Claim:
\[
\mathscr{H}_{m+1}\left[\left(\partial_r-\frac{m}{r}\right) f_m\right]=-k \hat{f}_m
\]
Proof:
first assuming the usual boundary condition terms vanish: \[
r J_n(k r) f_m(r) \longrightarrow 0 \text { at } r=0, \infty \tag{1}
\]
Calculating the integral
\[ \begin{aligned} \mathscr{H}_{m+1} & {\left[\left(\partial_r-\frac{m}{r}\right) f_m\right]=\int_0^{\infty}\left(\partial_r f_m-\frac{m}{r} f_m\right) J_{m+1}(k r) r d r } \\ & =\int_0^{\infty} r\left(\partial_r f_m\right) J_{m+1}(k r) d r-\int_0^{\infty} m f_m J_{m+1}(k r) d r \end{aligned}\tag{2} \]
Integrating the first term by parts
\[ \int_0^{\infty} r\left(\partial_r f_m\right) J_{m+1}(k r) d r=-\int_0^{\infty} f_m \partial_r\left(r J_{m+1}\right) d r\tag{3} \]
Therefore: \[ \begin{aligned} & \mathscr{H}_{m+1}\left[\left(\partial_r-\frac{m}{r}\right) f_m\right]=-\int_0^{\infty} f_m\left[\partial_r\left(r J_{m+1}\right)+m J_{m+1}\right] d r \\ & =-\int_0^{\infty} f_m\left[J_{m+1}+r \partial_r J_{m+1}+m J_{m+1}\right] d r \\ & =-\int_0^{\infty} r k J_m(k r) f_m d r \\ & =-k \hat{f}_m(k) \end{aligned}\tag{4} \]
Claim:
\[
J_{m+1}+r \partial_r J_{m+1}+m J_{m+1}=r k J_m(k r)\tag{1}
\]
Proof:
Let \(x:=k r\), then
\[ \begin{aligned} r\bigg(\partial_r & +\frac{m+1}{r}\bigg) J_{m+1}(k r) \\ & =r \cdot k J_{m+1}^{\prime}(x)+(m+1) J_{m+1}(x) \\ & =x J_{m+1}^{\prime}+(m+1) J_{m+1} \end{aligned}\tag{2} \]
Use the standard identify
\[ \frac{d}{d x}\left(x^\nu J_\nu(x)\right)=x^\nu J_{\nu-1}(x)\tag{3} \]
it says
\[ x J_\nu^{\prime}(x)+\nu J_{\nu}(x)=x J_{\nu-1}(x)\tag{4} \]
with \(\nu=m+1\), this gives
\[ x J_{m+1}^{\prime}(x)+(m+1) J_{m+1}(x)=x J_m(x)\tag{5} \]
substitute back \(x=k r\) :
\[ k r J_{m+1}^{\prime}(x)+(m+1) J_{m+1}(k r)=k r J_m(k r)\tag{6} \]
As given above \(x=k r\)
\[ k \partial_x=\partial_r \]
then
\[ \partial_x J_{m+1}(x)=\frac{1}{k} \partial_r J_{m+1}(k r) \tag{7} \]
substitute it back, rewrite (6)
\[ r \partial_r J_{m+1}(k r)+(m+1) J_{m+1}(k r)=k r J_m(k r)\tag{8} \]
Q.E.D.
Additional Results and Remarks
from (8), it readily to obtain
\[ \left(\partial_r+\frac{m+1}{r}\right) J_{m+1}(k r)=k J_m(k r) \]
similarly,
\[ \left(\partial_r-\frac{m-1}{r}\right) J_{m-1}(k r)=-k J_m(k r) \]