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Suppose a real function $g$ is differentiable at $a$, with $g(a)=0$ and $g^{\prime }(a)\ne 0$. The derivative definition gives $g(a+h)=g^{\prime }(a)h+o(|h|)$. The reverse triangle inequality then impliesThus $|g|$ has unequal one-sided derivatives, $|g^{\prime }(a)|$ on the right and $-|g^{\prime }(a)|$ on the left. More generally, multiplication by a function $w$ continuous at $a$ with $w(a)\ne 0$ gives right and left slopes $w(a)|g^{\prime }(a)|$ and $-w(a)|g^{\prime }(a)|$. This elementary expansion explains how a smooth oscillation can acquire infinitely many corners under the absolute value function, even after multiplication by a factor that makes it differentiable at an accumulation point.

Complex analysis studies functions of a complex number, especially holomorphic functions. Its distinctive feature is that a single complex derivative controls variation in every direction of the complex plane. The map $f(z)=z^2$ gives a first picture: straight coordinate lines become curved, yet small perpendicular arrows remain perpendicular wherever $f^{\prime }(z)\ne 0$.

The bottom panels show displacements rather than absolute positions. Multiplication by $f^{\prime }(1+i)=2+2i$ predicts the image of each small arrow; the exact image differs by $h^2$. The origin is exceptional because $f^{\prime }(0)=0$. This distinction between local conformal maps and a critical point is central: complex differentiability imposes strong geometric structure, without making every holomorphic function globally one-to-one.

The remarkable next step is that holomorphic functions are determined inside a circle by their values on its boundary. Precisely, let $R>0$, let $D=\{z:|z-a|<R\}$, and suppose $f$ is a holomorphic function on an open set containing the closed disc $\overline{D}$. Orient its boundary counterclockwise. The Cauchy integral formula states that, for every $w\in D$,The contour integral combines boundary values with a kernel depending on the requested interior point. It is a reconstruction formula, not merely a trick for evaluating an integral. The hypothesis about the closed disc ensures that all boundary values and the contour integral are well defined. ^[ref]

At the centre, substituting $z=a+R{e}^{i\theta }$ cancels the denominator against $dz=iR{e}^{i\theta }d\theta$:Thus the centre value is the angular average of the boundary values. For the polynomial $f(z)=z^2$, $a=1+i$ and $R=1/2$, expand the boundary data asBoth oscillating terms have zero average, leaving $f(a)=2i$. Away from the centre the weights change: on the unit circle,With $f(z)=z^2$ and $w=1/2$, expanding the denominator as a geometric series leaves only its second-power term after integration, giving $f(1/2)=1/4$. This also explains how the Cauchy integral formula recovers an entire Taylor series from boundary data.

When a holomorphic function has isolated singularities, a different boundary calculation detects their residues. Suppose $\Gamma$ is a counterclockwise simple closed contour and $f$ is holomorphic on a neighbourhood of it and its interior except at finitely many interior isolated singularities $a_j$. The residue theorem givesA residue is the coefficient of $(z-a_j{)}^{-1}$ in the Laurent series. A short proof idea is to deform the contour to small circles around the isolated singularities using the Cauchy integral theorem. Integrating the Laurent series on each circle kills every power except $-1$. ^[ref]

Here is how a complex pole determines a real improper integral. Set $g(z)=1/(1+z^2)$ and use the contour consisting of the segment from $-R$ to $R$, followed by the upper semicircle $A_R$, with $R>1$. Only the simple pole $i$ lies inside, andThe arc must actually disappear. On $A_R$, the reverse triangle inequality gives $|1+z^2|\ge R^2-1$, and its length is $\pi R$, soThe real integrand is bounded by $x^{-2}$ for $|x|\ge 1$, so both tails converge separately. Taking the limit therefore proves the ordinary improper integralFor example, at $R=10$ the real segment contributes $2\arctan (10)\approx 2.942255$, and the arc contributes $0.199337$, below the bound $10\pi /99\approx 0.317333$. The usefulness of the residue theorem is that the same strategy works for many rational functions whose real antiderivatives are cumbersome: find the enclosed poles, compute their residues, and control the connecting arc. ^[ref]

Historically, Cauchy's 1825 memoir on integrals with imaginary limits develops complex contour integration and its use in transforming and evaluating definite integrals. The opening sections define complex path integrals, discuss changing the path, and examine what happens near singular points. These are recognizable ancestors of the modern link between holomorphic functions, contours and residues. ^[ref]

For four points $x_1,x_2,y_1,y_2$ in a metric space,Both steps follow from the triangle inequality. The first also holds with an absolute value around the difference, by the reverse triangle inequality. This transports a difference of distances measured from one point into a sum measured from another.

If $\delta ={\min }_i(|a_{ii}|-{\sum }_{j\ne i}|a_{ij}|)>0$, then $A$ is invertible and its operator norm on ${\ell }^{\infty }$ satisfies the displayed bound. For $Ax=b$, choose $i$ maximizing $|x_i|$. The reverse triangle inequality gives $|b_i|\ge \delta \Vert x{\Vert }_{{\ell }^{\infty }}$. Taking $b=0$ proves injectivity and hence invertibility in finite dimensions; taking arbitrary $b$ proves the bound. The estimate is useful for converting strict diagonal dominance into quantitative matrix inverse stability.

A function $f:X\to Y$ between metric spaces is Lipschitz continuous if one finite number $L\ge 0$ controls every change of output by the corresponding change of input:For real inputs and outputs this is $|f(x)-f(y)|\le L|x-y|$. The same bound must work for every pair in the specified domain of a function. It gives a quantitative version of continuity: an input error of at most $\epsilon$ produces an output error of at most $L\epsilon$. The least possible bound is the Lipschitz constant. Lebl presents the real and metric-space definitions and their connection with uniform continuity. ^[ref]

Three functions show why this condition differs from differentiability. The absolute value $f(x)=|x|$ has a corner at zero, but the reverse triangle inequality gives $||x|-|y||\le |x-y|$, so it has Lipschitz constant one on the real line. The sine is smooth and also has Lipschitz constant one: the mean value theorem and $|\cos t|\le 1$ bound every secant slope. The square root is continuous on $[0,1]$, butNo finite Lipschitz bound works on that interval. At $h=1,1/4,1/16,1/64$, the ratios are $1,2,4,8$. Smaller input changes near zero therefore require larger amplification factors.

For a real function, Lipschitz continuity means that the absolute slopes of all secant lines are bounded. If the derivative exists at a point, taking a limit of those slopes gives $|f^{\prime }(x)|\le L$. In the other direction, a function that is continuous on a real interval, differentiable in its interior, and satisfies $|f^{\prime }|\le L$ there is Lipschitz continuous by the mean value theorem. The interval hypothesis matters: the theorem uses the segment connecting the two inputs.

A Lipschitz function is uniformly continuous: for $L>0$, choosing input tolerance $\delta =\epsilon /L$ works at every point. If $L=0$, the function is constant. The converse fails; the square root on $[0,1]$ is uniformly continuous, and satisfies the weaker Hölder continuity bound $|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}$, without a finite Lipschitz constant. These distinctions are developed in Lebl's worked examples. ^[ref]

The usefulness is controlled sensitivity. Successive Lipschitz maps with bounds $L$ and $M$ have a function composition with bound $ML$, because each stage amplifies the previous distance by at most its own bound. For an ordinary differential equation, Lipschitz continuity in the unknown state is the condition that makes nearby trial solutions controllably close in the Picard-Lindelöf theorem. For incomplete data, the McShane extension theorem fills in unobserved values while preserving the same bound. This one inequality connects error estimates, uniqueness, and interpolation.

Suppose $0<\Vert a\Vert <R$ in a Euclidean space. Among points outside or on the radius-$R$ ball centred at zero, the unique closest point to $a$ is $x_{\ast }=Ra/\Vert a\Vert$. Its squared distance is $(R-\Vert a\Vert {)}^2$.

Indeed, the reverse triangle inequality gives $\Vert x-a\Vert \ge \Vert x\Vert -\Vert a\Vert \ge R-\Vert a\Vert$. Equality requires $\Vert x\Vert =R$ and $x$ pointing in the same direction as $a$, giving the stated unique point. If $a=0$, every point on the sphere is a minimizer. If $\Vert a\Vert \ge R$, the point $a$ itself is feasible and the minimum distance is zero.

The exterior is not a convex set, so a direct geometric argument is useful: it certifies the global answer without treating every candidate satisfying the Karush-Kuhn-Tucker conditions as a minimum.

For $a,b\ge 0$ and $c\ge 1$,Apply this to the first component with $a=|z_2|$, $b=|w_2|$, $c=x^2+1$. The reverse triangle inequality givesFor the second component use $a=|z_1|$, $b=|w_1|$, $c=1$ to obtain $|F_2(x,Z)-F_2(x,W)|\le |z_1-w_1|$. Taking the maximum yieldsThus $F$ is globally Lipschitz continuous in the dependent vector, uniformly in $x$, with Lipschitz constant $1$. Differentiability of the absolute-value function at zero is unnecessary.

Periodicity reduces all the indicated vertical segments to the compact segment $1/2+i[-1,1]$, which misses every pole. Continuity therefore gives the first uniform bound. In fact an explicit identity gives a stronger estimate for every real $y$:Thus we can take $D_1=\pi$.

Write $L=m+1/2$. On the vertical sides of the square the same bound holds, since both real coordinates are half-integers. On either horizontal side $|\mathrm{Im}z|=L\ge 1/2$. For $\mathrm{Im}z=y>0$, put $w=e^{2\pi iz}$, so $|w|=e^{-2\pi y}$. Usinggives $|g(z)|\le \pi (1+e^{-2\pi y})/(1-e^{-2\pi y})\le \pi \coth (\pi /2)$. Conjugation gives the same estimate below the real axis. Hence an explicit bound valid on every square isFor $L>|b|$, the contour has length $8L$ and every point on it has $|z|\ge L$. The reverse triangle inequality yields $|z^2-b^2|\ge L^2-b^2$, so the ML inequality givesOrient the square positively. Its enclosed poles at the integers $-m,\dots ,m$ have residues $1/(n^2-b^2)$. Since $b$ is nonintegral, the additional poles $b$ and $-b$ are distinct from those integers and from one another. Their residues are $g(b)/(2b)$ and $g(-b)/(-2b)$, both equal to $g(b)/(2b)$ because $g$ is odd. Their combined contribution is therefore $\pi \cot (\pi b)/b$.

The residue theorem givesThe series is absolutely convergent, since its terms are $O(n^{-2})$. Taking the limit, then including the $n=0$ term, yieldsThis is cotangent contour summation. Keeping the $n=0$ term separate until the last step determines the sign of the $1/(2b^2)$ contribution.

The continuous dual space $X^{\prime }$ consists of all bounded complex-linear functionals $f:X\to \mathbb{C}$, with pointwise vector operations and norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. This is finite by boundedness. The scalar norm identity and triangle inequality follow under the supremum; if the norm is zero, scaling every nonzero vector into the unit ball shows that $f$ vanishes. Hence $X^{\prime }$ is a normed complex vector space.

For the dual of the convergent sequence space, take $\beta \in {\ell }^1$. The displayed formula is a bounded linear functional, since the series is absolutely convergent andIt is linear in $\beta$, so $\Vert \varphi (\beta )\Vert \le \Vert \beta {\Vert }_1$. For equality fix $N$, choose ${\alpha }_j$ for $j\le N$ of modulus at most one so that ${\alpha }_j{\beta }_{j+1}=|{\beta }_{j+1}|$, and make all later terms equal a constant $c$ of modulus at most one with $c{\beta }_1=|{\beta }_1|$. The choices for a zero coefficient can be arbitrary. This sequence lies in the unit ball of $c$ and converges to $c$. The reverse triangle inequality givesLetting $N\to \infty$ proves $\Vert \varphi (\beta )\Vert =\Vert \beta {\Vert }_1$, including injectivity.

For surjectivity let $F\in c^{\prime }$. By the permitted description of $c_0^{\prime }$, there is $\gamma \in {\ell }^1$ with $F(z)={\sum }_j{\gamma }_j{z}_j$ on $c_0$. Every $\alpha \in c$ decomposes as $\alpha -{\alpha }_{\infty }1\in c_0$ plus ${\alpha }_{\infty }1$. ConsequentlyTake ${\beta }_{j+1}={\gamma }_j$ and ${\beta }_1=F(1)-{\sum }_j{\gamma }_j$. This is an ${\ell }^1$ sequence and $F=\varphi (\beta )$, proving the stated isometric linear isomorphism onto $c^{\prime }$. No complex conjugation is inserted, since these are complex-linear dual functionals.

Assume $A$ is nonempty, as is necessary for an approximating point to exist. The reverse triangle inequality givesThus the distance from the fixed point $u$ is a continuous, indeed Lipschitz continuous, real-valued function on $A$. A continuous function on a nonempty compact set attains its minimum, so some $p\in A$ satisfiesIf empty compact sets are allowed, the printed statement needs this nonemptiness qualification.

Use the continuous representative of $\bar{u}$ furnished by the one-dimensional Sobolev embedding theorem, and define $u^{\ast }(x)=\bar{u}(|x|)$ on $A$. The one-dimensional Sobolev representative is absolutely continuous; Holder inequality applied to its fundamental theorem of calculus givesAlso $\Vert \bar{u}{\Vert }_{\infty }\le \Vert \bar{u}{\Vert }_6+\Vert {\bar{u}}^{\prime }{\Vert }_6$: select a point where $|\bar{u}|$ is no larger than its $L^6$ norm on the unit-length interval and apply the same fundamental theorem of calculus. Thus the one-dimensional Hölder norm is controlled by $\Vert \bar{u}{\Vert }_{W^{1,6}(1,2)}$.

The two Hölder seminorms must be compared using the correct distances. By the reverse triangle inequality,Conversely, take $x=re$ and $y=se$ for any fixed unit vector $e$. Then $|x-y|=|r-s|$, so taking suprema yields equality of the two Hölder seminorms. Their supremum norms are equal as well. Part (i) now givesThe representative agrees with $u$ almost everywhere, and the assignment is linear and injective on the radial Sobolev space. This proves the requested continuous embedding. The inner radius being positive makes the radial profile Sobolev norm comparable to its weighted radial version.

The definition of a convergent sequence isTo prove the product rule, convergence first gives an eventual bound $|u_n|\le |l|+1=:M$. ThenChoose $n$ sufficiently large that $|v_n-k|<ϵ/(2M)$ and $|u_n-l|<ϵ/[2(|k|+1)]$, as well as the bound on $u_n$. The right side is less than $ϵ$, proving $u_n{v}_n\to lk$ directly from the limit of a sequence definition.

If $l\ne 0$, eventually $|u_n-l|<|l|/2$, so the reverse triangle inequality gives $|u_n|>|l|/2$. ConsequentlyThus $1/u_n\to 1/l$. The condition $u_n\ne 0$ ensures the reciprocal sequence is defined at every index; the nonzero limit supplies the eventual uniform bound away from zero. Finally $\boxed{u_n=1/(n+1)\to 0}$ has every term nonzero, showing that this termwise property alone does not prevent a zero limit.

For order six, the support-midpoint sites are $x_i=i+3$, soBy part (b), the complete finite matrix is the symmetric five-diagonal matrixOnly indices from $0$ through $n$ are retained; the first and last rows are truncated, with no exterior coefficients or wraparound terms. For a row with all possible neighbors, the strict diagonal dominance margin is $(66-26-26-1-1)/120=1/10$. Boundary rows have an equal or larger margin.

Here is the appropriate inverse estimate with its proof. If a square matrix $A$ has $|a_{ii}|-{\sum }_{j\ne i}|a_{ij}|\ge d>0$, choose an index $i$ maximizing $|z_i|$. The reverse triangle inequality givesThus $Az=0$ implies $z=0$, and a square finite matrix is invertible. Applying the same estimate with $z=A^{-1}b$ proves $\Vert A^{-1}b{\Vert }_{{\ell }^{\infty }}\le d^{-1}\Vert b{\Vert }_{{\ell }^{\infty }}$. The inverse infinity-norm bound from diagonal dominance consequently yieldsThis also proves existence and uniqueness of the requested interpolating spline for every data vector and every finite $n$. The quintic cardinal spline midpoint collocation matrix includes the cases of one or two coefficients, where the dominance margin is even larger.

The entries in (b) satisfyThus $G$ has strict diagonal dominance with margin at least $1/3$. To prove the inverse infinity-norm bound from diagonal dominance directly, choose $i$ with $|a_i|=\Vert a{\Vert }_{{\ell }^{\infty }}$. The reverse triangle inequality givesThis also proves that the kernel is zero and hence that $G$ is invertible. Substituting $a=G^{-1}b$ and taking the operator norm givesCombined with (a), it yields $\Vert P_S{\Vert }_{\infty }\le 3$ for linear splines, independently of the knot spacings. The diagonal-dominance argument already proves the required estimate.

Suppose a connected subset $C\subseteq F$ contains distinct points $a,b$. The distance function $d_a(x)=|x-a|$ is a Lipschitz map with constant one, by the reverse triangle inequality. Its image $d_a(C)$ is connected and contains both $0$ and $|b-a|$, so it contains the whole interval $[0,|b-a|]$. Monotonicity and the Lipschitz inequality for Hausdorff dimension implya contradiction. Thus every nonempty connected subset of $F$ is a single point: $F$ is totally disconnected. The argument is the Hausdorff dimension of a connected set bound, and does not require $F$ to be closed.

Take $a=5$ and $x=\pi /2$. Since $5^k\equiv 1(\mathrm{mod}4)$, every cosine term vanishes at $x$, andSet $h=1/n$ and $m=\lfloor {\log }_{5}n\rfloor$. For $0\le k\le m$, the sine argument lies in $[0,1]$, where $\sin u\ge (\sin 1)u$. The initial part therefore has a common sign and magnitude at least $(m+1)(\sin 1)/n$. The remaining tail has absolute value at mostThe reverse triangle inequality now givesfor sufficiently large $n$, because $m+1>\ln n/\ln 5$. This establishes the sharp logarithmic lower boundThe constant can be reduced to include the finitely many small indices: for $n\ge 5$ the displayed lower bound is positive; for $n=2,3$ the first sine exceeds the tail bound $1/4$, and for $n=4$ the first two sines contribute $\sin (1/4)+\sin (5/4)/5$, exceeding the remaining tail bound $1/20$. At $n=1$ the logarithmic right side is zero.

Here $E_n(g)\asymp 1/n$ but $g$ is not Lipschitz continuous, so a first modulus of continuity bound $O(1/n)$ is sufficient by the first Jackson theorem for periodic approximation but is not necessary. Nor is the weaker necessary bound $O(\ln n/n)$ sufficient: an explicit comparison is the logarithmic cusp with slow trigonometric approximationextended periodically. On $[0,\pi ]$ it is increasing and concave, so its largest increment over distance $\delta$ is from zero, giving $\omega (h,\delta )=\delta \ln (e\pi /\delta )$ for $0<\delta \le \pi$.

For a concrete approximation lower bound, its cosine Fourier coefficients satisfyIndeed, integration by parts once replaces $h$ by $h^{\prime }(x)=\ln (\pi /x)$; integrating that against sine with primitive $(1-\cos jx)/j$ yields the stated positive integral. Its logarithmic growth follows because ${\int }_1^A\cos u/udu$ stays bounded by the Dirichlet test. All $a_j$ are negative and absolutely summable.

Let $V_n=2{\sigma }_{2n}-{\sigma }_n$, a de la Vallée Poussin sum. Its multiplier is one for $j\le n$, $2-j/n$ for $n<j<2n$, and zero for $j\ge 2n$; the Fejér summation is a uniform-norm contraction property gives $\Vert V_n\Vert \le 3$. Thus the functional $L_n(f)=f(0)-V_{n}f(0)$ has norm at most four and vanishes on every degree-at-most-$n$ trigonometric polynomial. Because the remaining multiplier weights are nonnegative and all $a_j<0$,The first Jackson theorem for periodic approximation gives the corresponding upper bound. The same first-modulus order $\delta \ln (1/\delta )$ occurs with both $E_n\asymp 1/n$ and $E_n\asymp \ln n/n$. Therefore first-modulus rates alone do not characterize the endpoint approximation class, unlike the matched direct and inverse rates for $0<\alpha <1$.

For the first additional construction, modify the rational/irrational example toAt zero, $|g(h)/h|=|h|\to 0$, so $g^{\prime }(0)=0$. At any nonzero point, density of the rational numbers and density of the irrational numbers give incompatible limiting values $x^2$ and $-x^2$. Thus $g$ is discontinuous, and hence not differentiable, there. Its differentiability set is exactly $\{0\}$.

For the second construction, use a summable absolute-value cusp series:On $[-R,R]$, each summand is bounded by $2^{-n}(R+1/2)$. The Weierstrass M-test therefore gives uniform convergence on every bounded interval, and the uniform limit theorem makes $G$ continuous everywhere.

To justify differentiability, each summand's difference quotient has absolute value at most $2^{-n}$, by the reverse triangle inequality. The sum of these bounds is finite, uniformly in the increment. Thus the tail of the difference quotient is uniformly small, and limits may be passed through the sum by first retaining finitely many terms and then making the tail small. If $x\ne 1/n$ for every $n\ge 2$, each individual summand is differentiable, givingIn particular $G^{\prime }(0)=-{\sum }_{n=2}^{\infty }{2}^{-n}=-1/2$. The accumulation of cusp locations at zero does not destroy the derivative, because their weights are summable.

At $x=1/m$, the same tail argument applies to both one-sided derivatives. Every term except the $m$th has the same derivative from both sides; the $m$th has derivatives $-2^{-m}$ and $2^{-m}$. ConsequentlyTherefore $G$ fails to be differentiable exactly at $1/2,1/3,1/4,\dots$, and is differentiable everywhere else.

Let $E_{\ast }={\inf }_{\mathrm{deg}q\le n}\Vert f-q{\Vert }_{\infty }$, and choose a minimizing sequence $p_m$ with $\Vert f-p_m{\Vert }_{\infty }\to E_{\ast }$. Since $E_{\ast }\le \Vert f{\Vert }_{\infty }$, discarding finitely many terms ensures $\Vert p_m{\Vert }_{\infty }\le 2\Vert f{\Vert }_{\infty }+1$.

Fix $n+1$ distinct nodes $a_0,\dots ,a_n$. The Lagrange interpolation formula expressesThe vector of node values is bounded in ${\mathbb{R}}^{n+1}$, so the Bolzano-Weierstrass theorem supplies a subsequence converging to $(v_0,\dots ,v_n)$. Put $p=\sum v_j{\ell }_j$, of degree at most $n$. Because the ${\ell }_j$ are fixed bounded functions,along that subsequence. The reverse triangle inequality implies $\Vert f-p{\Vert }_{\infty }=E_{\ast }$. Thus a best uniform approximation exists, with no appeal to an unproved abstract existence result.

Here ${\mathcal{P}}_n$ is the vector space of real polynomials of degree at most $n$.

A strict diagonal dominance argument proves the estimate even on a nonuniform strictly increasing spline knot sequence. Put $h_i=t_{i+1}-t_i>0$. Then $N_i$ has a rising piece of length $h_i$ and a falling piece of length $h_{i+1}$, while $M_i=2N_i/(h_i+h_{i+1})$. The same integrations as in part (b) give the linear-spline mixed Gram matrixOnly entries with indices in $1,\dots ,n$ are present. Thus ${\sum }_{j\ne i}|g_{ij}|\le 1/3$ in every row.

For $Ga=b$, choose an index $i$ with $|a_i|=\Vert a{\Vert }_{{\ell }^{\infty }}$. The reverse triangle inequality then yieldsHence $\Vert a{\Vert }_{{\ell }^{\infty }}\le 3\Vert b{\Vert }_{{\ell }^{\infty }}$. Applying this to every $b$ provesThis is the inverse infinity-norm bound from diagonal dominance; no total positivity theorem is needed. The proof also gives invertibility directly by taking $b=0$.

Apply the triangle inequality to the path from $x_1$ through $y_2$ to $y_1$:This is one direction of the reverse triangle inequality. A second application of the triangle inequality, this time through $x_2$, givesThe resulting four-point triangle inequality follows without any assumptions on coincidences among the four points.