= Solution
For a densely defined operator $A$ on a complex <Hilbert space>, <unbounded self-adjointness> means equality $A=A^*$ including equality of the <operator domains>. Symmetry alone asserts only $A\subseteq A^*$ and is insufficient. Use an <inner product> linear in its first argument. If $\operatorname{Im}z\ne0$, symmetry gives
$$
\|(A-z)u\|\,\|u\|\geq|\operatorname{Im}\langle(A-z)u,u\rangle|=|\operatorname{Im}z|\|u\|^2.
$$
Thus $A-z$ is bounded below. Since $A$ is closed, its range is closed; its orthogonal complement is $\ker(A^*-\overline z)=\ker(A-\overline z)=0$, so the range is also dense and therefore all of $H$. The inverse is bounded by $|\operatorname{Im}z|^{-1}$. The <nonreal resolvent estimate for a self-adjoint operator> proves \b[$\sigma(A)\subseteq\mathbb R$].
For bounded <self-adjoint> $T$, its <numerical range of an operator> and <numerical radius> are
$$
W(T)=\{\langle Tu,u\rangle:\|u\|=1\},\qquad w(T)=\sup_{\|u\|=1}|\langle Tu,u\rangle|.
$$
Every number in $W(T)$ is real. If $z\notin\overline{W(T)}$, its distance $\delta$ from that closed set is positive, and $\|(T-z)u\|\geq\delta\|u\|$. The same bound applies to $T-\overline z$, so the range is dense as well as closed and $T-z$ has a bounded inverse. Hence the <numerical-range spectral enclosure> is
$$
\boxed{\sigma(T)\subseteq\overline{W(T)}\subseteq\mathbb R.}
$$
The closure is necessary in infinite dimension: the endpoints of the numerical range need not themselves be attained.
Put $\beta=\sup W(T)$ and $B=\beta I-T$, a bounded <positive semidefinite operator>. Choose <unit vectors> $u_n$ with $\langle Bu_n,u_n\rangle\to0$. The <Cauchy-Schwarz inequality> for the positive form $\langle B\cdot,\cdot\rangle$ implies
$$
\|Bu_n\|^2=\sup_{\|v\|=1}|\langle Bu_n,v\rangle|^2\leq\|B\|\langle Bu_n,u_n\rangle\longrightarrow0.
$$
Therefore $\beta$ is an <approximate eigenvalue>. Apply the same argument to $T-\alpha I$, where $\alpha=\inf W(T)$, to get the other endpoint. This proves the <numerical-range endpoint approximate-eigenvalue theorem>. Each endpoint is in the <spectrum>: a bounded inverse would forbid <unit vectors> with vanishing residual.
If an <unbounded self-adjoint operator> $A$ is positive semidefinite, then for every real $z<0$, $\|(A-z)u\|\geq(-z)\|u\|$. The range argument above again proves invertibility, excluding negative values from the <spectrum>. For the converse, suppose $\sigma(A)\subseteq[0,\infty)$. Its <resolvent operator> $R=(I+A)^{-1}$ is bounded and self-adjoint. The <resolvent spectral mapping identity> puts $\sigma(R)\subseteq[0,1]$: for $w\ne0$, the relevant spectral parameter of $A$ is $1/w-1$, and nonreal values are excluded by self-adjointness. By the endpoint result just proved, $W(R)\subseteq[0,1]$. Positive-form <Cauchy-Schwarz inequality> then gives
$$
\|Rf\|^2\leq\langle Rf,f\rangle\sup_{\|v\|=1}\langle Rv,v\rangle\leq\langle Rf,f\rangle.
$$
For $u=Rf\in D(A)$ this yields $\langle Au,u\rangle=\langle f-Rf,Rf\rangle\geq0$. Thus the <spectral positivity criterion for a self-adjoint operator> is
$$
\boxed{A\geq0\quad\Longleftrightarrow\quad\sigma(A)\subseteq[0,\infty).}
$$
This resolvent proof uses the bounded numerical-range result to handle the full unbounded domain.
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