If $f\colon\mathbb{R}^2 \to \mathbb{R}$ is a continuous function, and $\partial_x f = 0$ in the weak sense, i.e. $$ \iint f(x,y) \partial_x \phi( x,y) dx dy =0 $$ for all $\phi \in C^\infty_c(\mathbb{R}^2)$, does it follow that $f(x,y)=c(y)$?
Many thanks in advance for any suggestions!
You can use functions $\phi \in C_\mathrm{c}^\infty (\mathbb{R}^2)$ of the form $\phi(x,y) = \chi(x) \psi(y)$ with $\chi, \psi \in C_\mathrm{c}^\infty (\mathbb{R})$ .
Since all functions are sufficiently nice, Fubini's theorem applies and we can conclude that $$ \int \limits_\mathbb{R} \psi(y) \int \limits_\mathbb{R} f(x,y) \chi' (x) \, \mathrm{d} x \, \mathrm{d} y = 0 $$ holds for every $\psi \in C_\mathrm{c}^\infty (\mathbb{R})$ . Thus the continuous function $y \mapsto \int_\mathbb{R} f(x,y) \chi'(x) \, \mathrm{d} x $ vanishes identically for every $\chi \in C_\mathrm{c}^\infty (\mathbb{R})$ by the fundamental theorem of the calculus of variations.
But then for fixed $y \in \mathbb{R}$ the function $x \mapsto f(x,y)$ must be constant (see for example this question) and equal to $c(y)$, say.
Since $f$ is continuous, we have $f(x,y) = c(y)$ for every $(x,y) \in \mathbb{R}^2$ with $c \in C^0 (\mathbb{R})$ .
