Non-compactness of the resolvent Consider a complete non-compact Riemannian manifold $M$ and the resolvent of the Laplacian $(-\Delta + \lambda I)^{-1}$. It is known that the resolvent is in general not a compact operator. I am trying to understand why not. I realize that the proof in the compact setting does not follow through for the non-compact case because of the failure of compact Sobolev embedding. I would be happy to see why the resolvent is not compact even for $M = \mathbb{R}^n$, and a justification for general non-compact $M$ would be excellent. Thanks.
For $M=\mathbb R^n$ take a smooth function with compact support $\phi\in C_0^\infty(M)$. 

Then 
$$
\phi = (-\Delta+\lambda I)^{-1}((-\Delta+\lambda I)\phi)
$$
obviously. Now consider translations $\phi_k(x):=\phi(x+kv)$ for $v\in \mathbb R^n$, $k\in \mathbb N$. Take the norm of $v$ large enough such that the supports of different $\phi_k$'s are empty.

The set
$$
X:=\{ (-\Delta+\lambda I)\phi_k,\ k\in\mathbb N\}
$$
is bounded in $L^2$, but
$$
 (-\Delta+\lambda I)^{-1}X
$$
is not relatively compact in $L^2$.
The simplest way to show that for a non-compact manifold the resolvent does not have to be compact is to give a counterexample. Following your (very good) choice of the manifold $M=\mathbb{R}^n$, lets show that $(-\Delta - \lambda I)^{-1}$ is not compact. For simplicity, assume $n=2$.



Assume $\vec{\mathbf{x}}, \vec{\mathbf{y}} \in \mathbb{R}^2$. Then the  fundamental solution $\mathit{\Phi}$  of 2D Laplacian $\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}  \ $  takes form

$$
\mathit{\Phi}\big(\vec{\mathbf{x}}, \vec{\mathbf{y}}\big) =- \frac{1}{2\pi}\ln \big|\vec{\mathbf{x}} - \vec{\mathbf{y}} \big| = 
- \frac{1}{4\pi}\ln \Big( (x_1 - x_2)^2 + (y_1-y_2)^2 \Big). \label{1}\tag{1}
$$ 

Recall that the fundamental solution of a linear PDE, (which is equivalent to the Green's function) is the kernel of resolvent of the corresponding differential operator. 


   For example, see  this link, page 4.


Denote the resolvent $R_\lambda = (-\Delta - \lambda I)^{-1}$. Then the resolvent of 2D Laplacian takes form
$$
R_\lambda[u](\mathbf{x}) = \int_{\mathbb{R}^2} \mathit{\Phi}\big(\mathbf{x}, \mathbf{y}\big) u (\mathbf{y}).\label{2}\tag{2} $$
Substituting $\eqref{1}$ into $\eqref{2}$, we get

$$
R_\lambda[u](\mathbf{x}) = - \frac{1}{2\pi}  \int_{\mathbb{R}^2} \bigg( \ln \big|\vec{\mathbf{x}} - \vec{\mathbf{y}} \big| \!\cdot\! u (\mathbf{y}) \bigg)\,\mathrm{d}^2 \mathbf y.\label{3}\tag{3}
$$
or, equivalently,
$$
R_\lambda[u]\big(x_1, x_2\big) =  
- \frac{1}{4\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \!\bigg(\! \ln\! \Big( (x_1 - x_2)^2 + (y_1-y_2)^2 \Big)\! \cdot\! u\big(x_1,x_2\big)\! \bigg)dy_1  dy_2, \label{4}\tag{4}
$$
where 
$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in \mathbb{R}^2$ and $\mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \in \mathbb{R}^2$.



Now, when we have the explicit expression  \eqref{3}, it is easy to see that  the resolvent $R_\lambda$ is not compact, since the integral kernel $\mathit{\Phi}\not\in L^2\big( \mathbb{R}^2\times  \mathbb{R}^2\big) $.