If $Lu=0$, then the Fourier transform of $u$ must have its support on the manifold where the symbol of $L$ is zero. Hence the Fourier transform of $u$ cannot be a function. This rules out $u\in L^p$ for $p\le 2$; it also rules out a compactly supported distribution. The Bessel function $J_0(\sqrt{x^2+y^2})$ satisfies $\Delta u+u=0$, and it is in $L^p$ for every $p>4$. You can generalize this example to $n$ dimensions, and you find that $u\in L^p$ for every $p>2n/(n-1)$.
If $Lu=0$, then the Fourier transform of $u$ must have its support on the manifold where the symbol of $L$ is zero. Hence the Fourier transform of $u$ cannot be a function. This rules out $u\in L^p$ for $p\le 2$; it also rules out a compactly supported distribution. The Bessel function $J_0(\sqrt{x^2+y^2})$ satisfies $\Delta u+u=0$, and it is in $L^p$ for every $p>2$.p>4$. 2 deleted 51 characters in body If$Lu=0$, then the Fourier transform of$u$must have its support on the manifold where the symbol of$L$is zero. Hence the Fourier transform of$u$cannot be a function. This rules out$u\in L^p$for$p\le 2$; it also rules out a compactly supported distribution. Clearly The Bessel function$u\in L^\infty$is possibleJ_0(\sqrt{x^2+y^2})$ satisfies $\Delta u+u=0$, as trivial examples with and it is in $u=1$ show. I suspect L^p$for every$p<\infty$is not possible, but I cannot immediately think of a proof.p>2$.