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Evaluate the smooth one-form on unit velocities to obtain $f(x,v)={\theta }_x(v)$. Its integral along a periodic orbit of the geodesic flow is exactly the one-form integral along the corresponding closed geodesic. Negative curvature makes this geodesic flow an Anosov flow; the geodesic flow of a connected closed negatively curved surface is also topologically transitive. Apply the smooth Livsic theorem to obtain $u\in C^{\infty }(SM)$ withIf there is more than one connected component, apply this argument to each component separately.

The vertical vector field of a surface unit tangent bundle differentiates rotation of $v$, soFor each $x$, writing ${\theta }_x(v)=A\cos \vartheta +B\sin \vartheta$ shows explicitly thatThe fibre disintegration of Liouville volume of a surface geodesic flow is $d\mu =d{A}_{g}d\vartheta$. Integrating the displayed equality over the base therefore gives $\Vert VXu{\Vert }_{L^2}^2=\Vert Xu{\Vert }_{L^2}^2$. Substitution into the supplied Pestov identity for a geodesic flow yieldsBoth terms are nonnegative and $-K\ge k_0>0$. Thus $Vu=0$ everywhere. Since every unit-circle fibre is connected, $u$ is independent of velocity and descends to a smooth function $h$ on $M$: $u=h\circ \pi$. Its geodesic derivative isComparison with $Xu={\theta }_x(v)$, first on unit vectors and then on all vectors by linearity, provesThus the one-form is an exact differential form. This proves the degree-one geodesic X-ray transform conclusion directly from the given integral identity and the stated smooth coboundary theorem.