Proof of an equivalence with respect to self-adjoint operators I am trying to solve this problem but I fail in the converse part...

Show  that  the  product of two  self-adjoint operators is  self-adjoint if and  only 
  if the  two  operators  commute.

($\Rightarrow$) If we suppose that $T,U,V$ are self-adjoint operators such that $T=UV$ and $U^*=U,V^*=V,T^*=T$ we have by Theorem that:
$$(UV)^*=V^*U^*=VU$$
but we know, by hypothesis, that: $$(UV)^*=T^*=T=UV$$.We already know that the adjoint operator is unique, so we hav that $UV=VU$ and we are done with this part.
So my troubles are this the converse, can anyone help please.  
 A: Assume $UV = VU$, and $U,V$ are self-adjoint.
$$(UV)^*=V^* U^* = V U = UV \implies (UV)^* = UV$$
i.e, $UV$ is self adjoint. Hopefully that helps.
A: The statement of the theorem could be drastically improved. It should read:

Let $U$ and $V$ be self-adjoint operators.
Theorem. The product $UV$ is self-adjoint if and only if $UV = VU$.

Proof. ($\impliedby$) Suppose that $U$ and $V$ commute. We aim to show that $(UV)^* = UV$. The defining characteristic of the adjoint is the following equation:
$$
\langle UV x,y\rangle = \langle x,(UV)^*y\rangle,\qquad\text{for all $x,y$.}
$$
By our assumption that $U$ and $V$ commute, we have
\begin{align*}
\langle UV x,y\rangle &= \langle VU x,y\rangle \\
&= \langle U x, V^*y\rangle \\
&= \langle x,U^*V^*y\rangle,
\end{align*}
where we have applied the defining characteristic of the adjoint twice. Using the fact that $U$ and $V$ are self-adjoint, so that $U^* = U$ and $V^* = V$, we deduce
$$
\langle UVx,y\rangle = \langle x,UVy\rangle,\qquad \text{for all $x,y$.}
$$
By uniqueness of the adjoint, $UV = (UV)^*$.$\qquad\square$
A: Suppose $UV=VU$. Then, for any $x,y$ in the given space
$$ 0=\langle UVx,y\rangle  - \langle VUx,y\rangle=\langle x,V^*U^*y\rangle - \langle x,U^*V^*y\rangle = \langle x, (V^*U^* - U^*V^*)y\rangle .$$
By the Zero-Operator Lemma (i.e., $T=0$ iff $\langle Tx,  y\rangle =0$ for all $x,y$), $V^*U^* - U^*V^*=0$. Since $U^*=U$ and $V=V^*$, we have $0=V U  - U^*V^*= VU - (VU)^*.$
