Is this space compactly contained in $L^p((0,\infty),rdr)$ for all $p\geq 2$?

Some Background: A typical problem in mathematical physics is the existence of positive radially symmetric solutions to a nonlinear Schrodinger type equation over $\mathbb{R}^{2+1}$. Such a problem typically reduces to the following nonlinear boundary value problem (or "$n$-vortex equation" under the $n$-vortex ansatz $f(x)=f(r)e^{in\theta+i\lambda t}$),

\begin{align} &f_{rr}(r)+\dfrac{1}{r}f_r(r)-\left(\dfrac{n^2}{r^2}+\lambda\right) f(r)+g(f^2(r))f(r)=0, \quad r\in(0,\infty),\label{vortexEq1}\\ &f(0)=0=f(\infty), \end{align} where $n\in\mathbb{Z}$, $\lambda>0$, and $g:\mathbb{R}\rightarrow\mathbb{R}$ is, say, a nice continuous function (e.g., $g(s)=s^2$ or $g(s)=s^2/(1+s^2)$).

Via a variational approach, solutions to the above problem may be found as critical points of a corresponding functional, say, $I$, over an appropriate Sobolev space, say, $H$. However, a good understanding of the space considered is necessary (my current difficulty).

Question: Let $H$ be the closure of $\mathcal{C}_0^{\infty}(0,\infty)$ with inner product \begin{align} (f,h)=\int_0^{\infty}\left\{rf_r(r)h_r(r)+\left(\dfrac{n^2}{r^2}+\lambda\right) rf(r)h(r)\right\}dr, \end{align} and norm $||f||_H^2=(f,f)$. Is $H$ compactly contained in $L^p((0,\infty),rdr)$ for all $p\geq 2$? (With particular interest in the case $p=2$.)

Further background and results: It is clear that $H$ is continuously embedded in $L^2((0,\infty),rdr)$ and $W^{1,2}((0,\infty),rdr)$, i.e., \begin{align} ||f||^2_{L^2((0,\infty),rdr)}=\int_0^{\infty}rf^2(r)dr\leq \int_0^{\infty}\left\{f_r^2(r)+f^2(r)\right\}rdr\leq\max\{1,1/\lambda\}||f||_H^2. \end{align} We can also show that $f(0)=0$ and $f(\infty)=0$ for every $f\in H$.

In a particular problem, contained in the paper https://doi.org/10.1515/jaa-2016-0010 by Carlo Greco published in the Journal of Applied Analysis, the author states that $H$ in compactly contained in $L^p((0,\infty),rdr)$ for all $p> 2$. Hence, my particular interest into the case $p=2$.

The embedding is not compact this can be observed by considering the sequence $f_k (r) = f (r/k)/k^2$ for some given function $f \in C^\infty_c (0, +\infty) \setminus \{0\}$. One has for every $k \ge 1$, by a change of variable $$\Vert f_k \Vert^2 = \frac{1}{k^2} \int_0^\infty \vert f' (r)\vert^2 r + \frac{n \vert f (r)\vert^2}{r} \,\mathrm{d}r + \lambda \int_0^\infty \vert f (r)\vert^2 r \,\mathrm{d}r$$ Since the sequence converges to $0$ almost everywhere, its limit in $L^2$ can only be $0$ but $$\int_0^\infty \vert f_k (r) - 0 \vert^2 r \,\mathrm{d}r = \int_0^\infty \vert f (r) \vert^2 r \,\mathrm{d}r > 0,$$ so that it cannot converge strongly to $0$.