This is a problem out of Logan's Applied Math book. Section 6.7, problem 2.
Show that for any locally integrable function f on $\mathbb{R}$ the function $u(x,y) = f(x-y)$ is a weak solution to the equation $u_x + u_y = 0$ on $\mathbb{R}^2$.
I've got a solution attempt that expresses $f_x$ and $f_y$ directly, then the equation cancels I get the integral of zero over $\mathbb{R}$ is zero. I'm just not sure if this derivative exists, it seems like thats the point of the weak solution.
I was more inclined to use the adjoint of the differential operator that gives the PDE to take the derivative of the test function and show that the integral of that over an unbounded domain is zero.
As Ian wrote, $u$ being a weak solution of $u_x+u_y=0$ means that $$\int_{\mathbb{R}^2} u(x,y) (v_x(x,y) + v_y(x,y))\, dx dy = 0\tag{1}$$ for every test function $v$. This follows by formal integration by parts of $$\int_{\mathbb{R}^2} (u_x(x,y)+u_y(x,y)) v(x,y) dx dy = 0\tag{2}$$
If $u(x,y)=f(x-y)$, then integral in $(1)$ becomes as $$\int_{\mathbb{R}^2} f(x-y) (v_x(x,y) + v_y(x,y))\, dx dy $$ After the change of variables $x-y=s$ and $x+y=t$, this is a multiple of $$\int_{\mathbb{R}^2} f(s) w_t(s,t)\,ds\,dt $$ where $w(s,t)=v(x,y)$. The latter integral is zero, since $\int w_t\,dt=0$.