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Suppose $B$ contains zero, is convex, and meets every line $\mathbb{R}v$, $v\ne 0$, in the finite symmetric open segment $\{tv:|t|<\lambda (v)\}$ with $\lambda (v)>0$. Then its Minkowski functional is a norm, with $p_B(v)=1/\lambda (v)$ and $B=\{x:p_B(x)<1\}$. Absolute homogeneity follows by rescaling the segment. If $s>p_B(x)$ and $t>p_B(y)$, convexity puts $(x+y)/(s+t)$ in $B$, giving $p_B(x+y)<s+t$; passing to the infima proves the triangle inequality. Conversely the open unit ball uniquely determines any norm by the displayed formula. No prior topology on the vector space is needed for these radial hypotheses.

A normed algebra is an algebra over a field with a norm satisfying $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. Its completion is a Banach algebra, since multiplication extends continuously. Completeness is additional structure, rather than part of the definition of a normed algebra.

A normed vector space is a vector space equipped with a norm, whose induced metric is $d(x,y)=\Vert x-y\Vert$. The norm must be positive on nonzero vectors, scale by the absolute value of a scalar, and satisfy the triangle inequality. These conditions make distance compatible with addition and scaling.

On $C([0,1])$, both the supremum norm $\Vert f{\Vert }_{\infty }={\max }_{0\le x\le 1}|f(x)|$ and the L2 norm $\Vert f{\Vert }_2=({\int }_0^1|f(x){|}^{2}dx{)}^{1/2}$ are genuine norms. A nonzero continuous function stays nonzero on an interval of positive length, so its squared integral cannot vanish. For $f(x)=x$, they give $1$ and $1/\sqrt{3}$, respectively. Always $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$, but the shrinking peaks below show that no uniform reverse bound exists. Unlike the finite-dimensional equivalence of norms, infinite-dimensional vector spaces can support inequivalent norm topologies.

For a bounded convex open set $U$ containing zero and invariant under $x\mapsto -x$, its Minkowski functional is a norm. Inner and outer Euclidean balls give positive finite bounds; symmetry and scaling give absolute homogeneity; convexity gives the triangle inequality. Openness makes $U=\{p_U<1\}$, while boundedness prevents nonzero vectors from having zero gauge.

For a nonzero vector in a real or complex normed vector space, a norming functional is a continuous linear functional satisfying the displayed conditions. The Hahn-Banach theorem extends the functional sending $av$ to $a\Vert v\Vert$ on its one-dimensional span. Consequently $\Vert w\Vert \ge \mathrm{Re}\ell (w)$, with equality at $v$, giving a supporting linear lower bound for the norm. At zero the zero functional can be used as a supporting functional.

A vector subspace $Z\subseteq X^{\ast }$ is norming if some $c>0$ satisfies the displayed inequality. It retains enough continuous linear functionals to control the norm of every vector. Every such vector subspace is dense for the weak-star topology: otherwise a nonzero evaluation linear functional would annihilate its weak-star closure, contradicting the inequality. In the real case its symmetric unit ball makes the absolute-value and signed supremum versions equivalent.

The radius is a scalar derived from the numerical range of an operator, rather than the range set itself.

For a bounded linear operator on a complex Hilbert space, the numerical radius is the supremum of the absolute values in its numerical range of an operator. It is a norm equivalent to the operator norm, with $w(T)\le \Vert T\Vert \le 2w(T)$. For a self-adjoint operator the two norms agree.

If two closed subspaces of a Hilbert space give $H=V\oplus W$, the projection onto $V$ along $W$ fixes $V$ and vanishes on $W$. It need not be an orthogonal projection. The bounded inverse theorem applied to the addition map $(v,w)\mapsto v+w$ on the complete component spaces proves boundedness of the component maps. With both complementary subspaces nonzero, its operator norm and that of $I-Q$ equal $\sec {\theta }_{V,W^{\perp }}$, with the directed subspace angle convention. Indeed, for fixed $v\in V$, the smallest possible norm of $v+w$ over $w\in W$ is $\Vert P_{W^{\perp }}v\Vert$, which proves the formula for $\Vert Q\Vert$. Equality with $\Vert I-Q\Vert$ follows from the block representation $Q=\left(\begin{array}{cc}I& B\\ 0& 0\end{array}\right)$ on $V\oplus V^{\perp }$: both squared operator norms are $1+\Vert B{\Vert }^2$. If a summand is zero, $Q$ is zero or the identity operator, and one complementary operator norm is zero; those cases must be treated directly.

For a tripartite three-uniform hypergraph on nonempty finite parts $X,Y,Z$, let $h:X\times Y\times Z\to \{0,1\}$ be its hypergraph edge indicator, let $p=\mathbb{E}h$ and put $g=h-p$. In the eighth-power convention, $\alpha$-quasirandomness meansAll averages are uniform and independent, with repeated sampled vertices of a hypergraph allowed. This expression is nonnegative because it equals ${\mathbb{E}}_{x_0,x_1,y_0,y_1}({\mathbb{E}}_z{\prod }_{i,j}g(x_i,y_j,z){)}^2$. It measures correlations across the eight faces of a tripartite octahedral configuration. If instead the norm itself is used as the quasirandomness parameter, the parameter in this definition is its eighth power.

Let $U\subseteq D$ be open. Regard $H_{\mathrm{supp}}=H_0^1(U)$ as functions on $D$ by zero extension of H01, and let $H_{\mathrm{harm}}$ consist of the weakly harmonic Sobolev functions on $U$ belonging to $H_0^1(D)$. For $h\in H_{\mathrm{harm}}$, the Dirichlet inner product $(h,\varphi {)}_{\nabla }$ vanishes for each test function supported in $U$. Approximate any $u\in H_{\mathrm{supp}}$ by those test functions in the gradient norm and use the Cauchy-Schwarz inequality. This gives $\boxed{(h,u{)}_{\nabla }=0}$. The proof also works with the inhomogeneous zero-boundary Sobolev space convention, and with a homogeneous completion realized as weak functions.

The orthogonal projection $P_F$ onto a closed subspace of a Hilbert space $F$ sends $x$ to its component in the orthogonal decompositionThe residual $r$ is orthogonal to every permitted direction. For any $z\in F$, the Pythagorean theorem givesThus $P_{F}x$ uniquely minimizes the distance to $F$. It is a linear map, is idempotent, and is nonexpansive: $\Vert P_{F}x-P_{F}y\Vert \le \Vert x-y\Vert$; taking $y=0$ gives $\Vert P_{F}x\Vert \le \Vert x\Vert$.

In ${\mathbb{R}}^2$, project $v=(1,3)$ onto the line spanned by $u=(2,1)$. The inner product givesThe check $⟨r,u⟩=-2+2=0$ confirms orthogonality, and $\Vert v{\Vert }^2=10=5+5=\Vert P_{F}v{\Vert }^2+\Vert r{\Vert }^2$. Projecting again changes nothing, since the result already lies on the line.

For a finite orthonormal basis $e_1,\dots ,e_m$ of $F$, the same operation is $P_{F}x={\sum }_j⟨x,e_j⟩e_j$, using the convention linear in the first argument. Replacing coordinate vectors by functions gives exactly the same approximation principle. For $f(x)=x^2$ in the real L2 space on $[-1,1]$ and $F=\mathrm{span}\{1,x\}$, the directions are orthogonal, with squared norms $2$ and $2/3$. Their inner products with $f$ are $2/3$ and zero, soDirect integration gives the squared error $\Vert x^2-1/3{\Vert }^2=8/45$. For any competing affine polynomial, the orthogonal residual yieldsEvery deviation from $a=1/3$, $b=0$ adds a nonnegative error. The best function need not pass through any selected data point: it minimizes an integrated error across the whole interval.

A finite-dimensional linear subspace $W$ of an inner product space has a unique closest vector to any given vector $x$. This is its orthogonal projection $P_{W}x$. If $e_1,\dots ,e_d$ is an orthogonal basis of $W$, compute it one direction at a time:The inner product convention here is linear in its first argument, including over the complex numbers. For an orthonormal basis, each denominator is one. When $W$ consists only of the zero vector, the empty sum gives $P_{W}x=0$. The finite-dimensional formula is described in Francis Narcowich, Inner Products and Norms: minimization problems.

Take $W=\mathrm{span}\{(1,2{)}^T\}$ and $x=(3,1{)}^T$, using the dot product. With $e=(1,2{)}^T$,The orthogonal projection residual has $r\cdot e=2-2=0$. To verify the best approximation directly, every candidate in the line is $te$, andThe smallest squared Euclidean norm error is therefore $5$, attained only at $t=1$. The diagram connects the perpendicular residual to this one-variable quadratic polynomial.

The general mechanism is the Pythagorean theorem in an inner-product space. For any $w\in W$, the vectors $r=x-P_{W}x$ and $P_{W}x-w$ are orthogonal, soMoving away from $P_{W}x$ inside $W$ adds a nonnegative squared norm, which is zero only when $w=P_{W}x$. This proves both optimality and uniqueness.

The useful feature is separation of preparation from evaluation. Construct an orthonormal basis once with the Gram-Schmidt process, then obtain each new best approximation using inner products instead of a fresh optimization problem. When $W$ is the column space of a matrix, this gives linear least squares; when $W$ is spanned by orthogonal polynomials, it gives a least-squares polynomial in an orthogonal-polynomial basis. The ambient inner product space need not be a Hilbert space for this finite-dimensional construction. In an infinite-dimensional target linear subspace, existence requires additional hypotheses, supplied by orthogonal decomposition by a closed subspace.

In a $2k$-dimensional normed vector space in John position, two orthogonal $k$-dimensional subspaces admit $\Vert x{\Vert }_E\le |x|\le 32\mathrm{vr}(E{)}^2\Vert x{\Vert }_E$. To see this, take $\epsilon$-nets in two orthogonal half-dimensional unit spheres, each of size at most $(1+2/\epsilon {)}^k$, and rotate them uniformly together. The volume ratio bound makes the probability that any net point has norm below $2\epsilon$ at most $2[4\mathrm{vr}(E{)}^2({\epsilon }^2+2\epsilon ){]}^k$. With $\epsilon =(32\mathrm{vr}(E{)}^2{)}^{-1}$ this is less than $1$. Since $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric, the successful nets give the stated inequality everywhere on both spheres. A dimension-independent constant therefore requires a bound on the volume ratio; it does not hold for arbitrary normed vector spaces.

An orthonormal set consists of vectors that have norm one and are pairwise orthogonal. Its Gram matrix is the identity matrix.

Here equivalent metrics means that the two metrics induce the same topology, and Lipschitz equivalent norms means that there are constants $c,C>0$ such that $cp(x)\le q(x)\le Cp(x)$ for every $x$.

Suppose first that the inequalities hold. Since they also hold for $x-y$, the identity maps between the two metric spaces are Lipschitz continuous. In particular, each is continuous, so the topologies agree.

Conversely, equality of the topologies makes the identity from the $p$-norm to the $q$-norm continuous at zero. There is therefore $\delta >0$ such that $p(x)<\delta$ implies $q(x)<1$. For nonzero $x$, apply this to $\delta x/(2p(x))$ and use the absolute homogeneity of both norms:The reverse identity is also continuous, so there is $\epsilon >0$ with $q(x)<\epsilon \Rightarrow p(x)<1$. Rescaling again gives $p(x)<2q(x)/\epsilon$. ThusThe inequalities also hold at zero. This is topology determines norm equivalence; it works in arbitrary dimension because the rescaling argument needs neither a basis nor compactness of a unit sphere.