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On the vector space of continuous functions $f:[0,1]\to \mathbb{R}$ of bounded variation with $f(0)=0$, the total variation of a function is a norm. Its triangle inequality and absolute homogeneity of a norm follow by applying the corresponding absolute value identities to every partition of an interval, then taking the supremum. The partition of an interval $(0,t,1)$ gives $|f(t)|\le V_0^1(f)$, so zero total variation of a function forces $f=0$. The same inequality gives $\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_V$, proving that convergence for this norm implies uniform convergence. Without $f(0)=0$, every constant function would have zero total variation of a function.

Suppose a perturbation $\delta x$ satisfies $\Vert \delta x\Vert \le \epsilon$, with $\epsilon \ge 0$. Absolute homogeneity of a norm and the dual norm definition giveFor nonzero $\delta x$, divide it by its norm to obtain an input in the unit ball; for zero perturbation the inequality is immediate. This is an exact error bound for a linear functional, for every perturbation in the stated input set.

For $p(x)=2x_1-x_2$ and a budget $\epsilon =0.1$, the sharp output-error bounds are $0.2$ for $\Vert \delta x{\Vert }_1\le 0.1$, $\sqrt{5}/10\simeq 0.224$ for $\Vert \delta x{\Vert }_2\le 0.1$, and $0.3$ for $\Vert \delta x{\Vert }_{\infty }\le 0.1$. Scale the maximizing vectors from the preceding example by $0.1$ to attain the positive endpoints; their negatives attain the negative endpoints. The L1 norm budget constrains a total absolute error; the maximum norm budget permits each coordinate to have that error separately. These budgets define nested sets and carry different information.

The coordinates must use the stated scale and units. Rescaling one coordinate changes both its coefficient in $a$ and the meaning of its input budget. For a nonlinear differentiable function with scalar values, its Fréchet derivative at the base point is a linear functional governing the first-order response. An exact finite-perturbation error bound additionally needs control of the Taylor remainder or of the Fréchet derivative along the path.

For measurable functions $f,g$ with values in the real numbers or complex numbers on the same measure space $(X,\mathcal{A},\mu )$, and $1\le p\le \infty$, membership in the same Lp space givesThus adding two functions cannot make their combined Lp norm exceed the sum of their separate Lp norms. The measure need not be finite or a probability measure. The hypotheses require finite Lp norms; $p=\infty$ uses the essential supremum of the absolute value, not an ordinary pointwise supremum. For complex numbers, the absolute value is the complex modulus. As usual, Lp spaces identify functions equal almost everywhere. ^[ref]

The Minkowski inequality makes the Lp norm a workable notion of size: it supplies the triangle inequality needed to turn $d_p(f,g)=\Vert f-g{\Vert }_p$ into a metric on Lp space. Applied to $(f-h)+(h-g)$, it says that travelling through an intermediate function $h$ cannot shorten the direct distance measured by this metric. Applied to two error functions, it gives a bound on the total error even when the errors reinforce one another; Minkowski inequality for error aggregation gives explicit examples.

The geometry is already visible in the Euclidean plane. Under counting measure on two coordinates, the closed unit balls for the L1 norm, Euclidean norm, and L-infinity norm are respectively $|x|+|y|\le 1$, $x^2+y^2\le 1$, and $\max (|x|,|y|)\le 1$. Every one is a convex set: if $u,v$ have Lp norm at most one, the Minkowski inequality and absolute homogeneity of a norm give $\Vert tu+(1-t)v{\Vert }_p\le t+(1-t)=1$ for $0\le t\le 1$. The Euclidean norm has no flat boundary segments, whereas the other two norms do. This is the distinction captured by a strictly convex normed space and developed in equality in Minkowski inequality.

A short Hölder's inequality proof outline handles $1<p<\infty$. First $|f+g{|}^p\le 2^{p-1}(|f{|}^p+|g{|}^p)$, so $f+g$ has finite Lp norm before any division. With conjugate exponents $p$ and $q=p/(p-1)$, set $h=|f+g{|}^{p-1}$; then $\Vert h{\Vert }_q=\Vert f+g{\Vert }_p^{p-1}$. The scalar triangle inequality followed by Hölder's inequality givesDivide if $\Vert f+g{\Vert }_p>0$; otherwise the desired bound is immediate. For $p=1$, integrate $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the result by the essential supremum. The same scalar triangle inequality works for real numbers and complex numbers.

For vectors $u,v\in {\mathbb{R}}^n$ or ${\mathbb{C}}^n$, counting measure turns the Minkowski inequality intoFor $p=\infty$, replace each displayed norm by the maximum of its coordinate absolute values. Take $u=(1,0)$ and $v=(0,1)$. Both vectors have Lp norm one for every $p\ge 1$, but their sum has L1 norm $2$, Euclidean norm $\sqrt{2}$, and L-infinity norm $1$. Thus the same vector addition gives equality $2=1+1$ at $p=1$, strict inequality $\sqrt{2}<1+1$ at $p=2$, and strict inequality $1<1+1$ at $p=\infty$.

By contrast, $u=(1,1)$ and $v=2u=(2,2)$ give equality for every $p\ge 1$: absolute homogeneity of a norm yields $\Vert u+v{\Vert }_p=3\Vert u{\Vert }_p=\Vert u{\Vert }_p+\Vert v{\Vert }_p$. At $p=2$, the three sizes are $\sqrt{2}$, $2\sqrt{2}$, and $3\sqrt{2}$. Opposite vectors behave very differently: $u+(-u)=0$, so the inequality is strict whenever $u$ is nonzero. Cancellation is allowed; the bound describes the largest possible total size.

Both terms in the graph norm are finite and nonnegative. Moreover, $\Vert x{\Vert }_T=0$ forces $\Vert x{\Vert }_X=0$, hence $x=0$; conversely $T0=0$ gives $\Vert 0{\Vert }_T=0$. This proves definiteness of a norm.

The linearity of $T$ and absolute homogeneity of a norm give, for each real $a$,Applying the triangle inequality in both normed spaces givesThus

Write $\psi ={\psi }_{x_0}$; its value at $x_0$, if that point is in $(0,1)$, may be assigned arbitrarily because Lp spaces identify functions equal almost everywhere. Its L-infinity norm is one.

The supplied estimate shows that $\psi$ cannot agree almost everywhere with any continuous function $q$: choosing $g=-q$ would give $0\ge |q(x_0)|+1$. Henceis a direct sum, so every element has a unique expression $g+c\psi$.

For any complex $a$ with $|a|\le 1$, define the linear functionalIf $c\ne 0$, the supplied estimate, applied to $g/c$, and absolute homogeneity of a norm giveFor $c=0$, the same bound follows from point evaluation. Thus $\Vert L_a\Vert \le 1$, and $L_a(1)=1$ makes its operator norm exactly one. The complex Hahn-Banach theorem extends $L_a$ to a norm-one linear functional on all of $X$, with value $a$ at $\psi$.

Conversely every norm-one extension satisfies $|{\ell }_{x_0}(\psi )|\le \Vert \psi {\Vert }_{\infty }=1$. We have therefore determined all possible values:In particular, taking $a=0$ and $a=1$ proves nonuniqueness of norm-preserving point evaluation extensions.

The integral is finite because a continuous function on a compact closed interval is bounded. Nonnegativity is immediate, and absolute homogeneity of a norm follows from $|cf(x)|=|c||f(x)|$. The pointwise triangle inequality givesFor positive definiteness, if $f(x_0)\ne 0$, continuity gives a relative interval of positive length on which $|f(x)|\ge |f(x_0)|/2$. Its integral is positive, including when $x_0$ is an endpoint. Thus $\Vert f\Vert =0$ implies $f=0$, and this is a norm on the given vector space.

For the given sequence,The last bound tends to zero as both indices tend to infinity, proving the Cauchy sequence property. Moreover, the sequence converges to the zero function in this norm:There is no conflict with the nonzero endpoint values: convergence in this norm does not imply pointwise convergence.

Uniform expectations require nonempty finite sets; assume this throughout the analytic argument. For real-valued functions on $X\times Y$, define the box norm byAll four variables are sampled as independent random variables with replacement, so repeated coordinates are included. The defining fourth power equalsIf it vanishes, every summand vanishes. In particular, the terms $y_0=y_1=y$ give ${\mathbb{E}}_{x}f(x,y{)}^2=0$ for each $y$, so $f=0$. Absolute homogeneity of a norm follows directly: $\Vert af{\Vert }_{\square }=|a|\Vert f{\Vert }_{\square }$.

For the triangle inequality, prove the mixed box Cauchy-Schwarz inequality. WriteSeparating the two $y$ averages and applying the Cauchy-Schwarz inequality in $(x_0,x_1)$ givesExpanding the square and reversing the order of finite sums,by a second Cauchy-Schwarz inequality. Thus $|\Lambda |\le {\prod }_{i,j}\Vert f_{ij}{\Vert }_{\square }$. Expand $\Lambda (f+g,f+g,f+g,f+g)$ into its sixteen multilinear terms and apply this bound to each term. Their upper bounds sum to $(\Vert f{\Vert }_{\square }+\Vert g{\Vert }_{\square }{)}^4$. Taking fourth roots yields the triangle inequality. Together with definiteness of a norm and absolute homogeneity of a norm, this proves the box norm is a norm. Complex functions require conjugates and are outside the printed real-valued convention.

For the bilinear correlation bound for the box norm, use the normalized L2 norm: let $B={\mathbb{E}}_{x,y}f(x,y)u(x)v(y)$ and $\Vert u{\Vert }_2^2={\mathbb{E}}_{x}u(x{)}^2$, $\Vert v{\Vert }_2^2={\mathbb{E}}_{y}v(y{)}^2$. The Cauchy-Schwarz inequality in $y$ givesThe remaining factor is ${\mathbb{E}}_{x,x^{\prime }}u(x)u(x^{\prime }){\mathbb{E}}_{y}f(x,y)f(x^{\prime },y)$. By the Cauchy-Schwarz inequality in $(x,x^{\prime })$, its magnitude is at mostTaking square roots proves

For the tripartite graph, assume its parts are nonempty and write $g_{XY},g_{YZ},g_{XZ}$ for its adjacency indicator functions. Let $f=g_{XY}-\alpha$, so $\Vert f{\Vert }_{\square }\le c$. The normalized triangle count isThe constant-degree hypothesis is exactly ${\mathbb{E}}_y{g}_{YZ}(y,z)=\beta$ for each $z$. Therefore the contribution of the constant $\alpha$ isFor fixed $z$, apply the bilinear correlation bound for the box norm with $u_z(x)=g_{XZ}(x,z)$ and $v_z(y)=g_{YZ}(y,z)$. Since these are indicator functions, $\Vert v_z{\Vert }_2^2=\beta$ and $\Vert u_z{\Vert }_2^2=d_X(z)$, the fraction of $X$ adjacent to $z$. Averaging and applying the Cauchy-Schwarz inequality in $z$ gives the sharper triangle counting with one box-uniform pair and constant opposite degree estimateMultiplying by $|X||Y||Z|$ yieldsThe middle bound also handles $\beta =0$ or $\gamma =0$, when the triangle count is zero.

To connect explicitly with the suggested expansion, put $f_2=g_{YZ}-\beta$, $f_3=g_{XZ}-\gamma$. Both have overall mean zero, and ${\mathbb{E}}_y{f}_2(y,z)=0$ for every $z$. Consequently ${\mathbb{E}}_{x,y,z}{f}_2(y,z)f_3(x,z)=0$; all terms with constant $\alpha$ reduce to $\alpha \beta \gamma$. The terms containing $f$ combine into the single error just estimated. This is where the exact degree assumption is used. If any part is empty the triangle count itself is zero, but the printed uniform expectations and edge density of a bipartite graph values would be undefined; the nonempty-parts convention must therefore be stated rather than dividing by zero.

First verify that $V_0$ is a vector space. Sums and scalar multiples preserve continuity and the condition at zero. For every partition of an interval $D$, the triangle inequality and absolute homogeneity of a norm of the absolute value implyTaking suprema yieldsFor $\lambda =0$ the second equality follows directly from $V(0)=0$. In particular, finite variation is preserved, so $V_0$ is closed under these operations.

Nonnegativity is immediate. For any $t\in [0,1]$, use the partition of an interval $(0,t,1)$, omitting repeated endpoints if necessary, to getThus $V(f)=0$ forces $f=0$, and conversely $f=0$ has zero total variation of a function. These observations prove that $\Vert f\Vert =V(f)$ is a norm on $V_0$.

Applying the same bound to $f_n-f$ givesTherefore convergence in the anchored variation norm implies uniform convergence on $[0,1]$. The anchoring condition $f(0)=0$ removes the nonzero constant functions which otherwise have zero total variation of a function.

A norm on a real vector space $X$ is a map $\Vert \cdot \Vert :X\to [0,\infty )$ such that, for $x,y\in X$ and $\lambda \in \mathbb{R}$,The last property is the triangle inequality; the middle property is absolute homogeneity of a norm. The induced metric isPositivity and separation follow from the norm axioms, symmetry from absolute homogeneity of a norm with $\lambda =-1$, and the triangle inequality from $x-z=(x-y)+(y-z)$.

For a real or complex polynomial $p(t)={\sum }_{k\ge 0}{a}_k{t}^k$, this finite sum defines a norm: definiteness follows from uniqueness of polynomial coefficients, absolute homogeneity of a norm follows termwise, and the triangle inequality follows by summing $|a_k+b_k|\le |a_k|+|b_k|$. On $[0,1]$, its supremum norm satisfies $\Vert p{\Vert }_{\infty }\le \Vert p{\Vert }_{\mathrm{coef},1}$, since $|t^k|\le 1$. The reverse comparison has no constant independent of the degree of a polynomial: for $p_n(t)=(1-t{)}^n$, the two norms are $1$ and $2^n$ respectively, by the binomial theorem.

Two norms $p,q$ on a vector space give the same topology if and only if $cp(x)\le q(x)\le Cp(x)$ for some positive constants. Indeed, continuity of the identity at zero gives $p(x)<\delta \Rightarrow q(x)<1$. For $x\ne 0$, apply this to $\delta x/(2p(x))$ to obtain $q(x)<2p(x)/\delta$. Reverse the roles of the norms for the lower bound. Conversely the two inequalities compare every translated open ball. This proof uses absolute homogeneity of a norm, not finite dimension.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.