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An additive composite expansion combines an inner expansion and an outer expansion by subtracting their common behavior in the overlap region:The subtraction removes the part counted twice. Near the boundary layer, the outer expansion cancels the overlap contribution; away from the boundary layer, the inner expansion cancels it. An error estimate is needed to establish accuracy across the whole domain.^[ref]

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$, $y(0)=0$, $y(1)=1$, the outer solution is one, the leading inner expansion is $1-e^{-x/ϵ}$, and their overlap is one. Thus $y_{\mathrm{comp}}=1-e^{-x/ϵ}$. With $q=e^{-1/ϵ}$, comparison with the exact normalized solution givesThe supremum norm error is therefore exponentially small on the entire interval, even though the outer solution alone has error one. This is an error bound for function values; errors in derivatives have additional powers of ${ϵ}^{-1}$.

The composite satisfies the differential equation exactly but has right boundary condition $y_{\mathrm{comp}}(1)=1-q$. Dividing it by $1-q$ restores that boundary condition and recovers the exact solution. This example makes both the strength and the limit of matching explicit: it gives a global approximation, while finite-$ϵ$ normalization can still carry information beyond every algebraic order.

A Banach star-algebra is a complex Banach algebra equipped with a continuous algebra involution. Thus the involution is conjugate-linear, reverses products, and squares to the identity. Some conventions additionally require $\Vert a^{\ast }\Vert =\Vert a\Vert$; this is the isometric convention. In the unital case the involution fixes the unit. The C-star identity is a stronger additional condition. The algebra of continuous complex-valued functions on a compact Hausdorff space, with pointwise operations, complex conjugation and the supremum norm, satisfies both the isometric condition and the C-star identity.

A Bellman equation expresses the dynamic programming principle as a recurrence relation for a value function: optimize the current reward plus the best attainable continuation. Let $s$ be a dynamic programming state, $A_t(s)$ the allowed actions at time $t$, $r_t(s,a)$ the immediate reward, and $F_t(s,a)$ the next dynamic programming state. With terminal time $T$ and terminal reward $g$, the deterministic finite-horizon Bellman equation isHere $V_t(s)$ measures total reward from time $t$ through the terminal reward. Start with $g$ and evaluate backwards; the terminal condition anchors the recurrence relation. For finite nonempty action sets, finite rewards, and finite terminal rewards, a maximum exists at each step. If an optimum need not be attained, use a supremum. A minimization problem instead uses a minimum or infimum. The 0-1 knapsack problem above is the same comparison between an immediate item value plus a continuation and the continuation after skipping. See Bertsekas, MIT lecture 2: the backwards recursion.

In a Markov decision process, the next dynamic programming state $S_{t+1}$ is random. If $r_t(s,a)$ denotes the expected immediate reward and $\beta$ is a discount factor, the finite-horizon Bellman equation becomesThe conditional expectation averages over the next dynamic programming state before choosing the current action; that action cannot use a future outcome that has not yet been observed. For a finite state space with transition probabilities $p_t(s^{\prime }\mid s,a)$, the continuation term is $\beta {\sum }_{s^{\prime }}{p}_t(s^{\prime }\mid s,a)V_{t+1}(s^{\prime })$. Taking $\beta =1$ gives the undiscounted finite-horizon problem. With a discount factor $0\le \beta <1$, rewards $k$ steps ahead carry a factor ${\beta }^k$, including the terminal reward. A maximizing action for each $(t,s)$ specifies a Markov policy, which contains decisions rather than just optimal scores.

For an infinite horizon with a finite state space, finite nonempty action sets, time-independent transition probabilities, bounded rewards $|r(s,a)|\le M$, and a discount factor $0\le \beta <1$, the optimal value function satisfiesIn this setting the maximizing dynamic programming operator is a contraction mapping in the supremum norm, so the Banach fixed-point theorem gives a unique bounded fixed point and convergence of repeated application to it. See Bertsekas, MIT lecture 15: bounded discounted problems. These hypotheses matter: with one state, one action, zero reward, and $\beta =1$, the Bellman equation is merely $V(s)=V(s)$, so every finite constant is a fixed point, although the actual total-reward value function is zero. The average-reward Bellman equation and Hamilton-Jacobi-Bellman equation address other settings with their own conditions.

Take the continuous function $f(x)=|x-1/2|$ on $[0,1]$. It is Lipschitz continuous with Lipschitz constant one, but is not differentiable at $x=1/2$. For $n=2$, the sample values are $1/2,0,1/2$, soThis is already a useful warning: the Bernstein polynomial is built from grid values but is not generally their polynomial interpolation.

Every finite approximant is differentiable, and symmetry makes its derivative zero at the center. The target instead has left derivative $-1$ and right derivative $1$. Smooth curves can therefore approach a corner in supremum norm while their center derivatives have no target value to approach. The Lipschitz error bound for Bernstein approximation quantifies the improvement in values.

At the sampling point $j/n$, the polynomial inside the Bernstein polynomial has value $n^{-m}j(j-1)\cdots (j-m+1)$. The falling factorial identity $j^{\underline{m}}\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n^{\underline{m}}\left(\genfrac{}{}{0ex}{}{n-m}{j-m}\right)$ and the binomial theorem prove the displayed formula. For fixed $m$, both the input and its scalar factor tend uniformly to $x^m$ as $n$ grows. The supremum norm contraction of the Bernstein polynomial therefore gives $\Vert B_n(x^m)-x^m{\Vert }_{\infty }\le m(m-1)/n$ for $n\ge m$.

Given $f$ in a normed vector space and a vector subspace $U$, an element $p_{\ast }\in U$ is a best approximation if $\Vert f-p_{\ast }\Vert ={\inf }_{p\in U}\Vert f-p\Vert$. Existence is automatic for a finite-dimensional vector subspace: a minimizing sequence can be bounded by the triangle inequality, and a subsequence converges by compactness in finite dimension. An arbitrary infinite-dimensional vector subspace need not contain a minimizer. The norm matters: best uniform approximation uses the supremum norm, while $L^1$ approximation minimizes integrated absolute error.

Bounded continuous functions on a topological space form a Banach space with the supremum norm. A uniform limit is continuous and bounded, proving completeness. This space differs from the bounded uniformly continuous functions: free-transport composition is jointly continuous in time and position, but need not be continuous in the global sup norm as time varies.

The bounded Lipschitz norm combines the Lipschitz constant and supremum norm of a bounded real-valued Lipschitz function on a metric space. Both summands satisfy the triangle inequality; the supremum term makes the sum definite even on constant functions. Its unit ball requires the sum of the two bounds to be at most one, a stronger requirement than bounding each separately by one.

For an arbitrary set $\Gamma$ and scalar field $\mathbb{F}$, this is the space of bounded functions $u:\Gamma \to \mathbb{F}$ with the supremum norm $\Vert u\Vert ={\sup }_{\gamma }|u(\gamma )|$. Uniform limits of Cauchy sequences prove it is a Banach space. For $\Gamma =\mathbb{N}$ it is the l-infinity sequence space; for the empty index set it is the zero space, with norm zero.

For distinct interpolation sites and an invertible B-spline collocation matrix, the interpolant is $P=B{A}^{-1}R$, where $R$ samples data and $Ba={\sum }_i{a}_i{N}_i$. Sampling and basis synthesis have operator norms at most one in the supremum norm. For the reverse inequality, realize the signs of a maximal absolute row sum of $A^{-1}$ as values of a continuous function of norm one, and apply uniform-norm stability of a B-spline basis. This yields both displayed bounds.

Extend the cylinder premeasure from a positive functional to a Borel probability measure and use cylinder-function density in Cantor space. The integral and functional agree on cylinder simple functions, and both are continuous in the supremum norm, so they agree on all of $C(\Omega )$. Uniqueness on the generating cylinder algebra gives uniqueness of the Borel measure.

For a nonnegative partition-normalized B-spline basis, the synthesis linear map $Tc={\sum }_j{c}_j{N}_j$ has operator norm one. Its condition number is consequently $\Vert T^{-1}\Vert$, with the supremum norm on the spline space and the maximum norm on its coefficients. The uniform-norm stability of a B-spline basis bounds this number independently of the spline knot sequence when the order is fixed. This is different from the condition number of a particular sampled B-spline collocation matrix.

The displayed supremum norm is finite on every entire function, positive on every nonzero one by the identity theorem, and submultiplicative under pointwise multiplication. It is not complete: partial geometric sums ${\sum }_{j=0}^n(z/2{)}^j$ are Cauchy on the disk but their limit there extends meromorphically with a pole at two. More generally, no complete algebra norm exists on all entire functions because the coordinate function has algebraic spectrum of an element equal to the whole complex plane.

Let $a_m\ge 0$ and $a_m\to 0$. The set $K=\{x=(x_m):x_m\in \mathbb{R},|x_m|\le a_m\}$ is a compact metric space for the supremum norm metric. Every member is a sequence tending to zero.

For any sequence in $K$, successively extract a convergent subsequence in each compact interval $[-a_m,a_m]$, then take diagonal indices. The coordinate limits define $x\in K$. For any $M$, coordinatewise differences in the tail are bounded by $2{\sup }_{m>M}{a}_m$, independently of the sequence index. This tends to zero as $M\to \infty$, while finitely many initial coordinates converge together. Hence the diagonal subsequence converges in the supremum norm. This proves sequential compactness, equivalent to compactness for metric spaces.

The image of $C([0,1])$ in the L1 space $L^1(0,1)$ is a meagre set. The canonical map $J$ is a bounded linear operator from the Banach space $C([0,1])$ with its supremum norm, since $\Vert Jf{\Vert }_1\le \Vert f{\Vert }_{\infty }$. It is not onto: the indicator of $(1/2,1)$ has no continuous function representative. Continuity would force values zero throughout $(0,1/2)$ and one throughout $(1/2,1)$, contradicting continuity at $1/2$. The operator range category dichotomy therefore proves the claim. Continuous ramps $f_n(t)=\min (1,\max (0,n(t-1/2)))$ also show directly that this image is not closed, since their L1 norm distance to that indicator is $1/(2n)$ for $n\ge 2$.

The continuous functions from the p-adic integers to the P-adic numbers form a Banach space over ${\mathbb{Q}}_p$ with supremum norm $\Vert f{\Vert }_{\infty }={\sup }_{x\in {\mathbb{Z}}_p}|f(x){|}_p$. Compactness of the domain bounds the norm, and uniform convergence in the complete codomain proves completeness. Translation and the forward difference operator $\Delta f(x)=f(x+1)-f(x)$ are bounded operators, with $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$.

A continuous function on Cantor space is uniformly continuous. Replacing it by one constant on each length-$n$ prefix cylinder set changes it by at most its oscillation on sets of diameter $2^{-n}$. This tends uniformly to zero. Each approximant is a finite linear combination of continuous indicator functions, proving density in the supremum norm.

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

For a function with bounded derivatives through order $k$, the derivative supremum norm is $|u{|}_{k;U}={\sum }_{j=0}^k{\sup }_U|D^{j}u|$, with any equivalent finite-dimensional norm for each derivative tensor. This is a classical $C^k$ norm; it is different from the Lp norm with exponent $k$.

The differentiation operator sends a differentiable function to its derivative; its domain and the norms must be specified. On $C^1([-1,1])$ with values in $C([-1,1])$, using the supremum norm on both, it is an unbounded linear operator: $f_n(x)=\sin (nx)$ satisfies $\Vert f_n{\Vert }_{\infty }\le 1$ and $\Vert D{f}_n{\Vert }_{\infty }=n$. Its graph norm is $\Vert f{\Vert }_{\infty }+\Vert f^{\prime }{\Vert }_{\infty }$, which is complete. Indeed, the kernel consists of constants and is complete, while $Sg(x)={\int }_0^{x}g(t)dt$ is a bounded right inverse by the fundamental theorem of calculus; apply completeness of a graph norm from a bounded right inverse.