PomeLemma 11.45.15.85744C3971DC Let $V>0$ be continuous on $\mathbb P^1$, and let $T$ be a finite signed measure such that
\[ \int _{\mathbb P^1}(\Delta f-Vf)\, \mathrm{d}T=0 \qquad (f\in C^\infty (\mathbb P^1)). \]
Then $T=0$.
Lemma 11.45.1585744C3971DC
Proof. Let $H_ t$ be the sphere heat operator. On spherical harmonics of degree $\ell $, it acts by $\mathrm{e}^{-t\ell (\ell +1)}$. The harmonic series defines a smooth heat solution for $t>0$; for smooth initial functions, the maximum principle gives positivity and contraction in the uniform norm. Density of smooth functions extends $H_ t$ to a positive, strongly continuous contraction on $C(\mathbb P^1)$. Choose $M\geq \max V$ and define
\[ R_ Mg=\int _0^\infty \mathrm{e}^{-Mt}H_ tg\, \mathrm{d}t. \]
Then $\| R_ M\| \leq M^{-1}$ and $(M-\Delta )R_ Mg=g$ in distributions. For any $g\in C(\mathbb P^1)$, the map
\[ f\longmapsto R_ M((M-V)f-g) \]
contracts by at most $1-\min V/M<1$. Its continuous fixed point satisfies $\Delta f=Vf+g$ in distributions, with a continuous right-hand side. The functions $f_\varepsilon =H_\varepsilon f$ are smooth and satisfy
\[ f_\varepsilon \longrightarrow f,\qquad \Delta f_\varepsilon =H_\varepsilon (Vf+g)\longrightarrow Vf+g \]
uniformly. Thus $(\Delta -V)f_\varepsilon \to g$ uniformly. Testing against the finite measure $T$ gives $\int g\, \mathrm{d}T=0$ for every continuous $g$, hence $T=0$. $\square$
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