[proofplan]
The proof is purely algebraic on the common invariant domain $D$. We expand each commutator using the definitions $J_+=J_x+iJ_y$ and $J_-=J_x-iJ_y$, then substitute the angular momentum commutation relations and their antisymmetric consequences. The first two computations show how $J_+$ and $J_-$ commute with $J_z$, and the final computation collects the two nonzero cross terms in $[J_+,J_-]$.
[/proofplan]
[step:Expand the commutator of $J_z$ with the raising operator]
Fix $\psi \in D$. Since $J_x,J_y,J_z,J_+:D \to D$, all compositions appearing below are defined on $D$. Using the definition of the commutator and the linearity of $J_z$, we compute
\begin{align*}
[J_z,J_+]\psi=J_z(J_x\psi+iJ_y\psi)-J_+(J_z\psi).
\end{align*}
By the definition of $J_+$ applied to $J_z\psi \in D$,
\begin{align*}
J_+(J_z\psi)=J_x(J_z\psi)+iJ_y(J_z\psi).
\end{align*}
Therefore
\begin{align*}
[J_z,J_+]\psi=[J_z,J_x]\psi+i[J_z,J_y]\psi.
\end{align*}
The hypotheses give $[J_z,J_x]\psi=i\hbar J_y\psi$. Also, from $[J_y,J_z]\psi=i\hbar J_x\psi$ and the definition of the commutator, $[J_z,J_y]\psi=-i\hbar J_x\psi$. Substituting these two identities yields
\begin{align*}
[J_z,J_+]\psi=i\hbar J_y\psi+i(-i\hbar J_x\psi).
\end{align*}
Since $i(-i)=1$, this becomes
\begin{align*}
[J_z,J_+]\psi=\hbar J_x\psi+i\hbar J_y\psi.
\end{align*}
Factoring out $\hbar$ and using $J_+\psi=J_x\psi+iJ_y\psi$, we obtain
\begin{align*}
[J_z,J_+]\psi=\hbar J_+\psi.
\end{align*}
[guided]
Fix $\psi \in D$. The reason the common invariant domain matters is that every operator product in the commutators must be defined. Because $J_x,J_y,J_z,J_+:D \to D$, the vectors $J_+\psi$, $J_z\psi$, $J_x\psi$, and $J_y\psi$ all lie in $D$, so applying the next operator is legitimate.
We begin from the definition of the commutator:
\begin{align*}
[J_z,J_+]\psi=J_z(J_+\psi)-J_+(J_z\psi).
\end{align*}
Now insert the definition $J_+\psi=J_x\psi+iJ_y\psi$. Linearity of $J_z$ gives
\begin{align*}
J_z(J_+\psi)=J_z(J_x\psi+iJ_y\psi)=J_z(J_x\psi)+iJ_z(J_y\psi).
\end{align*}
Since $J_z\psi \in D$, the definition of $J_+$ may also be applied to $J_z\psi$:
\begin{align*}
J_+(J_z\psi)=J_x(J_z\psi)+iJ_y(J_z\psi).
\end{align*}
Subtracting these expressions gives
\begin{align*}
[J_z,J_+]\psi=\bigl(J_z(J_x\psi)-J_x(J_z\psi)\bigr)+i\bigl(J_z(J_y\psi)-J_y(J_z\psi)\bigr).
\end{align*}
This is exactly
\begin{align*}
[J_z,J_+]\psi=[J_z,J_x]\psi+i[J_z,J_y]\psi.
\end{align*}
We now use the angular momentum commutation relations. One relation is already in the required order:
\begin{align*}
[J_z,J_x]\psi=i\hbar J_y\psi.
\end{align*}
For the other term, the hypothesis gives $[J_y,J_z]\psi=i\hbar J_x\psi$. Since commutators are antisymmetric by direct expansion,
\begin{align*}
[J_z,J_y]\psi=-[J_y,J_z]\psi=-i\hbar J_x\psi.
\end{align*}
Substituting both identities into the expanded commutator gives
\begin{align*}
[J_z,J_+]\psi=i\hbar J_y\psi+i(-i\hbar J_x\psi).
\end{align*}
The scalar identity $i(-i)=1$ converts the second term into $\hbar J_x\psi$, so
\begin{align*}
[J_z,J_+]\psi=\hbar J_x\psi+i\hbar J_y\psi.
\end{align*}
Factoring out $\hbar$ recovers the raising operator:
\begin{align*}
[J_z,J_+]\psi=\hbar(J_x\psi+iJ_y\psi)=\hbar J_+\psi.
\end{align*}
[/guided]
[/step]
[step:Expand the commutator of $J_z$ with the lowering operator]
Again fix $\psi \in D$. Using $J_-\psi=J_x\psi-iJ_y\psi$, the same commutator expansion gives
\begin{align*}
[J_z,J_-]\psi=[J_z,J_x]\psi-i[J_z,J_y]\psi.
\end{align*}
Substitute $[J_z,J_x]\psi=i\hbar J_y\psi$ and $[J_z,J_y]\psi=-i\hbar J_x\psi$:
\begin{align*}
[J_z,J_-]\psi=i\hbar J_y\psi-i(-i\hbar J_x\psi).
\end{align*}
Since $-i(-i)=-1$, this becomes
\begin{align*}
[J_z,J_-]\psi=-\hbar J_x\psi+i\hbar J_y\psi.
\end{align*}
Using $J_-\psi=J_x\psi-iJ_y\psi$, we conclude
\begin{align*}
[J_z,J_-]\psi=-\hbar J_-\psi.
\end{align*}
[/step]
[step:Expand the commutator of the two ladder operators]
Fix $\psi \in D$. Since $J_+,J_-:D \to D$, the commutator $[J_+,J_-]\psi$ is defined. Expanding from the definitions,
\begin{align*}
[J_+,J_-]\psi=(J_x+iJ_y)(J_x\psi-iJ_y\psi)-(J_x-iJ_y)(J_x\psi+iJ_y\psi).
\end{align*}
Using linearity of $J_x$ and $J_y$, the first product equals
\begin{align*}
J_x(J_x\psi)-iJ_x(J_y\psi)+iJ_y(J_x\psi)+J_y(J_y\psi).
\end{align*}
The second product equals
\begin{align*}
J_x(J_x\psi)+iJ_x(J_y\psi)-iJ_y(J_x\psi)+J_y(J_y\psi).
\end{align*}
Subtracting cancels the two square terms and leaves
\begin{align*}
[J_+,J_-]\psi=-2iJ_x(J_y\psi)+2iJ_y(J_x\psi).
\end{align*}
Factoring $-2i$ gives
\begin{align*}
[J_+,J_-]\psi=-2i\bigl(J_x(J_y\psi)-J_y(J_x\psi)\bigr).
\end{align*}
The expression in parentheses is $[J_x,J_y]\psi$, so
\begin{align*}
[J_+,J_-]\psi=-2i[J_x,J_y]\psi.
\end{align*}
Using $[J_x,J_y]\psi=i\hbar J_z\psi$, we obtain
\begin{align*}
[J_+,J_-]\psi=-2i(i\hbar J_z\psi)=2\hbar J_z\psi.
\end{align*}
Thus all three claimed commutation relations hold for the arbitrary vector $\psi \in D$.
[/step]
