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

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

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

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

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

Suppose for every $h$ we have that:


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

Then:


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

Proof
Aiming for a contradiction, assume that:


 * $\displaystyle \exists x_0,y_0 \in D : \map \alpha {x,y} > 0$

$\alpha$ is continuous in $D$.

Therefore, there exists a closed ball $B^-_{\epsilon}$ defined by:


 * $\map {B^-_{\epsilon}} {x_0, y_0} := \set {\tuple{x,y} \in D : \paren {x - x_0}^2 + \paren {y - y_0}^2 \le \epsilon^2}$

such that:


 * $\forall x,y \in B^-_{\epsilon} : \map \alpha {x,y} > 0$

Choose $\map h {x,y}$ in the following way:


 * $\map h {x,y} = \begin{cases}

0, \forall x, y \notin B^-_{\epsilon}\\

\sqbrk {\epsilon^2 - \paren {x - x_0}^2 - \paren {y - y_0}^2}^3, \forall x,y \in B^-_{\epsilon}

\end{cases}$

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

But then both $\alpha$ and $h$ are positive inside this ball.

Hence, the integral is positive.

This is contradicts assumptions of the lemma.

Hence:


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