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For $s>n/2$ and $u\in H^s({\mathbb{R}}^n)$, Fourier inversion and Cauchy--Schwarz giveThe last integral is finite exactly when $s>n/2$. Thus the sobolev embedding theorem gives

To see why the endpoint relevant to $H^1({\mathbb{R}}^3)$ fails, choose a smooth cutoff $\chi$ supported near the origin and equal to one there, and setNear zero, $|u{|}^2$ contributes ${\int }_0^1{r}^{3/2}dr$, while $|\nabla u{|}^2$ contributes a constant multiple of ${\int }_0^1{r}^{-1/2}dr$. Both are finite, so $u\in H^1({\mathbb{R}}^3)$, but $u$ is unbounded.

The real Stone-Weierstrass theorem says that if $X$ is compact Hausdorff and $A\subset C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$.

Let $\overline{A}$ be the uniform closure. Polynomial approximation to $\sqrt{t}$ on a bounded interval shows that $|f|\in \overline{A}$ whenever $f\in \overline{A}$. Hence $\overline{A}$ is closed underGiven $h\in C(X)$, separation and the constants provide, for each $x,y$, a function in $A$ agreeing with $h$ at $x$ and $y$. Compactness first combines finitely many such functions by minima to obtain one that agrees at $x$ and lies below $h+\epsilon$ everywhere; a second finite cover and maxima produces $f\in \overline{A}$ with $|f-h|<\epsilon$. Thus $h\in \overline{A}$.

The space $C_b(\mathbb{R})$ is Banach because a uniform Cauchy sequence converges uniformly to a bounded continuous function. Compactness is essential: the algebra of bounded continuous functions having finite limits at both $+\infty$ and $-\infty$ contains constants and separates points, for example using $\tanh x$, but its uniform closure has the same limiting property. It cannot uniformly approximate $\sin x$, so it is not dense in $C_b(\mathbb{R})$.

Because $K$ is a subcomplex, $\partial C_k(K)\subseteq C_{k-1}(K)$, sois well defined and squares to zero. The short exact sequence of chain complexesgives the long exact sequenceTaking $L=\partial {\Delta }^{n+1}$ and using the contractibility of ${\Delta }^{n+1}$ givesFor $0<k<n$, exactness makes $H_k(K)$ a quotient of a subgroup of $H_{k+1}(L,K)$, whose rank is at mostThe same stated inequality for $k=n$ follows from $H_n(K)\hookrightarrow H_n(L)\cong \mathbb{Z}$. Finally, an $n$-cycle in the boundary sphere has equal signed coefficients on all $n$-faces. A proper subcomplex omits an $n$-face, so that coefficient and hence every coefficient is zero. Therefore $H_n(K)=0$.

Write $N={\prod }_i{p}_i^{e_i}$ with odd primes $p_i$. The Jacobi symbol iswhere each factor on the right is a Legendre symbol. The supplementary law isFor coprime positive odd $m,n$, quadratic reciprocity givesApplying this with $m=3$ givesFinally, factor $d=2^{e}m$ with $m$ odd. The supplementary laws for $-1$ and $2$, followed by reciprocity on every odd prime factor of $m$, express $(-d/a)$ solely in terms of the residue class of the positive odd integer $a$ modulo $d$ when $d\equiv 0$ or $3(\mathrm{mod}4)$. If $a\equiv b(\mathrm{mod}d)$, all those factors, including the zero cases, agree; hence