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For $\mu >0$, the logarithmic barrier objective isThe domain of a function $D$ is a nonempty convex set by the same concavity argument as in part (a), with strict inequalities. Differentiating twice on $D$ givesHere ${\nabla }^{2}f⪰0$ by convexity, $-{\nabla }^2{c}_i⪰0$ by concavity, and each outer product is a positive semidefinite matrix. Hence ${\nabla }^2{f}_{\mu }⪰0$, proving convexity on $D$.

The domain qualification matters: the displayed real-valued logarithm expression is not defined on all of ${\mathbb{R}}^n$. The precise whole-space version is the extended-real convex function equal to $f_{\mu }$ on $D$ and $+\infty$ outside $D$.

Choose a strictly feasible point, an initial barrier parameter ${\mu }_0>0$, and $0<\theta <1$. At outer iteration $k$, approximately minimize the logarithmic barrier objective $f_{{\mu }_k}$ over its strictly feasible domain of a function. For example, use damped Newton methods, choosing an inner step size by backtracking line search so that every $c_i$ stays positive and the barrier objective function decreases. Solve to a tolerance $\Vert \nabla f_{{\mu }_k}(x^k)\Vert \le {\epsilon }_k$, where ${\epsilon }_k\to 0$. Set ${\mu }_{k+1}=\theta {\mu }_k$ and use $x^k$ as the next starting point.

The associated Lagrange multiplier estimates areReducing ${\mu }_k$ makes these perturbed complementary slackness equations approach the KKT conditions. This describes the generic interior-point method; existence of a solution to every inner minimization needs appropriate additional attainment assumptions and does not follow from strict feasibility alone.

By i and ii,The domain of a function of $S^{\prime }$ is $\mathrm{im}T$. Its rank-nullity theorem gives $r_{S^{\prime }}+n_{S^{\prime }}=\dim \mathrm{im}T=r_T$. Applying the rank-nullity theorem for $T$ now giveswhich is exactly the rank-nullity theorem for the function composition $ST$.

The domain of a function excludes $x=3$. Both branches are positive, with a vertical asymptote at $x=3$ approached from both sides with value $+\infty$, and horizontal asymptote $y=0$ at either infinity. The graph is symmetric about $x=3$ because it depends only on $|x-3|$.

For $x<3$, the formula is $2/(3-x)$ and the derivative is positive; for $x>3$, it is $2/(x-3)$ and the derivative is negative. It passes through $(0,2/3)$ and has no zeros.

Continuity. The function is continuous at $x_0$ when ${\lim }_{x\to x_0}f(x)=f(x_0)$. Equivalently, for every $\epsilon >0$ there exists $\delta >0$ such that $|x-x_0|<\delta$ within the domain of a function implies $|f(x)-f(x_0)|<\epsilon$; this is the epsilon-delta definition of continuity.

Differentiability. At an interior point $x_0$, the difference quotient must have a finite two-sided limit of a function:This defines differentiability and the derivative. At an endpoint one may instead state a one-sided definition relative to the interval. Differentiability implies continuity, because the increment is $h$ times a quotient with finite limit, but continuity alone does not guarantee differentiability.

On the real domain of a function $x>0$, the quotient rule gives

The domain of a function is $\mathbb{R}\setminus \{0\}$, because the denominator vanishes at zero. Near zero, $e^x-1\sim x$, so $x=0$ is a vertical asymptote, approached with value $-\infty$ from the left and $+\infty$ from the right. The horizontal asymptotes are $y=-1$ as $x\to -\infty$ and $y=0$ as $x\to +\infty$.

Moreover,Both branches define a strictly decreasing monotone function: the left branch lies below $-1$, and the right branch is positive. There are no zeros or vertical-axis intercept.

Let $o=(0,0)$ be the common point of the two circles. For sufficiently small $\epsilon >0$, the open neighbourhoodhas the following local structure: $W\setminus \{o\}$ consists of four disjoint open arcs, an upper and a lower branch from each circle. These are its four connected components, each open and closed relative to $W\setminus \{o\}$.

Suppose a coordinate chart of some dimension of a manifold $n$ existed at $o$. The case $n=0$ would make $o$ an isolated point, which it is not. For $n\ge 1$, shrink the chart inside $W$ so that its image is an open ball centred at the image of $o$. Its domain of a function $V$ is an open neighbourhood of $o$, so it meets all four arms of $W$. Intersecting those arms with $V\setminus \{o\}$ gives four disjoint nonempty clopen sets partitioning $V\setminus \{o\}$; hence there are at least four connected components.

But a punctured ball in $\mathbb{R}$ has exactly two connected components, and a punctured ball in ${\mathbb{R}}^n$ for $n\ge 2$ is path-connected. A homeomorphism preserves connected components, giving a contradiction in every dimension of a manifold. Thus $X$ is not a topological manifold. The argument uses arbitrarily small coordinate charts; the fact that the whole $X\setminus \{o\}$ has two pieces alone would not be a local obstruction.

A function $f:X\to Y$ is injective if, for all $x,x^{\prime }\in X$, $f(x)=f(x^{\prime })$ implies $x=x^{\prime }$. Thus distinct elements of the domain of a function have distinct images.

Not injective; surjective. The equal values $f(0)=f(2)=1$ disprove injectivity. For any $y>0$, take $x=1+y^{-1/2}>1$. This lies in the stated domain of a function, andConsequently every member of the codomain has a preimage.

Suppose first that $f=O(g)$. Choose $C>0$ and $N$ so that $|f(n)|\le C|g(n)|$ for $n\ge N$. There are only finitely many natural numbers below $N$. Because $g(n)\ne 0$ at each of them, all ratios $|f(n)|/|g(n)|$ in this finite prefix are finite. LetThen $|f(n)|\le c|g(n)|$ both in the prefix and in the tail, hence on the entire domain of a function. This proves the finite-prefix extension of a big O bound.

Conversely, a bound valid for every natural number is certainly valid eventually. The purported real constant cannot be negative because $g$ never vanishes; if it is zero, enlarge it to a positive constant. The definition of Big O notation then applies. The nonvanishing assumption is what permits the finite exceptions to be absorbed into one constant.

A periodic function repeats exactly after a nonzero shift of its input. For $f:\mathbb{R}\to Y$, a positive period of a function $T$ satisfiesOne interval of length $T$ therefore determines the whole function. For another domain of a function $D$, a period must also preserve the domain, $D+T=D$; equality only where two domains happen to overlap is weaker. Periodicity is an exact translation of a function symmetry, rather than merely two similar-looking pieces of its graph.

The sine $\sin (2\pi x)$ has fundamental period one. The function ${\sin }^2(2\pi x)$ repeats after half a unit: the sine changes sign under that shift, but squaring removes the sign. The mixed function $g(x)=2+\sin (2\pi x)+\frac{1}{2}\cos (4\pi x)$ again has fundamental period one. All positive integer multiples of a period are periods, so specifying one period need not identify the shortest repetition.

If the input is time and $T_0$ is the fundamental period, the frequency is $1/T_0$ and the angular frequency is $2\pi /T_0$. Repetition lets a Fourier series describe a cycle using sines and cosines; the period average separates the reference level from the oscillating terms. Strang develops this representation and its Fourier coefficients in §4.1 of Computational Science and Engineering by Gilbert Strang.^[ref]

This connection grew out of a physical problem. In his 1822 book Théorie analytique de la chaleur by Joseph Fourier, Joseph Fourier used periodic functions to extend profiles beyond their initial interval while treating heat conduction. Chapter III, §§233–234 combines even functions and odd functions into sine–cosine expansions and rescales their period.^[ref] A modern Fourier series interpretation still requires the appropriate convergence conditions.

A constant function has every real shift as a period and no fundamental period. Without continuity, even a nonconstant periodic function can lack a smallest positive period; periodic function without a least positive period gives a concrete example. The period group of a function organizes all these possibilities.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.

A pointwise limit assigns to each input the limit of a sequence of function values there, on the specified domain of a function. It is unique wherever it exists, by uniqueness of the numerical limit of a sequence. Its existence supplies no common approximation rate over the domain of a function. For $f_n(x)=x^n$ on $[0,1]$, it equals zero below one and one at one; restricting to $[0,a]$ with $a<1$ gives the zero function.

For a function $f:X\to Y$ and a subset $U\subseteq Y$, its preimage isIt is a subset of the original domain of a function. The notation does not require an inverse function: the preimage of a singleton set can have no members, one member, or many members. In the finite example, $f^{-1}(\{0\})=\{a,b\}$, $f^{-1}(\{1\})=\{c\}$ and $f^{-1}(\{3\})=\varnothing$. The preimage of a singleton set is also its fiber of a function.

For subsets $U,V\subseteq Y$, preimages preserve both unions and intersections:For the intersection, the same input $x$ belongs precisely when its single output $f(x)$ belongs to both $U$ and $V$. There is no opportunity to choose different witnesses, which is what made the image counterexample fail. The identities also hold for arbitrary families of output subsets.

In the finite example, choose $U=\{0,1\}$ and $V=\{0,2\}$. ThenThe intersection of the two preimages is exactly the same set $\{a,b\}$. Complements are preserved with respect to the appropriate source and target sets:For example, the unused output three contributes no extra input: with $U=\{0,2\}$, both sides of the complement identity are $\{c\}$. See Lebl, Proposition 0.3.15: preimages of unions, intersections and complements.

This makes preimages useful for transporting conditions. The solution set of $g(x)=0$ is $g^{-1}(\{0\})$, and a zero set is a preimage of zero. Requiring two output conditions takes the intersection of their preimages. In topology, a continuous function pulls every open set back to an open set; a measurable function pulls the specified measurable sets in its codomain back to measurable sets in its domain of a function. These definitions use preimages because their set operations remain consistent even when the function is noninjective.

An algebraic simplification proves equality only where both original and transformed expressions are defined. If a cancelled factor vanishes at a point, the simplified expression may instead describe an extension to that point. For example,holds only when $\cos x\ne 0$ and $\cos x+\sin x\ne 0$. The right side is zero at $x=\pi /2+k\pi$, but the left side is undefined there. A graph of the original function must retain those holes. The extension may be useful, but it is a different function with a larger domain of a function.

For functions $f$ and $g$ taking values in the same field, their quotient of functions is defined by $(f/g)(x)=f(x)/g(x)$ on the common domain of a function restricted to points where $g(x)\ne 0$. Thus $f(x)=1$ and $g(x)=x$ on the real line give a quotient of functions defined only for $x\ne 0$.

The radical of an integer $n>0$ is the product of its distinct prime factors, each included exactly once. If the prime factorization is $n=p_1^{a_1}\cdots p_r^{a_r}$ with distinct prime numbers $p_i$ and positive integer exponents $a_i$, thenThus the radical of an integer remembers which prime factors occur and forgets their prime valuations. At $n=1$, the empty product gives $\mathrm{rad}(1)=1$. The domain of a function here is the positive integers; zero is excluded. ^[ref]

For example, $72=2^3\cdot 3^2$ has distinct prime factors $2,3$, so $\mathrm{rad}(72)=2\cdot 3=6$. Similarly,using the prime factorizations $30=2\cdot 3\cdot 5$, $32=2^5$ and $36=2^2\cdot 3^2$. The prime omega function counts the retained prime factors, while the total number of prime factors counts their occurrences before repetitions are removed.

The radical of an integer is its largest squarefree divisor. Indeed, a squarefree divisor chooses at most one copy of each prime factor, so it divides $\mathrm{rad}(n)$; the radical of an integer itself is a squarefree divisor. Consequently,The last statement follows because every nonzero prime valuation is then one. It includes $n=1$, a squarefree integer. Applying the radical of an integer twice also changes nothing: $\mathrm{rad}(\mathrm{rad}(n))=\mathrm{rad}(n)$.

For any positive integers $a,b$, the exact product rule isTo prove it, a prime factor dividing just one input occurs once in the numerator and not in the denominator. A prime factor dividing both occurs twice in the numerator and once in the denominator, since it divides their greatest common divisor. Each prime factor of $ab$ therefore remains exactly once. For coprime integers, the denominator is $\mathrm{rad}(1)=1$, proving the radical of an integer is a multiplicative function. It is not a completely multiplicative function: $\mathrm{rad}(4)=2\ne 4=\mathrm{rad}(2{)}^2$.

For a worked example with shared prime factors, take $a=12$ and $b=18$. Their greatest common divisor is $6$, and $ab=216=2^3\cdot 3^3$, soThis explains why prime support of an integer is useful: questions about the presence of prime factors can discard their multiplicities. For example, the squarefree divisors of $72$ are exactly the integer divisors of $\mathrm{rad}(72)=6$, namely $1,2,3,6$. Also $\gcd (m,n)=1$ if and only if $\gcd (m,\mathrm{rad}(n))=1$, because coprimality depends only on shared prime factors.

A further characterization uses mutual divisibility of powers: for positive integers $a,b$,If their prime supports of an integer agree, sufficiently large $k,\ell$ make all the required prime valuations large enough. Conversely, these two divisibility relations force each prime factor of either input to divide the other. For $72$ and $36$, one can take $k=2$ and $\ell =1$: $72\mid 36^2$ and $36\mid 72$.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

A power law $q(t)=t^{\rho }$ responds to every fixed rescaling $t\mapsto ut$ by the same factor $u^{\rho }$. Regular variation keeps this scaling behaviour in the limit $t\to \infty$, while allowing departures from an exact power law. It is useful when the scaling exponent is stable but a smaller correction continues changing. The correction can even grow without bound.

An eventually positive measurable function $q$ has regular variation at infinity with index $\rho \in \mathbb{R}$ ifThe argument $ut$ is in the domain of a function for sufficiently large $t$. Each $u$ is held fixed as the limit is taken; allowing $u$ to change with $t$ is a different assertion. This is the scalar definition recalled by ^[ref]. The index is unique: taking $u=2$ determines $2^{\rho }$.

For example, on $t>1$ the function $q(t)=t^2\log t$ hasThus $q$ has regular variation of index two. Yet $q(t)/t^2=\log t\to \infty$: regular variation does not require asymptotic equivalence to a constant times a power law. Doubling $t$ gives $q(2t)/q(t)=4(1+\log 2/\log t)$. At $t=100,10^4,10^6$ this is approximately $4.60206,4.30103,4.20069$, respectively. The ratio approaches four slowly, even though the exponent is exactly two in the limiting scaling law.

The slowly varying function isolates this correction. Set $\ell (t)=q(t)/t^{\rho }$. ThenConversely, multiplying a slowly varying function by $t^{\rho }$ gives regular variation of index $\rho$. Hence the decomposition $q(t)=t^{\rho }\ell (t)$ is exact on its eventual domain of a function; it is the scaling relation, rather than the decomposition itself, that is asymptotic.

A Pareto distribution gives a decreasing example with a direct interpretation. For $x\ge 1$, take the tail probability $\mathbb{P}(X>x)=x^{-2}$. Its regular variation index is $-2$. Doubling a threshold divides its tail probability by four, andFor a tail probability with regular variation of index $-\alpha$, the same conditional probability tends to $2^{-\alpha }$. An exponential distribution behaves differently: its corresponding conditional probability is $e^{-\lambda t}\to 0$. This comparison explains why scaling limits are useful in extreme value theory; it does not establish that an observed dataset follows a particular tail model. The role of regular variation in heavy-tailed distributions is discussed in Lindskog, Resnick and Roy, introduction, p. 1.

An exponential function $q(t)=e^t$ is not regularly varying with any finite index: at $u=2$, $q(2t)/q(t)=e^t\to \infty$, rather than $2^{\rho }$. Boundedness alone is also insufficient, as the oscillating counterexample in slowly varying function shows. Checking a few rescaling ratios at finite arguments can illustrate a proposed law, but cannot prove its limiting definition.

Jovan Karamata developed this theory in his 1930 paper and returned to its fundamental results in 1933. The opening of his 1933 paper identifies the earlier work; pp. 57–58 connect the rescaling limit with a power law and a slowly varying factor. See J. Karamata, Sur un mode de croissance régulière. Théorèmes fondamentaux, Bulletin de la Société Mathématique de France 61 (1933), 55–62. This gives a precise language for “almost a power” without discarding corrections that remain large.