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Let $V$ be a vector space over a specified field $F$, and let $v_1,\dots ,v_n\in V$. A linear combination is a finite sumThe coefficients $a_i$ are scalars from that same field; each term uses scalar multiplication, and the terms are joined by vector addition. For coefficients in the real numbers, negative values reverse arrows and fractional values scale them by fractional amounts. The collection of attainable vectors is the linear span. The column picture in Strang's MIT notes interprets solving a linear system as choosing coefficients to reach a target vector.^[ref]

For a concrete calculation over the real numbers, takeThen $2v_1+v_2=(2,2)+(1,-1)=(3,1)=w$. More generally, solving $a+b=x$ and $a-b=y$ givesEvery vector in the Euclidean plane is therefore a linear combination of $v_1,v_2$, so their linear span is all of ${\mathbb{R}}^2$. The target $(3,1)$ has coefficients $(2,1)$; the target $(0,2)$ has coefficients $(1,-1)$, since $(1,1)-(1,-1)=(0,2)$. The displayed formula also makes the coefficients unique. Consequently $(v_1,v_2)$ is an ordered basis, with coordinate vector $[(x,y){]}_{(v_1,v_2)}=((x+y)/2,(x-y)/2{)}^T$.

The payoff is a reusable language for synthesis: linear combinations build vectors from chosen ingredients, linear span decides which targets are reachable, and linear independence decides whether the coefficients are unique. The ingredients can also be polynomials or functions, as long as they belong to a vector space over the specified field.

A scalar linear differential equation of order $n\ge 1$ has the formHere the coefficients $a_j$ and the forcing term $g$ are prescribed functions of the independent variable $t$. A classical solution of a differential equation is an $n$-times differentiable function satisfying the identity at every point of its interval. The unknown function $y$ and its derivatives enter to the first power, without products between them or coefficients depending on $y$. Equivalently, the differential operator $L$ is a linear map: for constant scalars $\alpha ,\beta$,The forcing term may be zero, giving a homogeneous linear differential equation, or nonzero, giving an inhomogeneous linear differential equation. Linearity of $L$ does not mean that every function graph of a solution of a differential equation is a straight line.

To use the usual order-$n$ initial value problem theory on an interval $I$, assume that $a_n(t)\ne 0$ throughout $I$ and that all $a_j$ and $g$ are continuous functions there. Dividing by $a_n$ normalizes the highest derivative; specifying $y(t_0),y^{\prime }(t_0),\dots ,y^{(n-1)}(t_0)$ at $t_0\in I$ then selects one solution of a differential equation throughout $I$. A zero of $a_n$ needs separate analysis: one cannot divide through it and retain these conclusions. The first-order normalization appears in OpenStax's definition and standard form of linear equations; the corresponding system theorem is in MIT's existence and uniqueness notes for linear systems.

The coefficients need not be linear functions of $t$. For example, $t^2{y}^{\prime }+y=0$ is a linear differential equation on $t>0$, although $t^2$ is quadratic. Its solution of a differential equation with initial condition $y(1)=1$ is $y(t)=e^{1/t-1}$: its derivative is $y^{\prime }=-y/t^2$, so substitution gives $t^2{y}^{\prime }+y=0$, and $y(2)=e^{-1/2}\approx 0.60653$. The zero of the highest coefficient at $t=0$ explains the domain restriction. In contrast, $y^{\prime }+y^2=0$, $y{y}^{\prime }+y=0$ and $y^{\prime }+\sin y=0$ are nonlinear ordinary differential equations because of their dependence on the unknown function. Variable coefficients and nonlinear ordinary differential equations are different issues; see MIT's definitions and examples of first-order linear equations.

The practical benefit of linearity is that it separates the effect of the initial condition from the effect of the forcing term. Once one particular solution $y_p$ satisfies $L[y_p]=g$, every solution of a differential equation isHere $\ker L$ is the kernel of a linear map, consisting of solutions of a differential equation to the associated homogeneous linear differential equation. Indeed, $L[y-y_p]=g-g=0$, and conversely $L[y_p+h]=g+0=g$. Thus the family for a fixed forcing term is an affine space obtained by translating $\ker L$. If $g$ is not identically zero, the zero function is absent, so this family is not a vector space under the usual operations on functions. This short argument explains why a general solution is a particular solution plus a homogeneous solution.

For integrable random variables $X_1,\dots ,X_n$ on the same probability space and fixed real numbers $a_1,\dots ,a_n$, linearity of expectation statesThe random variables can be dependent random variables. The formula follows by the linearity of integration; a finite linear combination of integrable random variables remains integrable. For example, if $X$ takes $-1,2,8$ with probabilities $1/2,3/10,1/5$, then $X$ and $2X$ are dependent random variables, yet $\mathbb{E}[X+2X]=3\mathbb{E}[X]=5.1$.

An especially useful application counts how many events occur. Put $I_i=1_{A_i}$, an indicator random variable, and $N={\sum }_i{I}_i$. Since $\mathbb{E}[I_i]=\mathbb{P}(A_i)$, we get $\mathbb{E}[N]={\sum }_i\mathbb{P}(A_i)$. There is no need to compute the whole probability distribution of $N$, or to assume independence. For a uniform random permutation of $n\ge 1$ objects, each object stays fixed with probability $1/n$, so the fixed point count of a uniform random permutation has expected value one for every $n$. ^[ref].

Linearity of expectation concerns addition and multiplication by fixed scalars. Products instead satisfy $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$ under the hypotheses of expectation of a product of independent random variables. For the example above, $\mathbb{E}[X^2]=14.5$, whereas $(\mathbb{E}[X]{)}^2=2.89$. Applying a nonlinear function before averaging can therefore change the answer substantially; use the law of the unconscious statistician to compute that average.

For integrable real-valued functions $f,g$ on a measure space $(X,\mu )$ and fixed real numbers $a,b$, their linear combination $af+bg$ is integrable, and the Lebesgue integral satisfiesThe same rule applies to any finite summation: ${\int }_X{\sum }_{j=1}^n{a}_j{f}_{j}d\mu ={\sum }_{j=1}^n{a}_j{\int }_X{f}_{j}d\mu$ for integrable functions $f_j$ and fixed scalars $a_j$. Finite signed values matter: integrability ensures all these integrals are finite, so no undefined $\infty -\infty$ occurs.

This is the distributive property of finite summation carried into integration: a fixed factor can be pulled out, and a finite sum can be integrated term by term. For simple functions, distribute the finite sums of value times measure; approximation extends additivity to nonnegative measurable functions, and splitting into a positive part and negative part gives the integrable signed case. See Lecture 12: Lebesgue Integrable Functions, the Lebesgue Integral and the Dominated Convergence Theorem (Proposition 122: linearity for integrable functions). For example, ${\int }_0^1(2x-3)dx=2{\int }_0^{1}xdx-3{\int }_0^{1}1dx=-2$. With a probability measure, linearity of integration gives linearity of expectation.

On a measure space of positive total measure, the L-infinity norm of an essentially bounded measurable function is $\Vert f{\Vert }_{\infty }={\mathrm{esssup}}_{\mu }|f|$. The absolute value matters: a constant function $f=-3$ has L-infinity norm three, even though its essential supremum is $-3$. The norm belongs to the equivalence class in Lp space; different representatives equal almost everywhere give the same answer. ^[ref].

For the function $g$ in essential infimum and essential supremum, $\Vert g{\Vert }_{\infty }=1$ under Lebesgue measure, whereas its supremum norm is four. Replacing the exceptional values by the ordinary values $x$ leaves every finite Lp norm unchanged as well: $\Vert g{\Vert }_p=({\int }_0^1{x}^{p}dx{)}^{1/p}=(p+1{)}^{-1/p}$. This tends to one as $p\to \infty$. That agreement explains why the endpoint with exponent $p=\infty$ uses an essential bound rather than a pointwise bound.

A finite probability space makes the distinction between zero and small measure explicit. Let $f(a)=1$, $f(b)=3$, and $f(c)=10$. Assign weights $\mu (a)=1/2-\epsilon$, $\mu (b)=1/2$, $\mu (c)=\epsilon$, where $0\le \epsilon <1/2$. For $1\le p<\infty$,When $\epsilon =0$, the singleton $\{c\}$ is a null set and $\Vert f{\Vert }_{\infty }=3$. For every $\epsilon >0$, however small, $\Vert f{\Vert }_{\infty }=10$. For example, $\epsilon =0.001$ gives $\Vert f{\Vert }_1=2.009$ and $\Vert f{\Vert }_2=\sqrt{5.099}$, yet the large-$p$ Lp norms approach ten. The ordinary pointwise supremum norm is ten in both cases.

Why can a rare value dominate? For a bounded function on a probability space, an almost everywhere bound $|f|\le M$ gives $\Vert f{\Vert }_p\le M$. If $0<t<\Vert f{\Vert }_{\infty }$, the set $A_t=\{|f|>t\}$ has positive probability, so $\Vert f{\Vert }_p\ge t\mu (A_t{)}^{1/p}$. As $p\to \infty$, the factor $\mu (A_t{)}^{1/p}$ tends to one. These two bounds squeeze the limit to $\Vert f{\Vert }_{\infty }$. The probability measure assumption keeps total measure finite; this explanation does not assert the same finite-$p$ integrability on arbitrary infinite-measure spaces.

The Lipschitz constant is the least nonnegative number that works in the defining Lipschitz bound, equivalently the displayed supremum of distance ratios. A constant function has constant zero; for a domain with at most one point we also set it to zero. An infinite supremum means there is no finite Lipschitz bound. The least constant need not be attained by one pair of inputs.

For $f(x)=3x-2$ on the real line, every distinct pair has ratio $3$, so the Lipschitz constant is $3$. For $f(x)=x^2$ on $[-R,R]$, with $R>0$,Pairs $x=R$, $y=R-\epsilon$ have ratio $2R-\epsilon$, approaching $2R$. Thus the Lipschitz constant is $2R$, although no distinct pair realizes that exact value. Reporting the domain is essential: the same function has no finite global Lipschitz constant on $\mathbb{R}$.

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.

A function between metric spaces is locally Lipschitz if every point has a neighbourhood on which one finite Lipschitz bound works for every pair. The neighbourhood and bound may depend on the point. A continuously differentiable real function on an open interval has this property: continuity of its derivative bounds that derivative on a sufficiently small closed interval, and the mean value theorem bounds every secant slope there.

The function $x^2$ is locally Lipschitz at every real point but has no single bound on the real line. The square root is locally Lipschitz on $(0,\infty )$, and has the bound $1/(2\sqrt{a})$ on $[a,b]$ for $0<a<b$. Adding zero to its domain changes the conclusion at that point: the ratios in the Lipschitz continuity example become unbounded, so it is not locally Lipschitz at zero. This distinction is what permits the delayed, nonunique ODE solutions illustrated under Picard-Lindelöf theorem.

A measurable function belongs to $L_{\mathrm{loc}}^2(U)$ when its squared absolute value has finite integral on every compact subset of $U$. Functions are identified if they agree almost everywhere. This is a local integrability condition, not a specified global Hilbert space norm. For example, $x$ belongs to $L_{\mathrm{loc}}^2(\mathbb{R})$ but its mean-square averages on expanding intervals diverge.

A real-valued function $f:D\to \mathbb{R}$, with $D\subseteq {\mathbb{R}}^n$, has a local maximum at $x_{\ast }$ if some $\delta >0$ satisfiesOnly a neighbourhood within the domain of a function is compared. Equality is allowed; a strict local maximum requires $f(x)<f(x_{\ast })$ for every other point in that neighbourhood. A constant function therefore has a local maximum and a local minimum at every input, with neither strict. This is a local form of an extremum of a real-valued function.

For $f(x)=x^4-2x^2$, zero is a strict local maximum because $f(x)=x^2(x^2-2)<0=f(0)$ when $0<|x|<\sqrt{2}$. It is not a global maximum: values far away grow without bound. This distinction makes a local maximum useful for describing a nearby peak without claiming that it is the highest value anywhere.

Negating the function exchanges local maxima and local minima, including their strict versions. At a smooth interior local maximum, the critical point argument forces zero gradient; for a function twice continuously differentiable nearby, its Hessian matrix is negative semidefinite. These necessary conditions alone do not establish a local maximum.