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Non-strict convexity allows many minimisers

Statement refuted

Counterexample. On the real Banach space $R^{2}$ consider $I (x, y) = x^{2}$ . Then $I$ is convex (Convex and strictly convex functionals on a convex subset of a real vector space) and ${arg min}_{R^{2}} I = {(0, y) : y \in R},$ an affine line of minimisers: $I \geq 0$ and $I (0, y) = 0$ for every $y$ , while for $x \neq 0$ , $I (x, y) > 0$ . The functional is not strictly convex, so the strictness hypothesis in Strict convexity gives uniqueness of a minimiser is genuinely needed: convexity alone does not force uniqueness of a minimiser.

Facts & Assumptions

Given: The functional $I : R^{2} \to R$ , $I (x, y) = x^{2}$ , on the real Banach space $R^{2}$ .

[F1]

A function $I$ on a convex set is convex when $I (λ u + (1 - λ) v) \leq λ I (u) + (1 - λ) I (v)$ for all $u, v$ and $λ \in [0, 1]$ , and strictly convex when the inequality is strict for $u \neq v$ and $λ \in (0, 1)$ with finite values (Convex and strictly convex functionals on a convex subset of a real vector space).

[F2]

A proper, strictly convex functional has at most one minimiser on a convex set; in particular at most one finite minimiser (Strict convexity gives uniqueness of a minimiser).

Counterexample

technique · direct verification

1.1F1algebra

$I$ is convex. For $u = (x_{1}, y_{1})$ , $v = (x_{2}, y_{2})$ and $λ \in [0, 1]$ one computes $I (λ u + (1 - λ) v) = (λ x_{1} + (1 - λ) x_{2})^{2}$ and $λ I (u) + (1 - λ) I (v) = λ x_{1}^{2} + (1 - λ) x_{2}^{2}$ , whose difference is $λ (1 - λ) (x_{1} - x_{2})^{2} \geq 0$ ; hence $I$ is convex by [F1].

1.2algebra

The minimisers. Since $I (x, y) = x^{2} \geq 0$ for every $(x, y)$ , and $x^{2} = 0$ exactly when $x = 0$ , the infimum of $I$ on $R^{2}$ is $0$ and the set of minimisers is exactly the affine line ${(0, y) : y \in R}$ .

1.3F1algebra

$I$ is not strictly convex. Take the distinct points $u = (0, 0)$ and $v = (0, 1)$ , which satisfy $I (u) = I (v) = 0 < + \infty$ , and $λ = \frac{1}{2}$ . Then $I (\frac{1}{2} u + \frac{1}{2} v) = I (0, \frac{1}{2}) = 0 = \frac{1}{2} I (u) + \frac{1}{2} I (v)$ , so the strict inequality required by [F1] fails; hence $I$ is not strictly convex.

2.1F2step 1.2step 1.3∎

Uniqueness genuinely needs strictness. The line ${(0, y) : y \in R}$ consists of pairwise distinct minimisers of the convex functional $I$ by step 1.2, so convexity alone does not force uniqueness; by [F2] the uniqueness conclusion requires the strict convexity hypothesis, which fails for this functional by step 1.3. This is exactly the sharpness recorded in the statement refuted.

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Sources

Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)
Riccardo Cristoferi, Calculus of Variations: Lecture Notes, Carnegie Mellon University 2016 (complete 133-page notes) (standard reference, not scraped)