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A contour deformation changes the path of a contour integral through a region where its integrand is a holomorphic function. The Cauchy integral theorem preserves the integral when the endpoints are fixed and the connecting boundary terms vanish. Crossing a pole instead contributes the appropriately oriented residue; a branch cut limits which deformations are permitted on one analytic sheet.

For a piecewise differentiable parametrized contour $z(t)$, the contour integral is $\int f(z(t))z^{\prime }(t)dt$. It depends on the orientation of the contour. For a holomorphic function, the Cauchy integral theorem allows a contour deformation through regions without singularities.

For the forward implication, finite order of a distribution makes $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$ entire. Use a smooth cutoff in an $\epsilon$-neighborhood of the support set, with derivatives of order $j$ bounded by $C_j{\epsilon }^{-j}$. Taking $\epsilon =(1+|\zeta |{)}^{-1}$ in the finite-order estimate gives the polynomial factor and adds at most $e$ to the desired exponential bound. This shrinking cutoff recovers the exact support indicator, instead of an arbitrarily enlarged one.

For the converse, the real restriction of $F$ has polynomial growth and defines an inverse tempered distribution. If a test function $\phi$ is supported in $x\cdot \omega \ge R+\eta$ for a unit vector $\omega$ and $\eta >0$, contour shifting givesRepeated integration by parts gives, for every integer $M$,Choosing $2M>N+n$ justifies the shift by Cauchy integral theorem and bounds the pairing by $C^{\prime }(1+t{)}^{N+2M}{e}^{-\eta t}$, which tends to zero. Half-spaces of this form cover the complement of the ball, and a partition of unity proves the support inclusion. For general compact convex $K$, replace $R$ by $H_K(\omega )$ in each separating direction. Fourier inversion gives uniqueness. A related proof outline appears in Richard Melrose's distribution-theory problem set.