On $L^2(\mathbb{R}^n)$ it is true that $\Delta$ has $\sigma(\Delta)=(-\infty,0].$ Also, there are no eigenfunction. Yet, even if one would not know this, negativity $\langle \Delta u,u \rangle \le 0$ does immediately imply that there could only be such functions satisfying $\Delta u = \lambda u$ for $\lambda \le 0.$


If we change the setting from $L^2$ to $C\_0(\mathbb{R}^n)$, the continuous functions tending to zero, is there a similar argument why $\Delta u = \lambda u$ is only possible for $\lambda \le 0$ and perhaps: Are there any such eigenfunctions?



Eigenfunction of Laplacian
We can find all the tempered distributions $u$ such that $\Delta u=\lambda u$ (thus, including continuous functions going to $0$ at infinity, since these are locally integrable): taking Fourier transform, $(r^2+\lambda)\widehat{u}=0$. Unless $\lambda$ is real and non-positive, this implies that the support of $\widehat{u}$ is empty, hence $u=0$. For $\lambda$ non-positive, this implies that the support of $\widehat{u}$ is on the sphere (spherical shell) $S$ of radius $\sqrt{-\lambda}$. For example, taking $\widehat{u}$ to be the distribution given by integration over that sphere, gives a continuous function
$$
u(x) \;=\; {1\over (2\pi)^n} \int\_S e^{i\xi\cdot x}\;d\xi
$$
Since this is rotationally invariant, take $x=(x\_1,0,\ldots,0)$, and, up to irrelevant constants $c$, 
$$
u(x\_1,0,\ldots,0)\;=\; c\cdot \int\_{-1}^1 (-\lambda-\xi\_1^2)^{{n-3\over 2}} \,e^{i\xi\_1x\_1}\;d\xi\_1
$$
For $n\ge 2$, this is the Fourier transform on $\mathbb R$ of the $L^1$ function that is $0$ outside $[-1,1]$ and $(-\lambda-\xi\_1^2)^{{n-3\over 2}}$ inside the interval. Thus, by Riemann-Lebesgue, it is continuous and goes to $0$ at infinity, as desired. Again, this succeeds for every $\lambda\le 0$. (And this spherically symmetrical solution is related to Bessel functions, I think.)


And we know that every distribution supported on the sphere is the composition of normal derivatives with distributions *on* the sphere. Distributions *on* the sphere have expansions in spherical harmonics (converging in Sobolev spaces).


EDIT: it doesn't change the conclusion, but I had the wrong exponent $(n-2)/2$, rather than $(n-3)/2$ (which I just changed it to), due to hastiness in converting the surface integral to a one-dimensional integral.


Also, the "obvious" way to evaluate Fourier transform of a compactly-supported distribution (by applying it to the exponentials) is provably correct (and this is standard, but should not be overlooked), because such distributions extend continuously to the larger (Frechet) space of smooth functions (from the smaller space of test functions, e.g.).


