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For a transient weighted graph, $H_0$ is the completion of the finitely supported functions in the norm $\Vert f{\Vert }_{H_0}=\mathcal{E}(f,f{)}^{1/2}$. Its elements have finite Dirichlet energy and vanish at infinity in the energy sense.

Every closed walkin a directed covering graph of an interval map has a point $x\in J_0$ with $F^j(x)\in J_j$ and $F^n(x)=x$. This follows by successively pulling $J_0$ back through the covering relations and applying the fixed-point property of a closed interval. If the itinerary has least period $n$ and avoids shared endpoints, $x$ lies on an $n$-cycle.

A component outside a longest cycle has at least $k$ attachment vertices in a $k$-connected graph, and no two attachments are consecutive, forcing cycle length at least $2k$.

The minimal subtraction scheme chooses counterterms that remove only the poles in the dimensional regulator $ϵ$, without additional finite terms.

The inverse function theorem answers a practical question: can nearby outputs be used to recover nearby inputs uniquely and smoothly? Its useful surprise is that a nonlinear function can be tested using its linear derivative. If the first-order changes preserve all input directions, then, on a sufficiently small region, the nonlinear map can also be reversed. This is the basis for treating a nonlinear map as a coordinate transformation.

Precisely, let $f:U\to {\mathbb{R}}^n$ be continuously differentiable on an open subset $U$ of Euclidean space, let $a\in U$, and put $b=f(a)$. If $Df(a)$ is an invertible linear map, there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $b$ such that $f:U_0\to V_0$ is a bijection with a continuously differentiable inverse function $g$. Thus this restriction is a diffeomorphism. In coordinates the test is $\det Df(a)\ne 0$, using the Jacobian determinant. The derivative of a local inverse map is obtained by inverting the Jacobian matrix, at the corresponding input. ^[ref]

For a one-dimensional example, $f(x)=x^2$ has $f^{\prime }(1)=2$. Near $x=1$, the inverse function is $g(y)=\sqrt{y}$, and the inverse function rule gives $g^{\prime }(1)=1/2$. But $x^2$ has no inverse function on the whole real line, since $x$ and $-x$ give the same output. The theorem supplies a local restriction, not global injectivity. At zero, the derivative test fails; the nonnegative branch still has a square root inverse, but that inverse has no finite derivative at zero. ^[ref]

Here is a two-dimensional example with an invertible derivative everywhere and infinitely repeated outputs:The exponential function turns $s$ into a positive radius, while $\theta$ is the angle in polar coordinates. Restricting, for example, to $|s|<0.35$ and $|\theta |<0.6$ produces the reversible patch pictured below. On its image, with Cartesian coordinates $(u,v)$ and $u>0$, the inverse function isThe arctangent selects the angle near zero. Without an angle restriction, $f(s,\theta +2\pi k)=f(s,\theta )$ for every integer $k$, so there is no single global inverse function. This map is a local diffeomorphism but is not injective.

The short proof idea is to correct a guessed input using the inverse linear response. With $A=Df(a)$, solving $f(x)=y$ is equivalent to finding a fixed point ofAt $a$, the derivative $D{T}_y(a)=I-A^{-1}Df(a)$ vanishes. Continuity makes $T_y$ a contraction mapping on a sufficiently small closed ball, which it maps into itself when $y$ is sufficiently close to $b$. The contraction mapping theorem then gives existence and uniqueness there. Establishing the smooth dependence on $y$ completes the inverse theorem. This explains why the nonlinearity can be tolerated: its correction has a small local slope.^[ref]

If the line segment from $x$ to $y$ lies in the domain of a differentiable map and $\Vert D{f}_z\Vert \le M$ along it, then

An invertible linear map is a linear bijection; its inverse is also linear.

Invertibility adds a bijection requirement to the definition of a linear map.

A function is continuously differentiable when its derivative exists and varies continuously.

Let $E,F$ be normed vector spaces, $U\subseteq E$ open, and $a\in U$. The Fréchet derivative of $f:U\to F$ at $a$ is a bounded linear map $D{f}_a:E\to F$ such thatThe approximation is uniform over directions: the error divided by the size of the whole increment must vanish, rather than merely vanishing on each fixed line. Boundedness of the linear map is part of the definition.

The Fréchet derivative is unique. If $L,M$ both satisfy the remainder condition, substitute $h=tv$ for each fixed vector $v$, divide by $|t|$, and let $t\to 0$ to obtain $(L-M)v=0$. Existence also implies continuity of $f$ at $a$, since $f(a+h)-f(a)=D{f}_a(h)+o(\Vert h\Vert )$ and $D{f}_a$ is bounded. In finite dimensions the Jacobian matrix represents this linear map; mere existence of directional derivatives need not supply one.

A spectral decomposition expresses a linear operator through its eigenvalues and projections onto its eigenspaces.

Eigenvectors belonging to distinct eigenvalues are linearly independent. Consequently, an $n$-dimensional operator with $n$ distinct eigenvalues is diagonalizable.

Suppose real matrices $A,B$ satisfy $AP=PB$ for an invertible complex matrix $P=X+iY$. Then $AX=XB$ and $AY=YB$. The real polynomial $\det (X+tY)$ is not identically zero because its value at $t=i$ is nonzero, so some real $t$ makes $X+tY$ invertible. This real matrix intertwines $A$ and $B$.

Let $A$ be invertible over an algebraically closed field. If $A^m$ is diagonalizable for an integer $m>0$ not divisible by the characteristic of a field (with no restriction on $m$ in characteristic zero), then $A$ is diagonalizable. Indeed, a nontrivial Jordan block $A{|}_E=\lambda I+N$ has $\lambda \ne 0$ andThe first nonzero nilpotent term prevents this power from being diagonalizable.

For a compactly supported nonzero function $\psi$, the translates $f_n(x)=\psi (x-n)$ converge uniformly to zero on every compact subset of $\mathbb{R}$, while $\Vert f_n{\Vert }_{\infty }=\Vert \psi {\Vert }_{\infty }$ prevents uniform convergence on all of $\mathbb{R}$.

Every consistent recursively axiomatized extension of elementary arithmetic is incomplete. If it were complete, its theorem set would be decidable; representing that decision procedure and diagonalizing against it yields a contradiction.

Every recursively enumerable set many-one reduces to the diagonal halting set. If a program halts exactly on $X$, the S-m-n theorem produces from $x$ an index $g(x)$ for a unary program that ignores its input and performs that computation on $x$. ThenConversely, a computable preimage of a recursively enumerable set is recursively enumerable.

An affine algebraic group is solvable when its derived series reaches the identity. A connected solvable affine algebraic group can be conjugated into an upper triangular matrix group by the Lie-Kolchin theorem.

A Lie algebra is solvable when its derived series eventually becomes zero.

A random field is finite-range dependent if collections indexed by sets farther apart than a fixed distance are independent. Sparse subsets of long paths then restore enough independence for path-counting arguments.

For a random field $(Y_x)$, level-set percolation studies the random vertex set $\{x:Y_x\ge a\}$ as the threshold $a$ varies.