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Is this true?

A weak minimum of a functional is a strong minimum

No — it is false.

The counterexample

dx at y ≡ 0

In the ball the integrand is non-negative so J ≥ 0; in the ball a steeply oscillating small y makes J negative.

The kind of mistake this is

Execution slip

The idea was right. The computation was not.

Drill statements like this
calculus of variations

From Calculus of Variations › Euler–Lagrange equation and standard functionals

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