Solution
= Solution
To prove (a) implies (c), shrink to a contractible coordinate neighborhood. The real <Poincare lemma> gives a real one-form $\eta$ with $\omega=d\eta$. Write $\eta=\eta^{1,0}+\eta^{0,1}$. Because $\omega$ has type $(1,1)$, $\bar\partial\eta^{0,1}=0$ and $\partial\eta^{1,0}=0$. The <Dolbeault-Poincaré lemma> supplies a <smooth function> $\psi$ with $\eta^{0,1}=\bar\partial\psi$. Reality of $\eta$ gives $\eta^{1,0}=\partial\overline\psi$. Thus
$$
\omega=\partial\bar\partial\psi+\bar\partial\partial\overline\psi
=\partial\bar\partial(\psi-\overline\psi)
=i\partial\bar\partial(2\operatorname{Im}\psi).
$$
Taking the real function $f=2\operatorname{Im}\psi$ gives the <local real potential for a closed (1,1)-form>. Conversely, $d(i\partial\bar\partial f)=0$, by $\partial^2=\bar\partial^2=0$ and anticommutation. Hence (a) and (c) are equivalent, completing all three conditions.
For the unheaded radial continuation, take the real potential $f$ as in (c). Rotation invariance makes $f$ constant on every circle of radius $r>0$, so
$$
\boxed{u(t)=f(e^{t/2}),\qquad t\in\mathbb R,}
$$
is well defined and smooth by composition. This avoids treating a <smooth function> as if it had a convergent Taylor series. At nonzero $z$, put $t=\log|z|^2$. The identities $\partial_z t=1/z$ and $\partial_{\bar z}t=1/\bar z$ give
$$
f_{z\bar z}(z)=\frac{u''(t)}{|z|^2}=e^{-t}u''(t).
$$
Let $q(z)=f_{z\bar z}(z)$, which is smooth on the whole plane. Then
$$
\omega=iq(z)\,dz\wedge d\bar z,
\qquad g=2q(z)(dx^2+dy^2).
$$
At the origin, rotation invariance gives the finite <Taylor expansion> $f(z)=f(0)+b|z|^2+O(|z|^4)$, with $b=q(0)$. Alternatively continuity of $q$ directly gives
$$
\lim_{t\to-\infty}e^{-t}u''(t)=q(0).
$$
The metric is positive away from the origin exactly when $u''(t)>0$ for all finite $t$, and it is positive at the origin exactly when this limit is positive. Smoothness at the origin is already supplied by the original smooth potential. Consequently the <positivity criterion for a radial Kähler potential> is
$$
\boxed{u''(t)>0\ \text{for every }t\in\mathbb R,
\qquad\lim_{t\to-\infty}e^{-t}u''(t)>0.}
$$
Closedness is automatic from the potential, and positivity completes the Kähler condition. No completeness of this metric is asserted. The <rotation-invariant Kähler potential on the complex plane> is understood as real; if a complex radial potential is initially allowed with real $\omega$, its smooth radial imaginary part is harmonic and hence constant, so that constant can be removed.