If Double Integral of a(x, y)h(x, y) vanishes for any C^2 h(x, y) then C^0 a(x, y) vanishes
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Theorem
Let $\map \alpha {x, y}$, $\map h {x, y}$ be functions in $\R$.
Let $\alpha \in C^0$ in a closed region $R$ whose boundary is $\Gamma$.
Let $h \in C^2$ in $R$ and $h = 0$ on $\Gamma$.
Let:
- $\ds \iint_R \map \alpha {x, y} \map h {x, y} \rd x \rd y = 0$
Then $\map \alpha {x, y}$ vanishes everywhere in $R$.
Proof
Aiming for a contradiction, suppose $\map \alpha {x, y}$ is nonzero at some point in $R$.
Then $\map \alpha {x, y}$ is also nonzero in some disk $D$ such that:
- $\paren {x - x_0}^2 + \paren {y - y_0}^2 \le \epsilon^2$
Suppose:
- $\map h {x, y} = \map \sgn {\map \alpha {x, y} } \paren {\epsilon^2 - \paren {x - x_0}^2 + \paren {y - y_0}^2}^3$
in this disk and $0$ elsewhere.
Thus $\map h {x, y}$ satisfies conditions of the theorem.
However:
- $\ds \iint_R \map \alpha {x, y} \map h {x, y} \rd x \rd y = \iint_D \size {\map \alpha {x, y} } \paren {\epsilon^2 - \paren {x - x_0}^2 + \paren {y - y_0}^2}^3 \ge 0$
Hence the result, by Proof by Contradiction.
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations: $\S 1.5$: The Case of Several Variables