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For fixed $m$, the reverse triangle inequality givesThe right side tends uniformly to zero as $n,k\to \infty$ because $(x_n)$ is a Cauchy sequence. Hence $(d(x_m,x_n){)}_n$ is a real Cauchy sequence. By completeness of $\mathbb{R}$, it has a finite, nonnegative limit:This conclusion uses completeness only of the real line, without assuming that $X$ is complete.

TakeBoth functions have values in $[1,\infty )$. The reverse triangle inequality gives $|g(x)-g(y)|\le |x-y|$, so $g$ is Lipschitz continuous and hence has uniform continuity.

For $x_n=n$ and $y_n=n+1/n$, the distances satisfy $|x_n-y_n|=1/n\to 0$, whereasIn particular tolerance $\epsilon =1$ defeats every proposed $\delta$: choose $n$ with $1/n<\delta$. Thus $f$ does not have uniform continuity on $\mathbb{R}$.

Put $S={\sum }_{j=0}^{n-1}|a_j|$ and choose $R>\max (1,2S/|a_n|)$. For $|z|>R$, the triangle inequality givesThe reverse triangle inequality therefore yieldsIf $P$ had no root, the function $h=1/P$ would be an entire function. The bound just proved makes $h(z)\to 0$ as $|z|\to \infty$. On the remaining closed disk, continuity and the extreme value theorem make $h$ bounded. Thus $h$ is a bounded entire function, so Liouville's theorem makes it constant. Its limit at infinity forces that constant to be zero, impossible for a reciprocal.

Consequently $P$ has a root in $\mathbb{C}$. This proves the existence part of the fundamental theorem of algebra. The invoked Liouville theorem follows from the Cauchy integral formula: the Cauchy estimate $|h^{\prime }(w)|\le M/r$ on arbitrarily large circles gives $h^{\prime }=0$ when $h$ is bounded by $M$.

Put $D(z)={\sinh }^{2}z+{\cosh }^{2}a$ and $S=\sinh (2a)>0$. Traverse the rectangle $-R\to R\to R+\pi i\to -R+\pi i\to -R$ with positive orientation of a contour, taking $R>a$. Part (i) shows that the only enclosed poles of $f(z)=z/D(z)$ areThey are simple poles becauseThe residue formula for a quotient at a simple denominator zero givesTheir imaginary terms cancel and their residues sum to $-2a/S$. Thus Cauchy's residue theorem gives

The choice of numerator $z$ makes the horizontal sides useful. The hyperbolic sine satisfies $\sinh (z+\pi i)=-\sinh z$, so $D(x+\pi i)=D(x)$. Consequently the lower and upper contour integrals, with their actual orientations, add toIt remains to justify discarding both vertical sides. On either side $z=\pm R+iy$, $0\le y\le \pi$, the reverse triangle inequality and part (b)(ii) implyThis lower bound is positive for sufficiently large $R$. Also $|z|\le R+\pi$ and each vertical side has length $\pi$. The estimation lemma bounds the absolute value of the sum $V_R$ of the vertical contributions byThe real improper integral converges absolutely, since its integrand is continuous on the real axis and decays like $4e^{-2|x|}$ at infinity. Combining all four sides and taking $R\to \infty$ givesThereforeAn independent real-variable check uses the hyperbolic tangent substitution $u=\tanh x$: the integral becomes ${\int }_{-1}^{1}du/({\cosh }^{2}a-u^2{\sinh }^{2}a)$. Its antiderivative uses the inverse hyperbolic tangent, giving $2\mathrm{arctanh}(\tanh a)/(\sinh a\cosh a)=4a/\sinh (2a)$. As $a\downarrow 0$, this tends to $2$, agreeing with ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{2}xdx=2$ and checking the factor of two.

Fix $0<r<1$ and $k\ge 1$. On $|z|=1$ and $|z_0|<r$, the reverse triangle inequality gives $|z-z_0|\ge 1-r$. Subtract the two Cauchy integral formula expressions and apply the estimation lemma, using circle length $2\pi$:Thus every fixed-order derivative converges uniformly on each smaller disc. The right-hand factor tends to zero by the assumed uniform convergence on $D$. Keeping a positive gap $1-r$ from the integration contour is the essential feature of this Cauchy derivative convergence bound. If $r\le 0$, the specified open disc is empty and the claim is vacuous.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

Suppose $w\in \mathbb{C}$ were omitted by $g$. Thenwould be an entire function. The reverse triangle inequality gives $|g(z)-w|\ge |g(z)|-|w|\to \infty$ as $|z|\to \infty$, so $F(z)\to 0$ there. On a sufficiently large closed disk, $F$ is bounded by continuity and the extreme value theorem; outside it, the limit gives a bound. The Liouville theorem therefore makes $F$ constant, and the limit forces $F\equiv 0$.

That is impossible, since $1/(g(z)-w)$ is nonzero at every finite $z$. No value $w$ can be omitted, and hence

For fixed $x\in X$, the function $a\mapsto d(x,a)$ on $A$ is continuous, indeed $1$-Lipschitz continuous by the reverse triangle inequality. A continuous real-valued function on a nonempty compact space attains its minimum by the extreme value theorem. Hence some $a_0\in A$ satisfiesThe metric is symmetric, so this is also $d(a_0,x)$. Compactness supplies attainment of the infimum.

On the upper semicircular contour, $|z|=R$ and the reverse triangle inequality gives $|z^4+1|\ge R^4-1>0$. Its arc length is $\pi R$, so the estimation lemma givesThe arc is open, so an ordinary integral sign is appropriate even though the source uses a closed-integral symbol there. Subtracting the vanishing arc contribution from the closed contour integral gives the real-line improper integral $\pi /\sqrt{2}$. Its integrand is an even function, and its tail is $O(x^{-2})$, ensuring convergence. Thus

Complete the square to write the objective asLet $z=(x,y)$ and $a=(1,2)$, so $\Vert a\Vert =\sqrt{5}$ while feasibility requires $\Vert z\Vert \ge \sqrt{20}=2\sqrt{5}$. The reverse triangle inequality givesThe objective is therefore at least zero. Equality is attained by $z=2a=(2,4)$, which lies on the boundary. Equality requires the vectors to have the same direction and $\Vert z\Vert =2\sqrt{5}$, so this minimizer is unique. HenceIt also satisfies the Karush-Kuhn-Tucker conditions with $\lambda =1/2$. The direct distance argument proves global optimality even though the exterior of a disk is not a convex set; this is nearest-point minimization outside a Euclidean ball.

Let $0<a<b<\infty$ and $1\le p<\infty$. A radial function in $W^{1,p}(\{a<|x|<b\})$ has a radial profile in $W^{1,p}(a,b)$; its first weak derivative is the profile of the radial derivative. To see the identity, sum the weak derivative tests against the components of $\varphi (r)x/r$ to obtain $\int g\varphi r^{n-1}=-\int \bar{u}(\varphi r^{n-1}{)}^{\prime }$. Substitute $\varphi =\zeta /r^{n-1}$ for arbitrary $\zeta \in C_c^{\infty }(a,b)$. The weighted Lp norms and $a>0$ give the displayed bound. Radiality of the radial derivative follows by applying a rotation to the weak derivative identity: if $u(Qx)=u(x)$, then $\nabla u(Qx)=Q\nabla u(x)$ almost everywhere, so the radial scalar product is invariant.

For $p>1$, the one-dimensional Sobolev representative obeys $|\bar{u}(r)-\bar{u}(s)|\le |r-s{|}^{1-1/p}\Vert {\bar{u}}^{\prime }{\Vert }_p$. The reverse triangle inequality transfers this to $u$, giving a continuous embedding of this radial Sobolev space into $C^{0,1-1/p}$ on the annulus. The two Hölder seminorms are equal: one inequality follows from $||x|-|y||\le |x-y|$, and the reverse follows by putting both points on the same ray. The restriction $a>0$ is essential for the unweighted profile bound.

Let $\Gamma$ be a piecewise smooth contour forming a simple closed curve, positively oriented: travel counterclockwise, with the interior on the left. Suppose $f$ is holomorphic on an open set containing $\Gamma$ and its interior, except at finitely many isolated singularities $a_1,\dots ,a_m$ strictly inside. ThenNo isolated singularity may lie on $\Gamma$: the ordinary contour integral must be defined there. Reversing the direction of the contour reverses the sign of the answer. The theorem applies to essential singularities as well as poles. ^[ref]

Return to $f(z)=1/(z^2+1)$. A counterclockwise circle $|z|=2$ encloses both simple poles, soYet $f$ is nonzero everywhere it is defined. A zero contour integral does not imply a zero function, or even absence of poles: the enclosed residues can cancel. By contrast, the counterclockwise circle $|z-i|=1/2$ encloses only $i$, givingThe picture compares these two selections. Changing the shape of a contour while keeping the same enclosed poles leaves the contour integral unchanged, provided the contour deformation stays where $f$ is holomorphic.

This is the practical appeal of the residue theorem: local Laurent coefficients determine a global contour integral. For a rational function, a few simple pole computations can replace a long parametrized calculation. It also makes real improper integrals accessible by treating the real line as part of a closed contour in the complex plane.

To compute $I={\int }_{-\infty }^{\infty }dx/(1+x^2)$, take $R>1$ and travel along the real line from $-R$ to $R$, then return along the upper semicircle $A_R$, parametrized by $z=R{e}^{it}$ for $0\le t\le \pi$. This contour is positively oriented and encloses the simple pole $i$ only; $-i$ is outside. HenceThe added arc must be controlled before taking a limit. On $A_R$, the reverse triangle inequality gives $|z^2+1|\ge R^2-1$, and the arc length is $\pi R$. The estimation lemma therefore yieldsThe denominator grows quadratically while the arc length grows only linearly. This quantitative comparison, rather than a sketch alone, explains why closing the contour does not alter the limiting real improper integral. It even supplies an error bound: the difference between $\pi$ and the finite real integral is at most $\pi R/(R^2-1)$.

For a numerical check, $R=10$ gives the finite real integral $2\arctan (10)\approx 2.94226$, leaving about $0.19934$ to reach $\pi$. The arc bound is $10\pi /99\approx 0.31733$, comfortably larger than that error. The graph shows both quantities approaching zero. The exact tail here can be checked with the antiderivative $\arctan x$; the residue theorem method remains useful for improper integrals where a convenient real antiderivative is unavailable.

Finally, both one-sided improper integrals converge absolutely: for $|x|\ge 1$, $0<1/(1+x^2)\le 1/x^2$, whose tails are integrable. Thus the symmetric limit computes the ordinary improper integral, not merely a Cauchy principal value, andClosing below instead would enclose the simple pole $-i$. With the same left-to-right traversal of the real line, that contour is clockwise, so its contour integral is $-2\pi i\mathrm{Res}(f,-i)=\pi$ again. This provides a useful sign check: the opposite residue is accompanied by the opposite direction around the enclosed region.

The same strategy extends to many other rational functions: choose a contour, identify its enclosed isolated singularities, compute their residues, and bound the added pieces. The choice of contour and the bound are as much part of the calculation as the residues. ^[ref]

Let the locations $a_n$ be distinct and bounded, and let $c_n>0$ with ${\sum }_n{c}_n<\infty$. The Weierstrass M-test gives locally uniform convergence of $G$, so the uniform limit theorem makes it continuous. The reverse triangle inequality bounds each term's difference quotient by $c_n$. Consequently a uniformly small tail permits passing each one-sided difference-quotient limit through the series. At every point outside the locations, $G^{\prime }(x)={\sum }_n{c}_n\mathrm{sgn}(x-a_n)$; at $a_m$, the right derivative minus the left derivative is $2c_m>0$. Thus the function has precisely the prescribed cusp locations, even when they accumulate at another point where the function remains differentiable.