Necessary Condition for Integral Functional to have Extremum/Two Variables/Lemma

Theorem
Let $R \subset \R^2$.

Let the boundary of $R$ be $\Gamma$.

Let $\alpha : R \to \R$ be a continuous mapping.

Let $h : R \to \R$ be a twice differentiable mapping such that $\map h \Gamma = 0$.

Suppose for every $h$ we have that:


 * $\displaystyle \iint_R \map \alpha {x,y} \map h {x,y} = 0$.

Then:


 * $\displaystyle \forall x, y \in R : \map \alpha {x,y} = 0$

Proof
Aiming for a contradiction, assume that:


 * $\displaystyle \exists x,y \in R : \map \alpha {x,y} > 0$

Since $\alpha$ is continuous in $R$, there exists $\epsilon > 0$ such that:


 * $\displaystyle \paren {x - x_0}^2 + \paren {y - y_0}^2 \le \epsilon^2$

where $x, x_0, y, y_0 \in R$, and $\epsilon$ is the radius of the corresponding circle.

Choose $\map h {x,y}$ be such that:


 * $\map h {x,y} = 0$

for $\tuple {x,y}$ outside the circle, and


 * $\map h {x,y} = \sqbrk {\paren {x - x_0}^2 + \paren {y - y_0}^2 - \epsilon^2}^3$

for $\tuple {x,y}$ inside the circle.

Such a choice for $\map h {x,y}$ satisfies the conditions of the lemma.

However,


 * $\iint_R \map \alpha {x,y} \map h {x,y}$

reduces to an integral over a circle and is positive.

This is a contradiction.

Hence:


 * $\displaystyle \forall x, y \in R : \map \alpha {x,y} = 0$