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Because the centered normal distribution density $g_t$ is nonnegative and has integral one, Fubini's theorem and translation invariance giveThus $f\ast g_t\in L^1$.

Use the angular-frequency conventionThe convolution theorem and the Gaussian transform giveSince $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:This proves the Fourier inversion theorem for the convolution directly.

Now additionally suppose $\hat{f}\in L^1$. For every $x$, the dominated convergence theorem givesFourier inversion identifies the right-hand side with $f(x)$ for Lebesgue almost everywhere $x$. Therefore the Gaussian approximate identity satisfies

For a divisor on an algebraic curve $D={\sum }_p{a}_{p}p$ over an algebraically closed field, its degree isFor a nonzero rational function $f$, its principal divisor iswhere zeros have positive order and poles have negative order.

Now let $D={\sum }_i{a}_i{p}_i$ have degree zero on the projective line. Choose a homogeneous linear form $L_i(X_0,X_1)$ whose zero is $p_i$. Since ${\sum }_i{a}_i=0$, the expressionis homogeneous of degree zero and therefore defines a rational function on ${\mathbb{P}}^1$. Each $L_i$ has one simple zero, and the cancellation of total degree removes any common scaling ambiguity, soThus every degree-zero divisor on ${\mathbb{P}}^1$ is principal. If $E$ and $E^{\prime }$ have the same degree, then $E-E^{\prime }$ is principal, whenceThis is the divisor class on the projective line.

Write $p_{\infty }=[1:0]$. The coordinate $t=X_0/X_1$ has no critical zero or pole in the finite chart. Near infinity use $s=X_1/X_0=t^{-1}$; thenHence the rational differential has one double pole at infinity and no other zero or pole:which represents the canonical divisor of the projective line.

If $D\sim m{K}_{{\mathbb{P}}^1}$, then $D\sim -2m{p}_{\infty }$, and linear equivalence of divisors gives $L(D)\cong L(-2m{p}_{\infty })$. A function in the latter space has no finite poles, so it is a polynomial, and its degree is at most $-2m$. ThereforeThis calculation uses only the rational functions on ${\mathbb{P}}^1$, rather than the Riemann-Roch theorem.

Finally, suppose distinct $p,q\in C$ satisfy $p-q=(f)$. The function $f$ has one simple pole and therefore defines a degree-one finite morphism $f:C\to {\mathbb{P}}^1$. A degree-one finite morphism between smooth projective curves is an isomorphism. It would follow that $C\cong {\mathbb{P}}^1$ and hence that $C$ has genus zero, contrary to the hypothesis. Thus the principal divisor with one simple zero and one simple pole cannot occur on $C$:

For $s\in \mathbb{R}$, the Sobolev space $H^s({\mathbb{R}}^n)$ consists of tempered distributions $u$ such thatFor a multi-index $\alpha$,Sincewe obtain the Sobolev derivative estimateThus differentiation is bounded and linear between the stated spaces.

Taking the Fourier transform ofgivesThe only possible solution is thereforeMoreover,This proves existence, uniqueness, and bounded dependence. Conversely, $u\mapsto (-\Delta +1)u$ maps $H^{s+2}$ boundedly to $H^s$, sois a linear isomorphism with inverse multiplier $(1+|\xi {|}^2{)}^{-1}$. This is the massive Laplacian isomorphism on Sobolev spaces.

Finally, the assumed estimate and Plancherel theorem implyThus $(u_j)$ is bounded in $H^2({\mathbb{R}}^n)$. The functions and all their first derivatives are supported in the fixed bounded set $\overline{\Omega }$. Their $H^1$ bounds give uniform translation estimatesfor $v=u_j$ and $v=D_{\ell }{u}_j$. Tight support and the Rellich-Kondrashov compactness theorem for H01, equivalently the Fourier compactness criterion, therefore provide a common subsequence for which $u_j$ and every first derivative converge strongly in $L^2$. HenceThis is compactness from bounded support and an H2 bound.