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The supremum norm of a bounded real- or complex-valued function on a set $X$ isFor a continuous function on a compact space, the extreme value theorem makes this supremum a maximum.

A continuous function on an open interval with positive symmetric derivative everywhere is strictly increasing. One proof takes an antiderivative $F$ using the fundamental theorem of calculus. For small positive $h$, the identityfollows from $f^s(x)>0$. If $F$ exceeded a chord, its difference from the chord would attain a positive interior maximum, contradicting this symmetric difference inequality. Thus $F$ is a convex function, and its derivative $f$ is nondecreasing. Equality at two different points would make $f$ constant between them and force an interior symmetric derivative to vanish, proving strict monotonicity.

The symmetric second derivative is the displayed limit when it exists. For a twice continuously differentiable function it equals the ordinary second derivative by Taylor expansion; the definition itself does not require ordinary differentiability. A continuous function with vanishing symmetric second derivative throughout an interval must be affine, by vanishing symmetric second derivative forces an affine function. For a twice-integrated Fourier series, sinc-squared weights express this quotient, connecting the operation to Riemann summation of a convergent trigonometric series.

This deliberately synthetic differential equation model illustrates response direction and lag. Every plotted number is chosen for teaching; none is a biological measurement or a fitted human parameter. Let externally prescribed, dimensionless $g>0$ represent a glucose stimulus relative to a reference. Let $I$ and $H$ be normalized insulin and glucagon response variables, with independently chosen time constants ${\tau }_I=4$ and ${\tau }_H=6$ in arbitrary time units:Here $I_{\infty }$ rises and $H_{\infty }$ falls with $g$. Their targets add to one by construction; the evolving $I+H$ need not equal one because their time constants differ. Start with $g=1$ and $I=H=1/2$. Set $g=1.6$ from $t=10$ to $40$, $g=0.65$ from $t=40$ to $70$, and $g=1$ otherwise. Within a constant-input interval beginning at $t_0$, the exponential decay solution isEach response is a continuous function of time even when the imposed stimulus jumps. The chosen starting values stay between zero and one, since each solution is a weighted arithmetic mean of its interval's starting value and target.

For an illustrative downstream comparison, define release tendency $R=0.3+0.6H/(1+I)$ and uptake tendency $U=(0.25+0.5I)g$, in the same arbitrary flux units. At the reference target, $R=U=0.5$. At the high-stimulus targets, $R\approx 0.343$ and $U\approx 1.094$; at the low-stimulus targets, $R\approx 0.742$ and $U\approx 0.212$. These chosen functions encode greater uptake with insulin and greater release with glucagon, moderated by insulin; they are not established enzyme kinetics. The $g$ trace is imposed externally, so the figure does not demonstrate self-correction of blood glucose. A closed model would also need a conservation of mass equation, nutrient input, finite glycogen stores, and additional tissue interactions. The qualitative restoring loops are shown separately in the insulin circuit.

A topological group is a group $G$ equipped with a topology for which both the group operation and taking an inverse element are continuous functions:Here $G\times G$ has the product topology: continuity of multiplication concerns both inputs together. Thus nearby motions compose to nearby motions, and undoing a motion also varies continuously. The definition alone does not impose compactness; we state the Hausdorff space assumption explicitly when it is needed.

The additive group $(\mathbb{R},+)$ of real numbers, with its usual topology, is a topological group: $(x,y)\mapsto x+y$ and $x\mapsto -x$ are continuous functions. Its identity element is $0$. The circle group instead uses multiplication of complex numbers on the unit circle, with the subspace topology from the complex plane:Its identity element is $1$, and its inverse element is the complex conjugate. Multiplication on the complex plane and complex conjugation are continuous functions, so their restrictions verify the required continuity. Angles add, but points multiply: a quarter-turn followed by another quarter-turn gives $i\cdot i=-1$, and the next two quarter-turns return to $1$. Ordinary addition would fail to stay on the unit circle, since $1+1=2$.

For any fixed $a\in G$, the translations $x\mapsto ax$ and $x\mapsto xa$ are homeomorphisms, with inverses given by translation by $a^{-1}$. Consequently the neighbourhoods of the identity element determine the entire topology. This is why local information near one motion can control a whole family of symmetries. See Tornier, Haar Measures, section 1.1 for these basic properties of topological groups.

A topological ring is a ring equipped with a topology making addition, additive inversion and multiplication continuous functions. An adic topology is obtained from powers of an ideal.

A topological semigroup is a semigroup with a topology for which multiplication is jointly continuous.

As a quadrature rule, the trapezoidal rule replaces an integrand by the linear interpolation joining its two endpoint values:For nonnegative endpoint values, this is the area of a trapezoid; in general it is a signed area. The quadrature nodes are $a,b$, each with quadrature weight $(b-a)/2$. Unlike the midpoint quadrature rule, it uses endpoint samples, which can be reused between adjacent panels.

Every affine function $f(x)=\alpha x+\beta$ agrees with its linear interpolation, so the trapezoidal rule integrates it exactly. On $[0,2]$, $f(x)=2x+1$ gives $T(f)=(2/2)(1+5)=6=I(f)$. On $[0,1]$, the polynomial $f(x)=x^2$ instead gives $T(f)=1/2$, exceeding $I(f)=1/3$ by $1/6$. Its degree of precision of a quadrature rule is exactly one.

If $f\in C^2[a,b]$, its signed approximation error satisfiesA convex function lies below its endpoint chord, so the trapezoidal rule overestimates its integral; a concave function gives the reverse inequality. The error formula assumes that the second derivative exists and is a continuous function throughout the closed interval.^[ref] A short derivation is the identityobtained by two integrations by parts. The nonnegative Peano kernel has integral $(b-a{)}^3/12$, and the weighted mean value theorem for integrals gives the formula.

The trapezoidal rule for an ordinary differential equation averages vector fields at the old and new states and is A-stable.

This time-stepping use applies the same trapezoidal rule to the integral of the derivative in an ordinary differential equation. That connection explains its place under implicit Runge-Kutta methods; its quadrature behavior and the nonlinear issue in trapezoidal rule fails B-stability concern different uses of the same endpoint averaging.

If $X\subseteq \mathbb{R}$ is not compact, there is an unbounded continuous function $X\to \mathbb{R}$. By the Heine-Borel theorem, either $X$ is unbounded, when $f(x)=x$ works, or it is not closed. In the latter case choose $a\in \overline{X}\setminus X$; the continuous function $f(x)=|x-a{|}^{-1}$ is unbounded because points of $X$ approach the missing limit point $a$. This proof uses the specific compactness characterization of real subspaces; boundedness of a general metric space does not imply compactness.

Suppose continuous functions $f_n$ converge by uniform convergence to $f$ on a domain of a function $X$, and $x_n\to x$ with all these inputs in $X$. The uniform limit theorem makes $f$ a continuous function, and the triangle inequality givesThe first term handles the changing approximation and the second handles the changing input. For $f_n(x)=x^n$ on $[0,1]$, the inputs $x_n=2^{-1/n}$ instead give $x_n\to 1$, $f_n(x_n)=1/2$, and $f(1)=1$. Pointwise convergence cannot supply the first uniform bound, and this pointwise limit also lacks continuity at one.

Let $f_n$ be continuously differentiable functions on an interval $I$. If $f_n^{\prime }\to g$ by uniform convergence on $I$ and $f_n(c)\to A$ at one fixed $c\in I$, then $f_n$ converge locally uniformly to a continuously differentiable function $f$ with $f^{\prime }=g$. If $I$ is bounded, the convergence is uniform on all of $I$.^[ref]

The anchor fixes the additive constant that a derivative cannot see. By the fundamental theorem of calculus, defineThe uniform limit theorem makes $g$ a continuous function, so the fundamental theorem of calculus gives $f^{\prime }=g$. The displayed estimate proves uniform convergence wherever $|x-c|$ has a common finite bound, and therefore proves locally uniform convergence on any interval.

For example, on $[0,2]$ take $f_n(x)=3+1/n+x^2/2+x/n$. Its derivative is $x+1/n$, with supremum error $1/n$ from $g(x)=x$, and $f_n(0)\to 3$. The reconstructed uniform limit is $f(x)=3+x^2/2$; the exact supremum error is $3/n$, equal to the bound above. Without an anchor, the constant functions $f_n(x)=n$ have zero derivatives but no finite pointwise limit. On the real line, $f_n(x)=x/n$ has convergent anchored values and uniformly convergent derivatives, but its supremum error from zero is infinite. The bounded-distance condition has a real role.

A uniform limit of continuous functions is a continuous function. No compactness assumption on the domain of a function is required.^[ref]

For a proof, fix an input $x$ and tolerance $\epsilon >0$. Uniform convergence supplies one $f_N$ with $\Vert f_N-f{\Vert }_{\infty }<\epsilon /3$. Its continuity at $x$ supplies $\delta >0$ with $|f_N(y)-f_N(x)|<\epsilon /3$ for allowed inputs satisfying $|y-x|<\delta$. The triangle inequality givesOne approximating function controls both endpoints of this estimate. Under pointwise convergence, selecting an index for $x$ supplies no common control at nearby $y$. This explains why the polynomial functions $x^n$ can have a discontinuous pointwise limit on $[0,1]$.

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.

A continuous function on a compact interval is uniformly continuous by the Heine-Cantor theorem. Choose a partition of an interval whose widths lie below the input tolerance for a desired output error $\epsilon$. On each half-open partition subinterval, use the value of the function at its left endpoint, and assign the rightmost endpoint its own value. This defines a step function whose error is strictly below $\epsilon$ at every input. Finitely assigned partition-point values are allowed and do not alter the Riemann integral.

Evaluation at zero is a bounded homomorphism on $C[0,1]$. Its kernel $I$ is a closed self-adjoint ideal and hence a C-star subalgebra. It has no identity: applying a proposed identity to $f(t)=t$ would force that identity to equal one at every positive point, contradicting continuity and its value zero at zero. The map $(f,\lambda )\mapsto f+\lambda 1$ from the unitization of a C-star algebra is a bijective star homomorphism, with inverse $h\mapsto (h-h(0),h(0))$. Pulling back the supremum norm verifies the C-star identity, so this is the prescribed unitization norm. Thus adjoining the missing constant direction recovers all continuous functions on the interval.

Let open sets $U_t$ be indexed by a dense subset of $[0,1]$, with $\overline{U_s}\subseteq U_t$ for $s<t$. Taking the infimum of membership indices, with value one outside their union, gives a continuous function. The strict sublevel sets are unions of $U_t$; strict superlevel sets are unions of complements of $\overline{U_t}$. These identities prove continuity.

Every second-countable space that is also a T3 space is a metrizable space. First regularity of a topological space and the Lindelöf property yield a normal space: cover each of two disjoint closed sets by countably many open sets whose closures avoid the other, and subtract the first $n$ opposite closures from the $n$th neighborhood. The resulting unions are disjoint open neighborhoods. Urysohn's lemma then gives, for each pair of basic sets ${\overline{B}}_n\subseteq B_m$, a continuous function equal to one on the smaller closure and zero outside the larger set. There are countably many such pairs. The shrinking form of regularity of a topological space shows these functions generate the topology, and $T_1$ makes them separate points. Their coordinate map embeds the space into the countable cube, with compatible metric $d(x,y)={\sum }_j{2}^{-j}|f_j(x)-f_j(y)|$. The bounded tails prove compatibility with the product topology. Thus a countable base converts separation axioms into a numerical distance.

Variation of parameters replaces the constants in a complementary solution by functions to construct a particular solution.

For the inhomogeneous linear differential equation $y^{\prime \prime }+p{y}^{\prime }+qy=r$, assume $p,q,r$ are continuous functions on an interval $I$, and $y_1,y_2$ are linearly independent homogeneous solutions there. With $W=W(y_1,y_2)$ and $x_0\in I$, variation of parameters constructs the particular solutionThe Wronskian has no zeros on $I$, so these quotients are defined. For $A{y}^{\prime \prime }+B{y}^{\prime }+Cy=F$, normalize first: $r=F/A$, not $F$.^[ref]

To derive the formula, write $y_p=u_1{y}_1+u_2{y}_2$ and impose $u_1^{\prime }{y}_1+u_2^{\prime }{y}_2=0$. This auxiliary choice removes extra terms from the first derivative. Substitution into the inhomogeneous linear differential equation leaves $u_1^{\prime }{y}_1^{\prime }+u_2^{\prime }{y}_2^{\prime }=r$. The two linear equations for $u_1^{\prime },u_2^{\prime }$ have determinant $W$, and Cramer's rule givesIntegrating from $x_0$ makes $y_p(x_0)=y_p^{\prime }(x_0)=0$. Arbitrary integration constants contribute only homogeneous solutions; adding $c_1{y}_1+c_2{y}_2$ supplies the general solution.

For a worked example, use the same homogeneous solutions $e^{-x},e^{2x}$ and the forcing term $r=6e^{3x}$ in $y^{\prime \prime }-y^{\prime }-2y=6e^{3x}$. With $x_0=0$ and $W=3e^x$,Consequently the particular solution selected by zero initial conditions isIts value and first derivative vanish at zero. Direct substitution confirms the forcing term: the differential operator $L[y]=y^{\prime \prime }-y^{\prime }-2y$ annihilates the last two terms and maps $e^{3x}$ to $4e^{3x}$. The two integrals therefore build a response from known homogeneous solutions, even when the forcing term has no convenient elementary guess.

For real continuous functions $f,g$ on a compact interval with $g\ge 0$, some $\alpha$ in the interval satisfies $\int fg=f(\alpha )\int g$. If $\int g>0$, the weighted average lies between the minimum and maximum of $f$, and the intermediate value theorem supplies $\alpha$. If $\int g=0$, the bound $|\int fg|\le \sup |f|\int g$ makes both sides zero. Nonnegativity of the weight is essential to this argument.

Let $G$ be a connected compact Lie group, $T$ a maximal torus, and $W$ its Weyl group. With normalized Haar measures and $D(t)={\prod }_{\alpha >0}(1-e^{-\alpha }(t))$, every continuous class function satisfies ${\int }_{G}f(g)dg=|W{|}^{-1}{\int }_{T}f(t)|D(t){|}^{2}dt$. For a general continuous function, replace $f(t)$ by its average ${\int }_{G/T}f(gt{g}^{-1})d(gT)$. Conjugation from $G/T\times T$ covers the regular elements $|W|$ times, and its real Jacobian determinant is $|D(t){|}^2$.