Is the following statement true? A weak minimum of a functional is a strong minimum

No. J[y] = ∫₀^π (1 − (y′)²)y² dx at y ≡ 0 In the C¹-ball the integrand is non-negative so J ≥ 0; in the C⁰-ball a steeply oscillating small y makes J negative.

Counterexample

Is this true?

A weak minimum of a functional is a strong minimum

No — it is false.

The counterexample

$J[y] = \int_{0}^\pi (1 - (y')^{2})y^{2}$ dx at y ≡ 0

Why it fails

In the $C^{1}-$ ball the integrand is non-negative so J ≥ 0; in the $C^{0}-$ ball a steeply oscillating small y makes J negative.

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Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

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calculus of variations

Go deeper· Explained in full in Euler–Lagrange, first integrals and null Lagrangians· Asked in CSIR NET June 2025· From Calculus of Variations › Euler–Lagrange equation and standard functionals

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

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