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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.

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.

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 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 Mahler coefficients tend to zero, so the displayed series defines a continuous function and has value zero at zero. The Pascal's identity and uniform convergence give $\Delta Jf=f$. Thus the forward difference operator is surjective on continuous functions on the p-adic integers. Its kernel consists of constants: period one implies agreement on the dense nonnegative integers, and continuity then implies constancy. This selects the unique discrete antiderivative vanishing at zero.

There is also a direct construction on locally constant functions. If $h$ has period $M=p^r$, put $a_j=h(j)$, $S={\sum }_{i=0}^{M-1}{a}_i$, and $P_j={\sum }_{i<j}{a}_i$. For $x=j+My$ with $0\le j<M$ and $y\in {\mathbb{Z}}_p$, setThis is a continuous function on each residue class. Increasing $j$ gives $Jh(x+1)-Jh(x)=a_j$; at $j=M-1$, the next point has residue zero and quotient $y+1$, giving the same identity. The ultrametric inequality gives $\Vert Jh{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }$, and $Jh(0)=0$. The construction is independent of the chosen period, since two normalized discrete antiderivatives agree on the nonnegative integers and then on the p-adic integers by continuity. It is linear on the locally constant functions, which form a dense subset in the supremum norm. Completeness therefore extends it to every continuous function, retaining $\Delta J=I$ and the norm bound. This proves surjectivity without first using the Mahler theorem.

The Fejér kernel is nonnegative and has unit mass in its convolution normalization. The integral triangle inequality therefore bounds the supremum norm of a Fejér sum by that of the original function. In the half-kernel convention $F_n=(2n{)}^{-1}|{\sum }_{j=0}^{n-1}{e}^{ijt}{|}^2$, its integral is $\pi$ and the convolution prefactor is $1/\pi$.

Every continuous $2\pi$-periodic function admits degree-at-most-$n$ trigonometric polynomials with supremum norm error bounded by a universal constant times its modulus of continuity at scale $1/n$, for $n\ge 1$. This first-order Jackson-type estimate transfers to algebraic polynomials through cosine substitution for polynomial approximation.

The full norm is $|u{|}_{k,\alpha ;U}=|u{|}_{k;U}+[D^{k}u{]}_{\alpha ;U}$, combining the derivative supremum norm with the top-derivative Hölder seminorm. At order zero it is the sum of the supremum norm and the Hölder seminorm.

The Hölder seminorm of exponent $0<\alpha \le 1$ isIt measures Hölder continuity without including the supremum norm of the function itself.

If $f_j\to f$ uniformly and the Hölder seminorms are uniformly bounded, each fixed difference quotient converges. Taking the supremum over point pairs gives $[f{]}_{\mu }\le {\mathrm{liminf}}_j[f_j{]}_{\mu }$. The supremum norms converge, so the full Hölder norm has the same lower semicontinuity of a Hölder seminorm.

Any submultiplicative algebra norm on all of $C(K)$ dominates its supremum norm, even when the new norm is incomplete. Every evaluation character extends continuously to the completion: the restrictions of the completion's character space form a closed subset of $K$, and an omitted point would give a nonzero function annihilated by an invertible function in the completion.

For a real positive linear functional on the space of continuous functions on a compact space, $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ gives $|\varphi (f)|\le \varphi (1)\Vert f{\Vert }_{\infty }$. Evaluation at $1$ gives equality of the operator norm and $\varphi (1)$. Thus positivity alone already implies continuity in the supremum norm.