Integrable solutions to an elliptic PDE on divergence form have a definite sign

I love this problem and have spent half the evening thinking about it.

Here is a rough sketch of an idea that could possibly work. Making it rigorous might be a bit of a chore due to the unboundedness of $mathbb{R}^n$, etc, but my guess is that it is doable.

Let $L$ denote the elliptic operator [ Lu : = -Delta u - mathrm{div}( f u ). ]

Suppose the operator $L$ has a principal eigenvalue $lambda_1$, which is the smallest number $lambda$ for which there exists a nonzero solution of the equation $Lu = lambda u$ in $mathbb{R}^n$. Then $lambda_1$ should be simple (!) and have a principal eigenfunction which does not change sign. Let $varphi in L^1(mathbb{R}^n)$ denote the principal eigenfunction, which we normalize to be positive.

Assume for now that $varphi$ and its derivatives tend to zero at infinity, and $f$ and its derivatives stay bounded. Then we may simply integrate the equation $Lvarphi = lambda_1varphi$ to get $lambda_1 int_{mathbb{R}^n} varphi dx = 0$. Well, that means that $lambda_1 = 0$.

Recalling the simplicity of $lambda_1=0$, we see that the equation $Lu = 0$ not only has a positive solution, its set of solutions is precisely ${ c varphi : cin mathbb{R} }$. This implies the result.

Now, you may have already noticed that there aren't really such principal eigenvalues in general, for example when $f=0$. But I think it is possible that the idea can still be made into a rigorous proof.

If we look at a really large domain, say the ball $B(0,R)$ with $R> 0$ very large, there is a principal eigenvalue $lambda_{1,R}$ of $L$ on $B(0,R)$ and it is going to be close to zero. This can be shown by considering the adjoint operator $L^*$, which has the same principal eigenvalue $lambda_{1,R}$, and it is easy to show that $0 < lambda_{1,R} leq CR^{-2}$, due to boundedness of $f$. This is where we use the special form of the equation.

If we have a solution to $Lu = 0$ on the whole space $mathbb{R}^n$, with $u(0)>0$, I believe it should be possible to prove that if we choose the normalization $varphi_{1,R}(0) = u(0)$, then $varphi_{1,R}$ converges to $u$ (at least locally uniformly) as $Rto infty$. In particular, $u$ must be positive everywhere.

There are obviously some details left to work out. It is very possible I am making a silly error and it doesn't work at all.