbb{N}$ with $n \ge 1$ , let $\mathbb{R}^{n}$ carry the norms of The $p$ -norms $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ and write $\iota$ for the canonical natural (The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

Every norm is dominated by the $1$ -norm. Let $N$ be a norm on $\mathbb{R}^{n}$ and put $C := \max\{\, N(e_k) : k<n \,\}$ , a maximum over a nonempty finite set of reals (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ , Every nonempty finite set of reals has a maximum and a minimum). Then $C \ge 0$ and $N(x) \;\le\; C\,\lVert x\rVert_1 \qquad \text{for every } x \in \mathbb{R}^{n}.$
The comparison chain. For every $x \in \mathbb{R}^{n}$ , $\lVert x\rVert_\infty \;\le\; \lVert x\rVert_2 \;\le\; \lVert x\rVert_1 \;\le\; \iota(n)\,\lVert x\rVert_\infty , \qquad \lVert x\rVert_1 \;\le\; \sqrt{\iota(n)}\;\lVert x\rVert_2 .$ In particular $\lVert\cdot\rVert_1$ , $\lVert\cdot\rVert_2$ and $\lVert\cdot\rVert_\infty$ are pairwise equivalent norms on $\mathbb{R}^{n}$ , with the constants displayed (Equivalent norms, and the dictionary with equivalent metrics).
Every norm is Lipschitz for the Euclidean metric. With $N$ and $C$ as in clause 2, $N : (\mathbb{R}^{n}, d_2) \to (\mathbb{R}, d_{\mathbb{R}})$ is Lipschitz with constant $C\sqrt{\iota(n)}$ (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it, The absolute value makes $\mathbb{R}$ 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 $\varepsilon$ - $\delta$ form).

Where $n \ge 1$ enters. Clauses 2 and 4 need the maximum defining $C$ to exist, and clause 3 mentions $\lVert\cdot\rVert_\infty$ ; at $n = 0$ each is a maximum over the empty index set and does not exist, exactly as in $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it and The $p$ -norms $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ . Clause 1 carries no hypothesis on the dimension and no hypothesis on the space.

Facts & Assumptions

Given: A vector space $V$ over $\mathbb{R}$ 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 $n \ge 1$ , the space $\mathbb{R}^{n}$ , a norm $N$ on it, and vectors $x, y \in \mathbb{R}^{n}$ .

[L1]

The norm axioms: $N(v) = 0$ exactly when $v = 0_V$ ; $N(\lambda v) = |\lambda|N(v)$ ; $N(u+w) \le N(u)+N(w)$ ; and $N(v) \ge 0$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L2]

Finite sums in a vector space: $\sum_{j<0}u_j = 0_V$ and $\sum_{j<p+1}u_j = \bigl(\sum_{j<p}u_j\bigr) + u_p$ (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ); and $(-1)v = -v$ (In any vector space $0_F v = 0_V$ , $\lambda 0_V = 0_V$ , $(-\lambda)v = -(\lambda v)$ , $(-1_F)v = -v$ , and $\lambda v = 0_V$ forces $\lambda = 0_F$ or $v = 0_V$ ).

[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, $\sum_{k<n}\lambda = \iota(n)\lambda$ , a sum of nonnegative terms is nonnegative, and every single term is at most such a sum.

[L5]

The standard basis: $e_i \in \mathbb{R}^{n}$ has $e_i(i) = 1$ and $e_i(j) = 0$ for $j \ne i$ , $e$ is an ordered basis of $\mathbb{R}^{n}$ , and every $x$ satisfies $x = \sum_{i<n}x_i e_i$ with coordinate list $i \mapsto x(i)$ (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ clauses 1 to 3, A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i<n} \lambda_i v_i$ for exactly one $\lambda : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

[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 $\lVert x\rVert_p$ for rational $p \ge 1$ , and $\lVert x\rVert_\infty$ , Each $\lVert\cdot\rVert_p$ is a norm on $\mathbb{R}^n$ , and the induced metrics are exactly $d_1$ , $d_2$ and $d_\infty$ of the published metric-spaces page): $\lVert x\rVert_1 = \sum_{k<n}|x_k|$ , $\lVert x\rVert_2 = \sqrt{\sum_{k<n}x_k^{2}}$ , $\lVert x\rVert_\infty = \max\{|x_k| : k<n\}$ , and each induces the correspondingly named published metric.

[L8]

Cauchy-Schwarz in root form (The Cauchy-Schwarz inequality for finite sums): $\bigl|\sum_{k<n}a_kb_k\bigr| \le \sqrt{\sum_{k<n}a_k^{2}}\sqrt{\sum_{k<n}b_k^{2}}$ .

[L9]

Square roots and squaring (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ , Squaring is monotone on the nonnegatives): every $c \ge 0$ has a unique $\sqrt{c} \ge 0$ with $(\sqrt{c})^{2} = c$ ; for $a,b \ge 0$ , $a \le b$ exactly when $a^{2} \le b^{2}$ .

[L10]

Absolute value (Basic properties of the absolute value): $|t| \ge 0$ , $|t|^{2} = t^{2}$ , $|st| = |s||t|$ , $|-t| = |t|$ , and $|t|$ equals $t$ or $-t$ .

[L11]

The canonical natural: $\iota(n) > 0$ for $n \ge 1$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing).

[L12]

Lipschitz maps and the regularity hierarchy (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, 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 $\varepsilon$ - $\delta$ form): a map with $d_Y(f(x),f(y)) \le L\,d_X(x,y)$ and $L \ge 0$ is Lipschitz, hence uniformly continuous, hence continuous; $d_{\mathbb{R}}(s,t) = |s-t|$ (The absolute value makes $\mathbb{R}$ 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).

Proof

technique · direct

1.1

The finite triangle inequality holds by induction on $p$ : at $p = 0$ both sides are $0$ , since $\sum_{j<0}u_j = 0_V$ and $N(0_V) = 0$ and the empty real sum is $0$ ; and if $N(\sum_{j<p}u_j) \le \sum_{j<p}N(u_j)$ , then $N(\sum_{j<p+1}u_j) = N(\sum_{j<p}u_j + u_p) \le N(\sum_{j<p}u_j) + N(u_p) \le \sum_{j<p}N(u_j) + N(u_p) = \sum_{j<p+1}N(u_j)$ .

L1L2L3L4

1.2

For $u, w \in V$ : $N(u) = N((u-w)+w) \le N(u-w) + N(w)$ , so $N(u)-N(w) \le N(u-w)$ ; 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 $N(w)-N(u) \le N(u-w)$ . 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 $j<n$ : $x_j^{2} \le \sum_{k<n}x_k^{2}$ , since every single term of a sum of nonnegative terms is at most the sum; taking nonnegative square roots and using $|x_j|^{2} = x_j^{2}$ gives $|x_j| \le \lVert x\rVert_2$ .

L4L7L9L10

1.4

For every $j<n$ : $|x_j| \le \sum_{k<n}|x_k| = \lVert x\rVert_1$ , again because a single term is at most the sum.

L4L7L10

1.5

$\sum_{k<n}|x_k| \le \sum_{k<n}\lVert x\rVert_\infty = \iota(n)\lVert x\rVert_\infty$ , since $|x_k| \le \lVert x\rVert_\infty$ for every $k<n$ and a constant list sums to $\iota(n)$ times its value; so $\lVert x\rVert_1 \le \iota(n)\lVert x\rVert_\infty$ .

L4L6L7L11

1.6

Instantiating [L8] at $a_k := |x_k|$ and $b_k := 1$ gives $\lVert x\rVert_1 = \bigl|\sum_{k<n}|x_k|\cdot 1\bigr| \le \sqrt{\sum_{k<n}|x_k|^{2}}\,\sqrt{\sum_{k<n}1} = \lVert x\rVert_2\sqrt{\iota(n)}$ .

L4L7L8L10

1.7

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

L1L5L6

1.8

$x = \sum_{i<n} x_i e_i$ , the coordinate list of $x$ with respect to the ordered basis $e$ being $i \mapsto x(i) = x_i$ .

2.1

$\lVert x\rVert_\infty$ is one of the numbers $|x_j|$ with $j<n$ , so step 1.3 gives $\lVert x\rVert_\infty \le \lVert x\rVert_2$ .

step 1.3L6L7

2.2

$\sum_{k<n}x_k^{2} = \sum_{k<n}|x_k|\,|x_k| \le \sum_{k<n}|x_k|\,\lVert x\rVert_1 = \lVert x\rVert_1\sum_{k<n}|x_k| = \lVert x\rVert_1^{2}$ , using step 1.4 termwise, monotonicity and scaling; taking nonnegative square roots gives $\lVert x\rVert_2 \le \lVert x\rVert_1$ .

step 1.4L4L7L9L10

2.3

Applying step 1.1 to the list $i \mapsto x_i e_i$ and then (N2): $N(x) = N\bigl(\sum_{i<n}x_ie_i\bigr) \le \sum_{i<n}N(x_ie_i) = \sum_{i<n}|x_i|\,N(e_i) \le \sum_{i<n}|x_i|\,C = C\lVert x\rVert_1$ , 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 $\iota(n) > 0$ and $\sqrt{\iota(n)} > 0$ , 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 $\mathbb{R}^{n}$ , then step 2.3, then step 1.6: $\bigl|N(x)-N(y)\bigr| \le N(x-y) \le C\lVert x-y\rVert_1 \le C\sqrt{\iota(n)}\;\lVert x-y\rVert_2$ .

step 1.2step 1.6step 2.3L4

4.1

Since $\lVert x-y\rVert_2 = d_2(x,y)$ and $\bigl|N(x)-N(y)\bigr| = d_{\mathbb{R}}(N(x),N(y))$ , step 3.2 says exactly that $N$ is Lipschitz with the nonnegative constant $C\sqrt{\iota(n)}$ , 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

Clause 2 is the half of norm equivalence that costs no compactness. It gives an upper bound for an arbitrary norm in terms of $\lVert\cdot\rVert_1$ , and hence in terms of $\lVert\cdot\rVert_2$ by clause 3, by a computation with the standard basis alone. The matching lower bound is where compactness of the unit sphere enters, and that is For $n \ge 1$ all norms on $\mathbb{R}^n$ are equivalent.
The constants of clause 3 are best possible, and the companion page shows it. Nothing here claims sharpness; the attaining vectors are exhibited on the companion page for $\mathbb{R}^{2}$ .
Clause 1 is stated for a general norm on purpose. It is used below for the Euclidean norm on $\mathbb{R}^{n}$ inside Steinitz's polygonal confinement theorem: finitely many vectors of norm at most $1$ summing to $0$ can be ordered so that every partial sum has norm at most $n$ and for an arbitrary $N$ in clause 2, and it is the only statement on this page that needs no hypothesis on the dimension at all.

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Sources

Norm (mathematics) (Wikipedia) (standard reference, not scraped)
Lipschitz continuity (Wikipedia) (standard reference, not scraped)
J. Demmel, MA221 Lecture 3: Vector Norms (standard reference, not scraped)
G. Zitelli, Math 641 Functional Analysis, Part I (standard reference, not scraped)

The finite and reverse triangle inequalities for a norm; and for n≥1n \ge 1n≥1 every norm NNN on Rn\mathbb{R}^nRn satisfies N(x)≤C∥x∥1N(x) \le C\lVert x\rVert_1N(x)≤C∥x∥1​ and is Lipschitz, hence continuous, for d2d_2d2​