The big ideas of mathematics, with the fine print

Statement

The exponentials $e_k(x)=e^{ikx}$ form an orthonormal basis of $L^2(\mathbb T)$ with normalized measure. For $f\in L^2$ with coefficients $\hat f(k)=\langle f,e_k\rangle$, the partial sums $S_Nf=\sum_{|k|\le N}\hat f(k)e_k$ satisfy

Fejér's theorem: the Cesàro means $\sigma_N f=\frac1{N+1}\sum_{n=0}^N S_nf$ converge uniformly to every continuous $f$. Carleson's theorem (1966): $S_Nf\to f$ almost everywhere for $f\in L^2$, and Hunt (1968) extended this to $L^p$, $p>1$.

Why it's true

$S_Nf=f*D_N$ with the Dirichlet kernel $D_N(x)=\sin((N+\tfrac12)x)/\sin(x/2)$, and $\sigma_Nf=f*F_N$ with the Fejér kernel $F_N=\frac1{N+1}\big(\sin(\tfrac{N+1}2x)/\sin(\tfrac x2)\big)^2$. Fejér's kernel is a positive approximate identity, so uniform convergence on continuous functions is a two-line argument. Density of trigonometric polynomials then gives completeness, and orthogonal projection gives $L^2$ convergence. The Dirichlet kernel is not positive, and $\|D_N\|_{L^1}\sim\frac4{\pi^2}\log N$ (the Lebesgue constants). That growth is the source of every pointwise pathology.

At a jump discontinuity of size $J$, the maximum of $S_Nf$ near the jump converges to

an overshoot of about $0.0895\,J$ that moves toward the jump but never shrinks.

Statement

Baire category theorem. In a complete metric space $X$, a countable intersection of dense open sets is dense. Equivalently, $X$ is not a countable union of nowhere-dense sets (it is not meagre). Consequences include the uniform boundedness principle, the open mapping theorem and the closed graph theorem.

Banach–Mazurkiewicz (1931). In $(C[0,1],\|\cdot\|_\infty)$ the set of functions differentiable at even one point (with finite one-sided derivative) is meagre. So "most" continuous functions are differentiable nowhere. An explicit example is Weierstrass's function

which is continuous (uniform convergence by the M-test) and differentiable nowhere. Weierstrass (1872) needed $b$ an odd integer with $ab>1+\tfrac{3\pi}2$. Hardy (1916) proved $ab\ge1$ suffices.

Why it's true

Let $E_n=\{f:\exists x,\ |f(x+h)-f(x)|\le n|h|\ \forall\,|h|\le 1/n\}$ (one-sided at the endpoints). Each $E_n$ is closed, by compactness of $[0,1]$. It is nowhere dense, because any $f$ is uniformly close to a polynomial plus a small, steep sawtooth, which escapes $E_n$. Any $f$ with a derivative at some point lies in some $E_n$, so these functions form a meagre set. For $W$: the $n$-th term has amplitude $a^n$ and wavelength $\sim b^{-n}$, so at scale $h\sim b^{-n}$ the difference quotient is $\gtrsim (ab)^n$, which doesn't stay bounded when $ab>1$. If instead $ab<1$, the differentiated series converges uniformly and $W\in C^1$.

Statement

There is a unique complete, translation-invariant, countably additive measure $\lambda$ on the Lebesgue $\sigma$-algebra of $\mathbb R$ with $\lambda([a,b])=b-a$. Its integral satisfies the convergence theorems that Riemann's lacks:

A bounded function on $[a,b]$ is Riemann integrable iff its set of discontinuities is $\lambda$-null (Lebesgue's criterion).

Why it's true

Define outer measure by countable covers with intervals, and restrict to the Carathéodory-measurable sets. These form a $\sigma$-algebra on which outer measure is countably additive. Uniqueness follows from the $\pi$–$\lambda$ theorem. Monotone convergence is built into the definition of the integral as a supremum over simple functions. Fatou's lemma follows from it, and dominated convergence is Fatou applied to $g\pm f_n$.

The middle-thirds Cantor set $C$ has $\lambda(C)=\lim(2/3)^k=0$, yet it is uncountable (ternary expansions without the digit 1 biject with $\{0,2\}^{\mathbb N}$) and has Hausdorff dimension $\log2/\log3$. Remove intervals of total length only $\tfrac12$ instead (the Smith–Volterra–Cantor set) and you get a nowhere-dense set of measure $\tfrac12$.

Statement

For a separable $f\in K[x]$ with splitting field $L$, the fundamental theorem of Galois theory gives an inclusion-reversing bijection between intermediate fields $K\subseteq E\subseteq L$ and subgroups $H\le G=\operatorname{Gal}(L/K)$, with $E/K$ normal iff $H\trianglelefteq G$. In characteristic 0, the roots of $f$ can be expressed by radicals iff $G$ is solvable. Since $S_5$ is not solvable ($A_5$ is simple), the general quintic has no formula in radicals.

Over $K=\mathbb C(c)$, the Galois group of $f(z)=z^5-z-c$ equals its monodromy group. Away from the branch points, the five roots are analytic functions of $c$. Carrying them around a loop in the $c$-plane permutes them, and the permutations from all loops form the group.

Why it's true

Radical extensions are built by adjoining $n$-th roots, and after adjoining roots of unity each step is cyclic (Kummer theory). A tower of cyclic steps corresponds to a subnormal series with abelian quotients, which is exactly what "solvable" means. For the monodromy picture: the branch points are where $f$ and $f'$ share a root. Here $5z^4=1$, which gives four branch values $c=-\tfrac45 z$ with $z^4=\tfrac15$. Near each one, exactly two roots collide like $\pm\sqrt{c-c_*}$, so a small loop around it swaps those two roots. If these transpositions connect all five roots, they generate all of $S_5$.

Statement

For a finite group $G$ acting on a finite set $X$, the orbit–stabilizer theorem says $|G\cdot x|\,|G_x|=|G|$. Summing over $X$ gives the Cauchy–Frobenius–Burnside lemma:

For $k$-colorings of an $n$-bead necklace under rotation ($G=C_n$), rotation by $j$ fixes exactly the colorings that are constant on its $\gcd(j,n)$ cycles, which gives

Pólya's enumeration theorem keeps track of how many beads get each color by weighting with the cycle index $Z_G$.

Why it's true

Count the pairs $\{(g,x):gx=x\}$ two ways. Summing over $g$ gives $\sum_g|X^g|$. Summing over $x$ gives $\sum_x|G_x|=\sum_x|G|/|G\cdot x|=|G|\cdot\#\text{orbits}$, because each orbit $O$ contributes $|O|\cdot\frac{|G|}{|O|}$.

Statement

Let $\psi(x)=\sum_{p^k\le x}\log p$ (Chebyshev's function), and let $\rho=\beta+i\gamma$ run over the nontrivial zeros of $\zeta$. For $x>1$ not a prime power,

with the sum over $\rho$ taken symmetrically, $|\gamma|<T$, $T\to\infty$. Because $\zeta$ has no zeros on $\operatorname{Re}s=1$, we get $\psi(x)\sim x$, which is equivalent to the prime number theorem $\pi(x)\sim x/\log x$. The Riemann hypothesis ($\beta=\tfrac12$ for all $\rho$) is equivalent to $\psi(x)=x+O(\sqrt x\log^2x)$.

Why it's true

Take the logarithmic derivative of the Euler product: $-\zeta'/\zeta(s)=\sum_n\Lambda(n)n^{-s}$. Perron's formula recovers $\psi(x)=\frac1{2\pi i}\int_{(c)}-\frac{\zeta'}{\zeta}(s)\frac{x^s}{s}\,ds$. Shift the contour to the left. The residues are $x$ from the pole at $s=1$, $-x^\rho/\rho$ from each zero, $-\log2\pi$ from $s=0$, and the trivial zeros at $-2,-4,\ldots$ sum to the last term. Each conjugate pair contributes $-2\operatorname{Re}(x^\rho/\rho)\approx -\frac{2\sqrt x}{|\rho|}\cos(\gamma\log x-\arg\rho)$, a wave in $\log x$ with frequency $\gamma$.

Statement

For a closed curve $\gamma$ in $\mathbb C\setminus\{a\}$, $\operatorname{wind}(\gamma,a)=\frac1{2\pi i}\oint_\gamma\frac{dz}{z-a}\in\mathbb Z$. It is a homotopy invariant, and it identifies $\pi_1(\mathbb C\setminus\{0\})\cong\mathbb Z$. The argument principle says that for $f$ meromorphic on a neighbourhood of a disc $D$, with no zeros or poles on $\partial D$,

Rouché's theorem follows: if $|g|<|f|$ on $\partial D$, then $f$ and $f+g$ have the same number of zeros in $D$.

Why it's true

Locally $f(z)=(z-z_0)^m h(z)$ with $h(z_0)\ne0$, so $f'/f=m/(z-z_0)+h'/h$, and the residues add up to the order. The fundamental theorem of algebra follows. For $p(z)=z^n+\ldots$ and large $R$, the image of $|z|=R$ is homotopic to $z^n$ through the straight-line homotopy (because $|p-z^n|<R^n$ on the circle), so it winds $n$ times. As $R$ shrinks to 0 the winding number drops to 0. It can only change when the curve passes through the origin, which happens exactly when the circle passes through a root. The lab shows this happening.

Statement

For a compact oriented Riemannian surface $M$ with boundary, Gaussian curvature $K$, and geodesic curvature $k_g$ along $\partial M$,

where the $\theta_i$ are interior angles at corners. For a geodesic triangle on the unit sphere ($K\equiv1$, $k_g=0$, $\chi=1$) this reduces to Girard's theorem: $\alpha+\beta+\gamma-\pi=\operatorname{Area}$. For a closed surface, $\int_MK\,dA=2\pi\chi(M)=2\pi(2-2g)$.

Why it's true

For Girard: the three great circles through the sides cut the sphere into lunes. Each pair of lunes with angle $\alpha$ has area $4\alpha$. Adding all three pairs covers the sphere once, plus the triangle and its antipode two extra times: $4(\alpha+\beta+\gamma)=4\pi+4\,\text{Area}$. In general, triangulate $M$, apply the local theorem to each triangle (Green's theorem plus the structure equation $d\omega_{12}=-K\,dA$), and sum. Interior edges cancel, and the angles at each vertex add up to $2\pi$. Euler's formula $V-E+F=\chi$ comes out of the bookkeeping.

Statement

For the logistic family $f_r(x)=rx(1-x)$ on $[0,1]$, let $R_k$ be the parameter at which the critical point $\tfrac12$ is periodic with period $2^k$ (the superstable parameters). Then

The same $\delta$ appears for every family of unimodal maps with a quadratic maximum, such as $r\sin\pi x$ and $1-rx^2$, and in fluid convection experiments. Chaos itself is measured by the Lyapunov exponent $\lambda(r)=\lim\frac1n\sum\log|f_r'(x_i)|$. It is positive on a set of parameters of positive measure (Jakobson, 1981), but periodic windows are dense (Graczyk–Świątek and Lyubich, 1997).

Why it's true

Renormalization: near $R_\infty$, the second iterate $f^2$ restricted to a small interval around $\tfrac12$, rescaled, looks like $f$ itself at a different parameter. The doubling operator $\mathcal Rg(x)=-\alpha\,g(g(-x/\alpha))$ has a fixed point $g^*$ with exactly one expanding eigenvalue, and that eigenvalue is $\delta$. Every family crossing the stable manifold transversally inherits the same scaling. Lanford (1982) gave a computer-assisted proof, and Sullivan, McMullen and Lyubich completed the conceptual proof in the 1990s.

Statement

Let $L_1$ be the size of the largest component of $G(n,p)$ with $p=c/n$. With high probability (Erdős–Rényi, 1960):

where $\beta(c)=0$ for $c\le1$ and $\beta(c)>0$ for $c>1$. For $c<1$, $L_1=\Theta(\log n)$. At $c=1$, $L_1=\Theta(n^{2/3})$. Connectivity has its own sharp threshold: if $p=(\log n+t)/n$, then $\Pr[G \text{ connected}]\to e^{-e^{-t}}$.

Why it's true

Explore a component by breadth-first search. Each newly reached vertex has about $\mathrm{Bin}(n,c/n)\approx\mathrm{Poisson}(c)$ new neighbours, so the exploration looks like a Galton–Watson branching process. That process survives with probability $\beta$, the positive root of $1-\beta=e^{-c\beta}$, which exists exactly when the mean offspring $c$ exceeds 1. A second-moment argument shows that the surviving explorations all merge into a single component. Friedgut–Kalai (1996) and Bourgain explain why monotone graph properties with enough symmetry have sharp thresholds at all.

Statement

Lévy's isoperimetric inequality on $S^{d-1}$ with uniform measure $\sigma$ implies that for 1-Lipschitz $f$ with median $M_f$,

The Gaussian analogue (Tsirelson–Ibragimov–Sudakov) says $\Pr(|f(G)-\mathbb Ef(G)|>t)\le2e^{-t^2/2}$ for $G\sim N(0,I_d)$, with no dependence on $d$. Corollaries include: most of the sphere's area lies within $O(1/\sqrt d)$ of any equator, most of a ball's volume lies near its surface, and independent random points are nearly equidistant.

Why it's true

Among sets of a given measure, a spherical cap has the smallest $t$-neighbourhood. A hemisphere's $t$-neighbourhood leaves out a cap of measure $\int_t^{\pi/2}\cos^{d-2}\theta\,d\theta\big/\!\int_{-\pi/2}^{\pi/2}\cos^{d-2}\theta\,d\theta\lesssim e^{-(d-2)t^2/2}$. Apply this to $\{f\le M_f\}$. Other proofs use log-Sobolev inequalities and the Herbst argument, or Talagrand's transportation inequality. For distances: if $X,Y$ are uniform in $[0,1]^d$, $\|X-Y\|^2$ is a sum of $d$ i.i.d. terms with mean $1/6$, so $\|X-Y\|/\sqrt d\to1/\sqrt6$ with fluctuations of order $d^{-1/2}$.

Statement

Gödel I (1931). If $T$ is consistent, recursively axiomatizable, and interprets Robinson arithmetic $\mathsf Q$, then some sentence $G_T$ is neither provable nor refutable in $T$. Gödel II. If, in addition, $T$ proves enough induction, then $T\nvdash\mathrm{Con}(T)$. Turing (1936). The set $\{e : \varphi_e(e)\downarrow\}$ is computably enumerable but not decidable. Matiyasevich–Robinson–Davis–Putnam (1970). There is no algorithm that decides whether a Diophantine equation has an integer solution, which answers Hilbert's tenth problem.

Why it's true

Arithmetize syntax: formulas and proofs become numbers, and "$p$ is a proof of $\varphi$" becomes a primitive recursive relation, representable in $\mathsf Q$. The diagonal lemma gives a fixed point for any formula $\psi(x)$, a sentence $\sigma$ with $T\vdash\sigma\leftrightarrow\psi(\ulcorner\sigma\urcorner)$. Take $\psi=\neg\mathrm{Prov}_T$. Halting is the same diagonal argument: if $h$ decided halting, the program "on input $e$, loop iff $h(e,e)$ says $\varphi_e(e)$ halts" contradicts itself on its own index. This is Cantor's argument from the intro edition, idea 12, applied to programs instead of sequences.