= Solution
Write $E_0=\hbar\omega/2$. Repeated application of the <annihilation operator> lowers the <energy eigenvalue> in steps of $\hbar\omega$. The lower bound from part (ii) forces this process to stop. To see why it also quantizes the starting energy, write
$$
\frac{E-E_0}{\hbar\omega}=n+\delta,\qquad n\in\{0,1,2,\ldots\},\quad 0\leq\delta<1.
$$
The first $n$ lowering steps have positive squared <norms>. If $\delta>0$, the next step is also nonzero by part (ii), and its energy would be $E_0+(\delta-1)\hbar\omega<E_0$, contradicting the bound. Thus $\delta=0$, and
$$
\boxed{E_n=\left(n+\frac12\right)\hbar\omega,\qquad n=0,1,2,\ldots.}
$$
Existence of the <ground state> can also be checked directly in the position representation. The equation $a_-\psi_0=0$ is
$$
-i\hbar\psi_0'-im\omega x\psi_0=0,
$$
whose solution is proportional to $e^{-m\omega x^2/(2\hbar)}$. This <Gaussian function> is <square-integrable>, so it gives an actual <ground state>. Its normalization can be chosen using the <Gaussian integral>.
All the displayed levels occur: for a nonzero state at $E_n$,
$$
\|a_+\psi_n\|^2=\langle\psi_n,a_-a_+\psi_n\rangle=2m\hbar\omega(n+1)\|\psi_n\|^2>0.
$$
Thus repeated raising creates a nonzero <energy eigenstate> at every $E_n$.
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