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Assume $\alpha <\beta$ and abbreviateThe integral of the nonnegative square expands to $A+2\lambda B+{\lambda }^{2}C\ge 0$ for every real $\lambda$. If $C>0$, completing the square, or choosing $\lambda =-B/C$, gives $A-B^2/C\ge 0$. Hence the Cauchy-Schwarz inequality isIf $C=0$, continuity and nonnegativity imply $g=0$ throughout the interval, so the same inequality is immediate.

When $C>0$, equality holds exactly when $\int (f-(B/C)g{)}^{2}dx=0$. The integrand is continuous and nonnegative, so it must vanish everywhere, giving $f=(B/C)g$. Including the case of a zero function, equality is equivalent to linear dependence of $f$ and $g$ as elements of the real vector space of continuous functions on the interval.

A square matrix $A$ is diagonalizable over a specified scalar field if an invertible matrix $P$ exists such thatEquivalently, the underlying vector space has a basis of eigenvectors of $A$: those basis vectors are the columns of $P$, and their eigenvalues are the diagonal entries of $D$. Here the field is $\mathbb{C}$ unless specified otherwise.

A square matrix $A$ is a diagonalizable matrix over a specified field if some invertible matrix $P$ makes $P^{-1}AP$ diagonal. Equivalently, the underlying vector space has a basis of eigenvectors of $A$.

For example,The first is already diagonal. The second has only the eigenvalue one, of algebraic multiplicity two, but its eigenspace consists of $(x,0{)}^T$ and has dimension one. It cannot supply a two-vector eigenbasis.

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.

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 $x-z=(x-y)+(y-z)$.

Use the vector-space axioms: addition is an abelian group operation with identity zero; ${\lambda }_1(x)=x$; ${\lambda }_a{\lambda }_b(x)={\lambda }_{ab}(x)$; ${\lambda }_a(x+y)={\lambda }_a(x)+{\lambda }_a(y)$; and ${\lambda }_{a+b}(x)={\lambda }_a(x)+{\lambda }_b(x)$, with ${\lambda }_0(x)=0$. Group inverses may be expressed by ${\lambda }_{-1}$, so no extra inverse symbol is needed.

Add the dimension-unbounded schemeOver a finite field, a finite-dimensional space has finitely many Elements, while every infinite-dimensional space is infinite. Hence this scheme axiomatizes exactly the infinite underlying vector spaces requested. In this finite-field case the vector-space axiom list itself is finite; the infinitude of the additional dimension scheme is substantive.

No. If a finite conjunction $\varphi$ axiomatized the infinite spaces, it would follow from the vector-space axioms together with all the $D_n$. Compactness would make it follow from finitely many of those axioms and dimension requirements. Let $N$ be at least every dimension mentioned there. The finite-dimensional space $K^N$ satisfies all vector-space axioms and those finitely many $D_n$, hence would satisfy $\varphi$. But it is a finite set and so cannot be a model of the purported axiomatization. This contradiction proves failure of finite axiomatizability, including finite axiomatizability relative to vector spaces.

Every vector space has a basis and is classified up to linear isomorphism by its dimension. Over a finite field, a vector is a finite linear combination of basis Elements. If the basis has infinite cardinality $\lambda$, the set of finite such combinations has cardinality $\lambda$: there are at least $\lambda$ basis vectors and at most a countable union of finite powers of $\lambda$, each of size $\lambda$. Thus the underlying cardinality of an infinite space equals its dimension.

Any two models of cardinality $\kappa \ge {\aleph }_0$ therefore have bases of cardinality $\kappa$. A bijection between their bases extends uniquely to an $L$-isomorphism. A space with that basis size exists, giving the required existence clause as well. Hence $T$ is categorical in every infinite cardinal.

For quantifier elimination for infinite underlying vector spaces, we first justify that formulas cannot make finer distinctions than their linear-equation data. In any infinite underlying vector space, eliminate an existential variable from a disjunction of conjunctions of linear equations and inequations. If an equation has a nonzero coefficient of that variable, solve for it and substitute. If none does, the inequations forbid only finitely many values, leaving an available value in the infinite domain; the remaining equations concern only the other variables. Terms are linear combinations, so induction gives quantifier elimination for $T$ and completeness, since closed terms reduce to zero.

For one variable there are two possible complete types, zero and nonzero. Both are isolated, and every definable set is their union. ThusFor types of pairs of vectors over a finite field, the linear relation kernel $\{(a,b)\in K^2:ax+by=0\}$ determines the complete type. If $q=|K|$, the possibilities are one rank-zero pair, $q+1$ rank-one directions, and one rank-two independent type. Concretely these are $(0,0)$; $x\ne 0$ with $y={\lambda }_a(x)$ for each $a\in K$; $x=0,y\ne 0$; and the independent pairs. Their isolating formulas are finite because $K$ is finite. HenceEvery class is the union of a selection of these atoms. Basis extension also shows that pairs with the same relation kernel are conjugate by a linear automorphism within any infinite-dimensional model.

The cardinality of a vector space is central: here “infinite” must continue to refer to the underlying set. Over a countably infinite field, every nonzero finite-dimensional space is already infinite. Thus the model class now consists of all nonzero $K$-vector spaces, not only infinite-dimensional ones.

Since independence of a pair is not definable over an infinite field, for (i), independence of a single vector is still $x\ne 0$, but independence of tuples of length at least two is not definable by one formula in this one-sorted language. Quantifier elimination follows from the linear-equation argument in the preceding part for all these infinite domains. Any candidate two-variable formula therefore involves only finitely many linear equations. Take an independent pair, and compare it with $(x,{\lambda }_{a}x)$, where $x\ne 0$ and $a\in K$ avoids the finitely many ratios making one of those nontrivial equations vanish. Both pairs satisfy the same listed atomic formulas, contradicting a proposed definition of independence. Longer tuple independence would define pair independence by existentially completing a pair inside an infinite-dimensional model, giving the same obstruction.

For (ii), the vector-space axiom schema together with $\exists x(x\ne 0)$ axiomatizes the infinite underlying spaces. For (iii), there is again no universal axiomatization, because the zero subspace is a substructure.

For finite axiomatizability with infinitely many scalar symbols, in (iv) there is no finite axiomatization in the full language: any finite axiom set mentions only finitely many scalar symbols. In a nonzero vector space alter an unused nonzero ${\lambda }_a$ to send zero to a nonzero vector, retaining all mentioned interpretations. The finite sentences keep their truth values, but a vector-space axiom is now false. Relative to the already imposed vector-space axioms, the extra condition is finitely axiomatized by the single nonzero-vector sentence. These absolute and relative answers must be distinguished.

For (v), the one-variable algebra still has two atoms and four Elements. In two variables the isolated atoms are $(0,0)$ and the countably many rank-one directions, represented by $x\ne 0,y={\lambda }_{a}x$ and by $x=0,y\ne 0$. There is also a nonisolated complete type of an independent pair. A formula using finitely many linear equations either picks finitely many isolated atoms, or contains that independent type and all but finitely many isolated atoms. Therefore $F_2/E_2(T)$ is the finite-cofinite Boolean algebra on a countably infinite set of atoms, with the independent type the extra nonisolated point in its Stone space. In particular this algebra is countably infinite, rather than the full power set of its atoms.

For (vi), dimension can be $1,2,3,\dots$ or ${\aleph }_0$ while the domain is countable, producing pairwise nonisomorphic countable models. Thus $T$ is not countably categorical. For an uncountable $\kappa$, a vector space over this countable field has cardinality $\kappa$ exactly when its dimension is $\kappa$, since in general a nonzero space has size $\max (|K|,\dim V)$ when that maximum is infinite. Basis bijections then give categoricity in every uncountable cardinal. Quantifier elimination also shows that these different positive dimensions are elementarily equivalent; independence of an arbitrarily long tuple is not being silently treated as definable over this infinite field.

Yes. As an Abelian group, the reals are a nontrivial torsion-free divisible group, equivalently a vector space over the rationals. Rational scalar multiplication is uniquely definable by equations such as $ny=mx$, so adding those operations is a definitional expansion.

In this language, quantifier elimination for torsion-free divisible Abelian groups reduces formulas to Boolean combinations of rational linear equations. Over finitely many parameters, a complete one-variable type either fixes a value in their linear span or excludes every value in that span. The first kind is realized in the span itself; the second is realized by any real number outside it. That span is countable, while the reals are uncountable. Thus every such type is realized and the additive real group is an omega-saturated model.

For finite tuples, the same argument successively supplies any required finite number of new rationally linearly independent elements. The order and multiplication of the real field are absent from this language, so they cannot impose additional type conditions.

Fix $i$, write $\alpha =x_{i-1}$, $\beta =x_i$, and let $E=T-S$ on this interval. The Hermite interpolation conditions give bothTwo applications of integration by parts therefore giveOn this single cubic piece, the same calculation also givesHere $S^{(4)}=0$ inside the interval, and all four endpoint values of $E,E^{\prime }$ are zero. Since $T^{\prime \prime }=E^{\prime \prime }+S^{\prime \prime }$, we obtain the exact identityThis is an integral inequality, with no assertion that the integrand on the left has one sign pointwise. The stronger nodal derivative conditions make the boundary terms vanish on each interval separately; part (i) only needed their cancellation after summing.

The stated Hermite interpolating polynomial exists uniquely in ${\Pi }_{2n+1}$. Indeed, a polynomial in the kernel of a linear map of the value-and-derivative data map has a double root of a polynomial at each of $n+1$ distinct interpolation nodes. A nonzero such polynomial would have polynomial degree at least $2n+2$, which is impossible. The map is therefore an injective function and a linear map between vector spaces of the same finite dimension of a vector space, $2n+2$, and hence is a linear isomorphism. No higher smoothness of the original function $f$ is required to construct it.

An explicit illustration uses interpolation nodes $0,1,2$, values $0,1,0$, and zero derivatives at every interpolation node. The clamped cubic spline is $S(x)=3x^2-2x^3$ on $[0,1]$ and its reflection on $[1,2]$. Its Hermite interpolating polynomial is the quartic polynomial $T(x)=x^2(2-x{)}^2$. On $[0,1]$, $T-S=x^2(1-x{)}^2$, and the corresponding reflected formula holds on $[1,2]$. Direct integration gives $J(S)=24$, $J(T)=128/5$, and $J(T-S)=8/5$, exactly as the energy identity predicts.

Write ${\mathcal{P}}_n$ for the vector space of polynomials of polynomial degree at most $n$, and put $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$. As the question calls the weighted integral an inner product, its finiteness is implicit; for example, a positive integrable weight $w$ suffices for all the continuous functions involved here.

For every $q\in {\mathcal{P}}_n$, differentiating the squared weighted norm $\Vert f-p_n-tq{\Vert }_w^2$ at its minimum $t=0$ gives the orthogonality equationsIn particular, the requested linear system isIts Gram matrix $G_{kj}={\int }_{-1}^{1}w(x)x^{k+j}dx$ is a symmetric positive-definite matrix: for any nonzero vector $a$,Indeed a nonzero polynomial is nonzero on some open interval, and the weight is positive. Therefore $G$ is an invertible matrix, proving uniqueness. Conversely, if these equations hold, the Pythagorean theorem in an inner-product space gives, for every $q\in {\mathcal{P}}_n$,so the solution really is the unique best approximation.

On the vector space of nodal values with zero endpoints define the discrete Sobolev normIt is a norm because it is the Euclidean norm of the scaled list of nodal values and first finite differences, and vanishes only for the zero list. It is the discrete counterpart of the Sobolev space $H^1$ norm. Here $W=U-u$ has zero endpoints, so this definition applies directly.

Set $M_4={\max }_{0\le x\le 1}|u^{(4)}(x)|$, finite by the assumed continuity. Taylor's theorem with fourth-order remainders gives, for suitable ${\xi }_{i,+}\in (x_i,x_i+h)$ and ${\xi }_{i,-}\in (x_i-h,x_i)$,Adding the two expansions cancels the odd derivatives, soConsequently $\Vert \phi {\Vert }_h^2\le h(N-1)M_4^2{h}^4/144\le M_4^2{h}^4/144$. Doubling the bound in (c) gives $2\Vert D_x^{-}e]{\Vert }_h^2+\Vert e{\Vert }_h^2\le \Vert \phi {\Vert }_h^2$, and thereforeThus $M_4/12$ is a valid nonnegative constant. To satisfy the request for a strictly positive constant even when $u^{(4)}\equiv 0$, take $C=\max \{1,M_4\}/12$, independent of $h$. This proves second-order convergence of a numerical method in a discrete Sobolev norm, including control of the backward difference operator applied to the error.

The inclusion $F\subseteq K$ makes $K$ a vector space over $F$ by its own addition and multiplication by elements of $F$. The extension is finite when this vector space has a finite basis. Its degree of a field extension isBeing a finite field extension concerns vector-space dimension, not the cardinality of the underlying field.

Since $K$ is an $F$-vector subspace of the finite-dimensional vector space $L$, it has a finite $F$-basis $u_1,\dots ,u_m$. Any finite $F$-spanning set for $L$ also spans $L$ over $K$, so choose a finite $K$-basis $v_1,\dots ,v_r$ of $L$.

Every $\ell \in L$ can be expanded first in the $v_j$ and then in the $u_i$, showing that the products $u_i{v}_j$ span $L$ over $F$. If ${\sum }_{i,j}{c}_{ij}{u}_i{v}_j=0$ with $c_{ij}\in F$, independence of the $v_j$ over $K$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$; independence of the $u_i$ then gives every $c_{ij}=0$. The products form an $F$-basis, proving the tower law for field extensions:

A field extension $K/F$ is finite when $K$, regarded as a vector space over $F$, has finite dimension. Its degree of a field extension isThus a finite basis $u_1,\dots ,u_m$ means that every $k\in K$ is uniquely expressible as ${\sum }_i{c}_i{u}_i$ with coefficients $c_i\in F$.

The degree of a field extension is the dimension of the larger field, regarded as a vector space over the smaller field:The scalar multiplication is the multiplication of $K$ restricted to scalars in $F$. A finite extension has finite dimension; otherwise the degree is infinite.

The degree of a field extension is the dimension of the larger field as a vector space over the smaller one:The field extension is a finite field extension precisely when this dimension is finite, equivalently when finitely many elements of $K$ span $K$ over $F$. Finiteness here concerns the dimension, not the number of elements of the fields; for example, $\mathbb{C}$ has degree two over $\mathbb{R}$ although both fields are infinite.