# Showing $f$ is zero from $f = a f_x + b f_y$

Suppose $f:\mathbb {R^2}\to \mathbb{R}$ has continuous partial derivatives and satisfy:

$$f = a f_x+ b f_y$$ for some constants $a$ and $b$. Prove if $f$ is bounded, then $f$ is identically zero.

This seems obvious but I'm not too sure how to start approaching this question. I've tried splitting it to two different cases,where $f$ is constant and $f$ is not. There seems to be a relation I am missing and I have not gotten anthere since.

If $a$ or $b$ is zero, then the question can be answered without any knowledge of partial differential equations (PDE). Otherwise, both $a$ and $b$ are nonzero, and the equation $f=af_x + bf_y$ is a PDE. Writing the solution as $f(x,y) = e^{x/a} g(x,y)$, one obtains $$a g_x + b g_y = 0 \, .$$ This PDE is an advection equation with solution $g(x,y) = G(y - bx/a)$ for any continuously differentiable function $G$. Since $f$ is assumed bounded, there is $A \geq 0$ such that $\left|f(x,y)\right|\leq A$ for all $(x, y)$, i.e. $$e^{x/a} \left| G(y - bx/a)\right| \leq A\, .$$ The change of variable $\xi = y-bx/a$ gives $$\left| G(\xi)\right| e^{-\xi/b} \leq A\, e^{-y/b}\, .$$ for all $(\xi,y)$. The left-hand side is a nonnegative function of $\xi$, while the right-hand side is a function of $y$ with lower bound equal to zero. Thus, one concludes that $\left| G(\xi)\right| e^{-\xi/b} = 0$ for all $\xi$. Finally, $f$ is identically zero.