Test Function with Vanishing Partial Derivative

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Theorem

Let $\phi \in \map \DD {\R^2}$ be a test function such that:

$\tuple {x, y} \stackrel \phi {\longrightarrow} \map \phi {x, y}$

Suppose $\phi$ is a solution to the following partial differential equation:

$\ds \dfrac {\partial \phi}{\partial x} = 0$


Then $\phi$ is identically $0$.


Proof

$\ds \dfrac {\partial \phi}{\partial x} = 0$ implies that:

$\forall x \in \R : \map \phi {x, y} = \map C y$

where $C : \R \to \C$ is a complex-valued function.

By definition, $\phi$ is a test function.

Hence, $\phi$ must have a compact support $\Omega \subset \R^2$.

Let $\map {B^-_\epsilon} 0 \subset \R^2$ be a closed ball in Euclidean space such that:

$\Omega \subseteq \map {B^-_\epsilon} 0$

By Closed Ball in Euclidean Space is Compact, $\map {B^-_\epsilon} 0$ is a compact.

Then $\map {B^-_\epsilon} 0$ also qualifies as a compact support of $\phi$.

By definition of a test function:

$\forall y \in \R : \size y > \epsilon \implies \map \phi {x, y} = 0$

But for each $y \in \R$ we have that $\map \phi {x, y} = \map C y$ is a constant.

By smoothness of test functions, this constant has to be the same for all $y \in \R$.

Hence, $\map \phi {x, y} = 0$.

$\blacksquare$


Sources