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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.

Cut the square model of the Klein bottle along the midline parallel to the pair of sides whose identification reverses orientation. Each half becomes a Möbius band, and the two new boundary circles are identified. If $a$ and $b$ are the core loops of these two bands, each boundary travels twice around its core. The Seifert-van Kampen theorem therefore gives

Writing $x=a^{-1}b$ and $y=a$ converts this into the standard Klein-bottle presentationThe orientation homomorphism sends both $a$ and $b$ to the nonzero element of $C_2$. Its kernel is the index-two subgroupBy the classification of connected covering spaces, $H$ determines the orientation double cover of the Klein bottleChoosing torus generators $\alpha ,\beta$ along the two translation directions, we may takeThese commute: the standard relation says that $a$ conjugates $a^{-1}b=x$ to $x^{-1}$, so $a^2$ centralizes $x$.

The space $Y$ is the mapping cylinder of $p$ with its unused end omitted. Sliding $(x,t)$ to $(x,0)$ gives a deformation retraction of $Y$ onto the bottom quotient $T^2/\sim \cong K$. Hence

Finally suppose $U\subseteq X$ is open and homeomorphic to $Y$, and identify the bottom copy of $K$ inside $U$. This $K$ is compact and therefore closed in the Hausdorff space $X$. Apply van Kampen, in its groupoid form if intersections are disconnected, toThe intersection $U\setminus K$ deformation retracts to a positive-height torus, and its map to ${\pi }_1(U)\cong {\pi }_1(K)$ has image preciselyThe orientation homomorphismis zero on this intersection, so it is compatible with the trivial homomorphism from ${\pi }_1(X\setminus K)$ to $C_2$. The pushout property in van Kampen extends it to a surjectionThereforeThis is the Klein-bottle mapping-cylinder obstruction to simple connectivity.

Two nonzero fractional ideals $I,J$ of ${\mathcal{O}}_K$ are in the same ideal class whenfor some $\alpha \in K^{\times }$, equivalently when $I{J}^{-1}$ is principal. Multiplication is well defined on classes:The identity is $[{\mathcal{O}}_K]$, the inverse of $[I]$ is $[I^{-1}]$, and commutativity comes from ideal multiplication. Nonzero fractional ideals are invertible because ${\mathcal{O}}_K$ is a Dedekind domain. Thus these classes form the abelian ideal class group $\mathrm{Cl}(K)$.

For finiteness, begin with a nonzero integral ideal $I$. By hypothesis choose $0\ne \alpha \in I$ withSince $(\alpha )\subseteq I$, the idealis integral, represents $[I{]}^{-1}$, and satisfiesHence every class has an integral representative of bounded norm. There are only finitely many integral ideals of bounded norm, so $\mathrm{Cl}(K)$ is finite. This is the bounded-norm ideal representatives argument.

Now take $K=\mathbb{Q}(\sqrt{-33})$. Since $-33\equiv 3(\mathrm{mod}4)$,The imaginary-quadratic Minkowski bound for ideal classes isEvery class is therefore represented by an integral ideal of norm at most seven.

The ramified prime ideals above two and three arewithNeither is principal, since the norm equationhas no integer solution. Their product is also nonprincipal, since $a^2+33b^2=6$ has no solution. Thusare four distinct classes, all of order at most two.

The prime five is inert, so it contributes no ideal of norm five. The prime seven splits asSincecomparison of norms, all equal to $42$, confirms the factorization and givesThe conjugate prime has the inverse class, which is the same because this class has order two. Ideals of norms four and six yield respectively the principal class and $[{\mathfrak{p}}_2{\mathfrak{p}}_3]$. The Minkowski bound now shows that the four displayed classes exhaust the group. ThereforeThis is the Ideal class group of Q of square root of minus thirty-three.