CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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A norm need not satisfy the parallelogram law

Statement refuted

Every norm on a real vector space containing two linearly independent vectors is induced by an inner product, equivalently satisfies the parallelogram law.

Facts & Assumptions

[A1]

A norm is induced by an inner product if and only if it satisfies the parallelogram law $∥ x + y ∥^{2} + ∥ x - y ∥^{2} = 2 ∥ x ∥^{2} + 2 ∥ y ∥^{2}$ (Jordan–von Neumann: a norm is induced by an inner product exactly when it satisfies the parallelogram law).

[A2]

For $1 \leq p < \infty$ the space $ℓ^{p}$ is the quotient $L^{p} (#)$ of counting measure on $N$ , whose norm is $∥ f ∥_{p} = (\int ∣ f ∣^{p} d #)^{1 / p} = (\sum_{k} ∣ a_{k} ∣^{p})^{1 / p}$ , while $∥ f ∥_{\infty} = \sup_{k} ∣ a_{k} ∣$ ; almost-everywhere equality is equality everywhere ( $ℓ^{p}$ is the $L^{p}$ space of counting measure, The space $L^{p} (μ)$ as the quotient by null functions, Counting measure on an arbitrary set).

[A3]

Rational powers of a fixed base $b > 1$ are strictly increasing in the exponent, and satisfy $(b^{r})^{s} = b^{r s}$ and $b^{r + s} = b^{r} b^{s}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , Laws of rational exponents, Rational powers $a^{r}$ of a positive base).

[A4]

Every inner-product norm satisfies the parallelogram law (Real and complex inner-product spaces and their induced length, Jordan–von Neumann: a norm is induced by an inner product exactly when it satisfies the parallelogram law).

Counterexample

technique · direct

Given: A rational $p$ with $1 \leq p < \infty$ , $p \neq 2$ , and the coordinate vectors $e_{1}, e_{2}$ of the sequence space $ℓ^{p}$ , together with the case of the supremum norm $∥ \cdot ∥_{\infty}$ .

1.1A2A3

In $ℓ^{p}$ the function $f = e_{1} + e_{2}$ has $∣ f ∣^{p} = 1$ at the two indices $0, 1$ and $0$ elsewhere, so $∥ e_{1} + e_{2} ∥_{p} = (1 + 1)^{1 / p} = 2^{1 / p}$ , and likewise $∥ e_{1} - e_{2} ∥_{p} = 2^{1 / p}$ because $∣ f ∣ = 1$ at the same two indices; in $ℓ^{\infty}$ the two norms are $∥ e_{1} \pm e_{2} ∥_{\infty} = 1$ .

2.1step 1.1A3algebra

The parallelogram law in $ℓ^{p}$ would therefore read $2 \cdot 2^{2 / p} = 4$ , that is $2^{2 / p} = 2 = 2^{1}$ , which by strict monotonicity of the rational powers of base $2$ forces $2 / p = 1$ , that is $p = 2$ ; for $p \neq 2$ it fails, and in $ℓ^{\infty}$ the two sides are $2$ and $4$ , so it fails there too.

3.1step 2.1A1A4∎

By the Jordan–von Neumann characterisation, the norm of $ℓ^{p}$ with $p \neq 2$ , and the supremum norm of $ℓ^{\infty}$ , are therefore not induced by any inner product on a space containing the two linearly independent coordinate vectors: the parallelogram law fails on $e_{1}, e_{2}$ , and it would hold on every pair of vectors if an inducing inner product existed.

Depends on

Used by

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §1.3.3, p.39 and §2.3.6 (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Lecture 15 (standard reference, not scraped)