= Solution
The given <angular momentum commutation relations> imply $[L_3,L_1]=i\hbar L_2$ and $[L_3,L_2]=-i\hbar L_1$. By <linearity>,
$$
[L_3,L_\pm]=i\hbar L_2\pm i(-i\hbar L_1)=\boxed{\pm\hbar L_\pm}.
$$
Similarly,
$$
[L_+,L_-]=-2i[L_1,L_2]=\boxed{2\hbar L_3}.
$$
For the remaining <commutators>, the <commutator product rule> gives, for each $j$,
$$
[L^2,L_j]=\sum_i\left(L_i[L_i,L_j]+[L_i,L_j]L_i\right)=i\hbar\sum_{i,k}\epsilon_{ijk}(L_iL_k+L_kL_i)=0.
$$
The last sum vanishes because the <Levi-Civita symbol> is antisymmetric under $i\leftrightarrow k$, while $L_iL_k+L_kL_i$ is symmetric. Taking $j=3$ and then linear combinations with $j=1,2$ proves
$$
\boxed{[L^2,L_3]=0=[L^2,L_\pm].}
$$
If $L^2\psi=\lambda\hbar^2\psi$ and $L_3\psi=m\hbar\psi$, these identities yield
$$
L^2L_+\psi=L_+L^2\psi=\lambda\hbar^2L_+\psi,\qquad L_3L_+\psi=(L_+L_3+\hbar L_+)\psi=(m+1)\hbar L_+\psi.
$$
One must also prove that this candidate <eigenstate> is nonzero. Since the <angular momentum operators> are <self-adjoint operators>, $L_+^*=L_-$, and expanding the product gives
$$
L_-L_+=L_1^2+L_2^2+i[L_1,L_2]=L^2-L_3^2-\hbar L_3.
$$
Therefore
$$
\|L_+\psi\|^2=\hbar^2\bigl[\lambda-m(m+1)\bigr]\|\psi\|^2.
$$
This is nonnegative and, by the assumed inequality, nonzero. Hence it is positive. The <angular momentum raising operator> really does produce a nonzero simultaneous <eigenstate> with the claimed <eigenvalues>.
To prove that raising terminates, let $v$ be any nonzero state reached after $r$ steps, with projection $q=m+r$. Positivity gives
$$
\lambda\hbar^2\|v\|^2=\sum_{i=1}^3\|L_iv\|^2\geq\|L_3v\|^2=q^2\hbar^2\|v\|^2,
$$
so $q^2\leq\lambda$. Thus the sequence cannot remain nonzero for every integer $r\geq0$. At its last nonzero state $v=(L_+)^k\psi$, one has $L_+v=0$, and the squared <norm> identity gives
$$
\boxed{\lambda=(m+k)(m+k+1)\quad\text{for some integer }k\geq0.}
$$
Under the paper's initial inequality the first step is nonzero, so in fact $k\geq1$. An <angular momentum ladder endpoint> must therefore be reached: bounded projection follows from positivity, and the final raising <norm> fixes the total <angular momentum> <eigenvalue>. Writing $j=m+k$, the complementary product $L_+L_-=L^2-L_3^2+\hbar L_3$ gives $\|L_-v\|^2=2j\hbar^2\|v\|^2$, also showing $j\geq0$.
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