Zeroth-order term in elliptic estimates

Question

When solving an elliptic equation $$ Lu = f \ \text{in} \ \Omega $$ $$ u = 0 \ \text{on} \ \partial \Omega $$ for an elliptic operator $L$ of order $m$ on a bounded open set $\Omega$, one has the a priori estimates $$ \|u\|_{H^{k}(\Omega)} \leq C(\|{f}\|_{H^{k - m}(\Omega)} + \|{u}\|_{L^2(\Omega)}). $$ The zeroth-order term on the right hand side, by which I mean $\|u\|_{L^2(\Omega)}$, can be removed in this estimate if there are no nontrivial solutions to the homogeneous equation, i.e. if $Lu = 0$ in $\Omega$ and $u = 0$ on $\partial \Omega$ implies $u = 0$.

With $\Omega = \mathbb{R}^n$, one has a similar estimate. Here, boundary conditions are to be interpreted "at infinity", i.e. $u = 0$ on $\partial \Omega$ means that $u$ vanishes at infinity. My question is: can the zeroth-order term be removed in this case, under the hypothesis that the homogeneous equation has no nonzero solutions?

Answer 1

No, the inequality without the zeroth-order term is seen to be false just by scaling. Already for the case $-\Delta u = f $ in $\mathbb{R}^n$ (that satisfies your hypothesis, since every $u\in H^1(\mathbb{R}^n)=H^1_0(\mathbb{R}^n)$ harmonic is necessarily identically zero) plugging $u_\lambda(x):=u(x/\lambda)$ in the inequality without the zeroth-order gives a contradiction.

Note that, among other things, the use of weighted Sobolev spaces (as in the post you cite) exactly solves this scaling issue, since every seminorm $[u]^2_{H^{s,\delta}(\mathbb{R}^n)}$ essentially scales independently of $s$ and precisely as $\lambda^{n+2\delta}$. So, for an operator of order $m$, trying to do the same trick on the weighted inequality $ \| u \|^2_{H^{s,\delta}(\mathbb{R}^n)} \le C \|Lu\|^2_{H^{s-m,\delta+m}(\mathbb{R}^n)}$ would give you $\lambda^{n+2\delta} \lesssim \lambda^{n+2(\delta+m)} \cdot \lambda^{-2m} = \lambda^{n+2\delta}$ which is, after all, not a contradiction.