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For the converse, let $f$ be bounded and $\Vert f\Vert =f(1)=c$. If $c=0$, it is zero and positive. Suppose $c>0$. For any Hermitian $h$, the Banach algebra exponential $e^{ith}$ is unitary for real $t$, so its norm is one by the C-star identity. ConsequentlyWriting $f(h)=u+iv$ and squaring moduli gives $|f(e^{ith}){|}^2=c^2-2cvt+O(t^2)$. The bound for both signs of small $t$ forces $v=0$, so $f(h)$ is real.

For a Hermitian element $a$, repeated applications of the C-star identity give $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula therefore gives $\Vert a\Vert =r(a)$. Take $h=x^{\ast }x$; by the permitted spectral positivity assumption, $\sigma (h)\subseteq [0,\Vert h\Vert ]$. If $M=\Vert h\Vert >0$, affine spectral mapping and the just-proved Hermitian norm identity give $\Vert 1-h/M\Vert \le 1$. Hence $|c-f(h)/M|\le c$. Since $f(h)$ is real, this implies $f(h)\ge 0$. If $M=0$, $h=0$ and the result is immediate. Thus a bounded functional with $\Vert f\Vert =f(1)$ is positive.

For the spectral-value request, let $\lambda \in \sigma (x)$ and define a functional on the linear span of $1,x$ by $g(a1+bx)=a+b\lambda$. If $b\ne 0$, the scalar $a+b\lambda$ belongs to $\sigma (a1+bx)$: subtracting it from that element gives $b(x-\lambda 1)$, which is not invertible. The same norm bound is immediate for $b=0$. ThereforeThis also proves well-definedness if that span is one-dimensional. We have $g(1)=1$ and $\Vert g\Vert =1$. The complex Hahn-Banach theorem extends it to a functional $f$ on all of $A$ with norm one and $f(1)=1$. The converse just proved makes that extension positive. Thus there is a state with $f(x)=\lambda$, the spectral values attained by C-star states result.

Finally, for $x\ne 0$, write $x=h+ik$ with $h,k$ Hermitian. At least one of them is nonzero. Its norm equals its spectral radius, so compactness of its nonempty spectrum supplies a nonzero spectral value. Apply the preceding state construction to that Hermitian element. Part (i) makes $f(h),f(k)$ real; the selected one is nonzero, so $f(x)=f(h)+if(k)\ne 0$. This proves positive functionals separate every nonzero element, the states separate elements of a C-star algebra property. Looking only at $\sigma (x)$ would not prove this last step, since a nonzero quasinilpotent $x$ can have spectrum $\{0\}$.

Write $x=h+ik$ with both terms Hermitian. At least one is nonzero; its norm equals its spectral radius, so it has a nonzero spectral value. Spectral values attained by C-star states gives a state taking that nonzero value on the chosen term. Both state values on $h,k$ are real, so the corresponding complex value on $x$ cannot vanish. This also detects nonzero quasinilpotent elements, which cannot be separated by looking only at their own spectrum.