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Here $\varphi$ is a finite real convex function on the whole real vector space; no topology or continuity is required. We prove the convex domination form of the Hahn-Banach theorem by extending across one vector. Suppose a linear functional $f$ is already dominated on a vector subspace $W$ and choose $v\notin W$. For $w\in W$ and positive $t,s$, the proposed extension $g(w+\tau v)=f(w)+\tau c$ must satisfy the boundsTo prove compatibility, take $w_1,w_2\in W$ and $t,s>0$. The convex combination of $w_1-tv$ and $w_2+sv$ with weights $s/(s+t)$ and $t/(s+t)$ lies in $W$. Domination there and convexity giveRearranging proves every lower candidate is at most every upper candidate. Also the candidates $w=0,t=s=1$ showThus both endpoints are finite, and some $c\in [L,U]$ exists. For $\tau >0$ the upper bound gives domination of $f(w)+\tau c$; for $\tau <0$ the lower bound gives it; and for $\tau =0$ domination was assumed. This defines a dominated linear functional on $W+\mathbb{R}v$ extending $f$.

Order all dominated extensions by inclusion of their domains and agreement of values. A chain has the union extension, still linear and dominated. By Zorn's lemma there is a maximal extension. The one-vector construction would enlarge any proper domain, so its domain is $V$. Hence there is a full linear extension with $\boxed{g{|}_W=f,g\le \varphi }$. A convex function here need not be positively homogeneous; this is why the bounds included all positive $s,t$.

Now fix $y\in V$. The translated convex function ${\phi }_y(z)=\varphi (y+z)-\varphi (y)$ vanishes at zero. Apply the just-proved extension theorem to the zero linear functional on $\{0\}$ to obtain a linear functional ${\ell }_y$ with ${\ell }_y(z)\le {\phi }_y(z)$ for every $z$. SetThis is an affine function, satisfies $a_y(x)\le \varphi (x)$ for every $x$, and has $a_y(y)=\varphi (y)$. Let $A=\{a_y:y\in V\}$. Every member is an affine minorant, while the member indexed by $x$ attains $\varphi (x)$ at $x$. Thus the exact representation isThis is the finite convex function as supremum of affine minorants property. The linear functionals are algebraic; without a topology no assertion of continuous affine minorants is intended.

The weak topology $\sigma (E,E^{\ast })$ is the coarsest topology making every continuous linear functional in $E^{\ast }$ continuous. A basic neighborhood of $x$ imposes finitely many inequalities $|f_j(z-x)|<\epsilon$, and a sequence converges weakly precisely when each $f(x_n)$ converges to $f(x)$. The Hahn-Banach theorem makes the dual separate points, so the weak topology is Hausdorff.

Here is a proof of the requested weak subsequence property, the compact-to-sequential direction of the Eberlein-Šmulian theorem. Let $Y$ be the closed linear span of the countable set $\{x_n:n\ge 1\}$. It is a separable Banach space. The Hahn-Banach theorem separates points outside a closed vector subspace from that subspace, so $Y$ is weakly closed in $E$. Therefore $C=K\cap Y$ is weakly compact.

Choose a countable dense set $(u_j)$ in the unit sphere of $Y$, and choose norming functionals $f_j\in Y^{\ast }$ with $\Vert f_j\Vert =1$ and $f_j(u_j)=1$. They separate points: for $y\ne 0$, a sufficiently close $u_j$ to $y/\Vert y\Vert$ gives $|f_j(y)|>0$. This is a countable norming family. Extend the functionals to $E$ by the Hahn-Banach theorem. The evaluation mapis weakly continuous and injective. Its domain is compact and its target Hausdorff, so it is a homeomorphism onto its image: images of closed subsets are compact, hence closed. A countable product of real or complex lines is metrizable, for example by ${\sum }_j{2}^{-j}\min (1,|a_j-b_j|)$. Thus $C$ has a compact metrizable weak topology. Compact metric spaces are sequentially compact, giving a subsequence with $\boxed{x_{n_j}\rightharpoonup x\in C\subseteq K}$. If $Y=\{0\}$ the result is immediate. We did not assert that a separable Banach space has a separable dual or that its weak topology is generally metrizable.

A vector space over the real numbers is a set $V$ with an addition making it an abelian group and a scalar multiplication by $\mathbb{R}$ satisfying, for scalars $a,b$ and vectors $u,v$, the identities $a(u+v)=au+av$, $(a+b)v=av+bv$, $a(bv)=(ab)v$ and $1v=v$. A vector subspace is a nonempty subset closed under addition and scalar multiplication; a proper vector subspace is one unequal to $V$. A spanning set is a set whose finite linear combinations give every vector of $V$. A basis is a linearly independent spanning set. The dimension is the number of vectors in a basis, or its cardinality when the basis is infinite.

The sum of vector subspaces and their intersection of vector subspaces areBoth are vector subspaces: addition and scalar multiplication preserve the displayed forms or simultaneous membership. The intersection of vector subspaces is never empty because $0\in U\cap W$.

For the two given hyperplanes, adding and subtracting their defining equations gives $x_1=x_2$ and $x_3=x_4$. HenceThese two nonzero vectors are orthogonal, and consequently form an orthogonal basis of the intersection of vector subspaces. Put $u=(1,-1,-1,1)$ and $w=(1,-1,1,-1)$. Direct substitution gives $u\in U$, $w\in W$. Their dot products with $b_1,b_2$ vanish, and $u\cdot w=0$. Each defining equation of $U$ or $W$ is one nonzero linear constraint in four variables, so each vector subspace has dimension three. Thus each space has the indicated orthogonal basis:The last four vectors are nonzero and mutually orthogonal, so they are a basis of ${\mathbb{R}}^4$. They lie in $U+W$, proving $U+W=V$. The dimension identity is therefore verified numerically:

True. Both vector subspaces contain zero, and if $u,v\in U\cap W$, every linear combination $\alpha u+\beta v$ belongs to each of $U$ and $W$. Thus it belongs to their intersection. This proves the intersection of vector subspaces is a vector subspace.

False. Distinct vector subspaces may be nested: in $V={\mathbb{R}}^2$, take $U=\{0\}$ and $W=\mathrm{span}(1,0)$. Their union is $W$, a subspace. More precisely, $U\cup W$ is a subspace exactly when one is contained in the other. If neither contains the other, choose $u\in U\setminus W$ and $w\in W\setminus U$. Were $u+w$ in the union, subtracting $u$ or $w$ in the appropriate subspace would contradict those choices.

The additive Abelian group $G$ is divisible because it is a real vector space. Each convex subgroup $C_n$ is divisible too: for $c\ge 0$ in $C_n$ and a positive integer $m$, its unique additive quotient $c/m$ satisfies $0\le c/m\le c$, hence belongs to $C_n$. Negative elements follow by taking inverses. Thus all $C_n$ are rational vector subspaces, independently of whether the given real scalar multiplication respects the order.

Choose an ordered-group isomorphism ${\theta }_n:C_{n+1}/C_n\to \mathbb{R}$. It is rational-linear, since additive homomorphisms between divisible torsion-free groups preserve division by integers. Extend a rational basis of $C_n$ to one of $C_{n+1}$; equivalently choose a rational-linear section $s_n:\mathbb{R}\to C_{n+1}$ of ${\theta }_n$ composed with the quotient map. ThenSince $C_1=0$ and the union is $G$, each $g\in G$ has a unique expression $g={\sum }_{j=1}^N{s}_j(r_j)$ with only finitely many nonzero coefficients. Define $\Phi (g)=(r_1,r_2,\dots )$. This is an injective additive group homomorphism. If $r_N\ne 0$ is the largest active coefficient, the image of $g$ in the ordered quotient $C_{N+1}/C_N$ has sign equal to that of $r_N$. Convexity ensures that this is also the sign of $g$.

The maximum-support convention for the Hahn group $V({\mathbb{Z}}_{>0},\mathbb{R})$ uses precisely finite supports and the sign of the largest active coefficient. Hence $\Phi$ preserves and reflects the total order, and therefore preserves join and meet. This proves the finite-support Hahn embedding along a countable convex chain. In fact every finite coefficient sequence is obtained, so $\Phi$ is an ordered-group isomorphism in this convention; the requested embedding follows.

Use the dominated form of the real Hahn-Banach theorem. Let $p:V\to \mathbb{R}$ be a sublinear functional on a real vector space, meaning $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for $t\ge 0$. If a linear functional $f$ on a vector subspace $M$ satisfies $f\le p$ there, then it has a linear extension $F$ to $V$ with $F\le p$ everywhere.

Here is the one-dimensional extension step. For $z\notin M$, the value $a=F(z)$ must satisfyThis interval is nonempty: for any $m,n\in M$,so every candidate lower bound is at most every candidate upper bound. The candidates with $m=n=0$ also bound the endpoints by finite numbers, so a real $a$ can be chosen. Define $F(m+tz)=f(m)+ta$. For $t>0$, the upper bound, applied to $m/t$, gives $F(m+tz)\le p(m+tz)$ after multiplication by $t$. For $t<0$, the lower bound applied to $m/(-t)$ gives the same conclusion after multiplication by $-t$; $t=0$ is the original hypothesis. Thus the extension remains dominated.

Order all dominated extensions by extension of their domains and values. Every chain has the dominated linear union as an upper bound. The Zorn lemma supplies a maximal extension. If its domain were not all of $V$, the step just proved would enlarge it, a contradiction. This proves the real Hahn-Banach extension theorem. In particular, taking $p(x)=\Vert f\Vert \Vert x\Vert$ on a normed vector space and applying the domination to both $x$ and $-x$ yields the usual norm-preserving extension of a bounded linear functional.

For generalized limits, take $V={\ell }^{\infty }(\mathbb{N};\mathbb{R})$, the l-infinity sequence space, and $M=c$, the convergent sequence space. Let $f(x)={\lim }_n{x}_n$ on $c$ and $p(x)={\mathrm{limsup}}_n{x}_n$ on $V$. This $p$ is finite and sublinear, and agrees with $f$ on $c$. The dominated extension $L$ therefore satisfiesThus a generalized limit exists, withIt is positive, agrees with the ordinary limit on every convergent sequence, and satisfies $|L(x)|\le \Vert x{\Vert }_{\infty }$. Since $L(1)=1$, its operator norm is exactly one.

If generalized limits are additionally required to be shift invariant, the same theorem gives the stronger Banach limit. Replace the dominating functional byIt is again sublinear and agrees with $f$ on $c$, by the Cesàro theorem for convergent sequences. The resulting $L_C$ lies between the lower and upper limits of the Cesàro means. These in turn lie between the ordinary lower and upper limits of a bounded sequence, so $L_C$ is a positive, norm-one extension of the ordinary limit. For the left shift on bounded sequences $S$, the means of $Sx-x$ equal $(x_{N+1}-x_1)/N\to 0$. Applying domination to both signs forces $L_C(Sx-x)=0$. HenceThis Cesàro construction of a Banach limit proves existence even under the stronger convention, rather than leaving shift invariance implicit.

The two diagonal $2\times 2$ blocks allow the characteristic polynomial to factor immediately:Thus the characteristic equation isSolving the two block eigenvalue equations gives the following nonzero eigenvectors and corresponding eigenvalues:For example, the lower block maps $(1,i{)}^T$ to $(2i,-2{)}^T=2i(1,i{)}^T$, and maps $(1,-i{)}^T$ to zero. The four eigenvalues are distinct. Alternatively, the determinant of the matrix with columns $a,b,c,d$ is $(-2)(-2i)=4i\ne 0$. Hence they form a basis and span the complex vector space ${\mathbb{C}}^4$. Their nonzero scalar multiples are all the corresponding eigenvectors.

For any eigenvector $v$ with eigenvalue ${\lambda }_v$, its one-dimensional vector subspace satisfies $A(sv)={\lambda }_{v}sv$. If ${\lambda }_v\ne 0$, every $tv$ has preimage $(t/{\lambda }_v)v$, so the restricted linear map is onto that same subspace. ConsequentlyThe last line is the kernel of a linear map direction: multiplication by $A$ collapses that line to the zero-dimensional subspace.

Define the linear map $A:{\mathbb{R}}^5\to {\mathbb{R}}^2$ by the two prescribed left-hand sides. Then $V=\ker A$, so it is a vector subspace: it contains zero, and linear combinations of vectors with zero image still have zero image.

Subtract the first equation from the second to obtain $a_2=-2a_3-3a_4-4a_5$. The first equation then gives $a_1=a_3+2a_4+3a_5$. ConsequentlyEach of the three displayed vectors satisfies both equations, and the decomposition proves they span $V$. If their linear combination is zero, its third, fourth and fifth coordinates respectively force all three coefficients to vanish. They are therefore linearly independent, giving the basis

Let $T$ be the isometry. Since $T(0)=0$, it preserves norms, and the polarization identity applied to distances gives $⟨T(x),T(y)⟩=⟨x,y⟩$. For an orthonormal basis $e_1,e_2,e_3$, the vectors $T(e_i)$ are again an orthonormal basis, andExpanding in that basis proves $T(x)={\sum }_i⟨x,e_i⟩T(e_i)$. Hence $T$ is a linear map represented by an orthogonal matrix $Q$.

A reflection in a hyperplane with nonzero normal $v$ is $H_v(x)=x-2⟨x,v⟩v/⟨v,v⟩$. If $Q{e}_1\ne e_1$, choose $v=Q{e}_1-e_1$. Because $\Vert Q{e}_1\Vert =\Vert e_1\Vert$, substitution shows $H_{v}Q{e}_1=e_1$. If they agree, no reflection is needed. The resulting orthogonal transformation fixes $e_1$ and preserves its orthogonal complement. Apply the same construction there to fix $e_2$; the second normal is perpendicular to $e_1$, so the first vector remains fixed. Finally the restriction to the remaining line is either the identity or negation; in the latter case a third reflection, normal to $e_3$, fixes it. Reversing these steps writes $Q$ as a product of at most three reflections, each in a plane through the origin. This proves the three-dimensional case of the Cartan–Dieudonné theorem.

Three reflections are sometimes necessary. The map $-I$ is the product of the three coordinate-plane reflections. Its determinant is $-1$, excluding zero or two reflections. A single plane reflection fixes a two-dimensional vector subspace, whereas $-I$ fixes only zero, excluding one. Thus the bound is sharp.

For a symplectic vector space $(V,\omega )$ and a vector subspace $W$, the symplectic orthogonal complement isNondegenerate bilinear form identifies $V$ with its dual space, so $\dim W^{\omega }=\dim V-\dim W$. If $W$ is a symplectic subspace, $W\cap W^{\omega }=0$, and hence $V=W\oplus W^{\omega }$; the restriction to $W^{\omega }$ is also nondegenerate.

Here is an inductive construction of a symplectic basis. Choose $e_1\ne 0$. Nondegenerate bilinear form supplies $f_1$ with $\omega (e_1,f_1)=1$. The plane $P=\mathrm{span}(e_1,f_1)$ is a symplectic subspace, and its symplectic orthogonal complement gives $V=P\oplus P^{\omega }$. Repeat on $P^{\omega }$. The process ends in dimension zero, since each step removes two dimensions and a one-dimensional alternating form is degenerate. We obtain $e_1,f_1,\dots ,e_n,f_n$ withThe linear map taking the coordinate vectors of ${\mathbb{R}}^{2n}$ to this basis pulls $\omega$ back to ${\sum }_{j}d{x}_j\wedge d{y}_j$, proving the asserted normal form.

Now split $V=C\oplus D$ with $D=C^{\omega }$, and choose symplectic bases on both planes. Parametrize ${\Gamma }_A$ by $v\mapsto (Av,v)$. Orthogonality removes the cross terms, and a two-dimensional alternating form transforms by the determinant, soConsequently the correct nondegenerate bilinear form criterion isThe strict inequality printed in the PDF requires an extra orientation condition: it says that the symplectic orientation on the graph agrees with the orientation transported from $D$. It is not equivalent to being a symplectic subspace. For example, $A=\mathrm{diag}(-2,1)$ has determinant $-2$ and gives a nondegenerate restriction $-{\omega }_D$. This is the orientation criterion for a graph of symplectic planes.

This same map supplies the requested obstruction. Orient $C$ and ${\Gamma }_A$ by their restricted symplectic forms. If $(d_1,d_2)$ is a positive basis of $D$, then $(A{d}_1+d_1,A{d}_2+d_2)$ is negatively oriented in ${\Gamma }_A$. The coordinate transformation from a basis of $C$ followed by these graph vectors to a basis of $C\oplus D$ has determinant $1$. Reversing the graph basis to give its positive orientation therefore makes the oriented direct sum $C\oplus {\Gamma }_A$ negative relative to the ambient symplectic orientation.

Two distinct complex lines in standard ${\mathbb{C}}^2$ have the opposite behaviour: complex bases of the two lines give a complex-linear isomorphism $\mathbb{C}\oplus \mathbb{C}\to {\mathbb{C}}^2$, whose real determinant is the positive number $|{\det }_{\mathbb{C}}{|}^2$. Their complex orientations are their symplectic orientations. A symplectic linear map preserves both the ambient and each plane's symplectic orientation, so it cannot change this sign. Thus the pair with $A=\mathrm{diag}(-2,1)$ cannot be the image of a pair of complex lines. Interchanging the planes does not change the sign, since both dimensions are even.

Here is Witt's lemma with the hypotheses appropriate to the classical groups. Let $V$ be finite-dimensional with a nonsingular alternating bilinear form, a nonsingular symmetric bilinear form in characteristic different from two, or a nonsingular Hermitian form with nontrivial involution. Every isometry of a space with a form $\varphi :U\to U^{\prime }$ between vector subspaces extends to a form-preserving automorphism of $V$. For quadratic forms, the same statement holds in every characteristic when the polar bilinear form of a quadratic form is nonsingular and the isometry of a space with a form preserves the quadratic form itself. The restricted forms on $U,U^{\prime }$ may be degenerate. We prove these versions, including the characteristic-two quadratic case needed later.

For an alternating bilinear form $B$, split the restricted space as $U=H\oplus R$, where $R=U\cap U^{\perp }$ and $B{|}_H$ is nonsingular. Choose a symplectic basis for $H$ and a basis $r_1,\dots ,r_t$ of $R$. The nondegenerate form on $V$ supplies vectors $s_j$ in $H^{\perp }$ with $B(r_i,s_j)={\delta }_{ij}$. Adding linear combinations of the $r_i$ to the $s_j$ makes $B(s_i,s_j)=0$: for each pair $i<j$ one may cancel its pairing by adding an appropriate multiple of $r_i$ to $s_j$, without disturbing the dual pairings or previously treated pairs. This uses no division by two. Thus $H$ together with the pairs $(r_i,s_i)$ is nonsingular and its orthogonal complement for a sesquilinear form has a symplectic basis.

Apply the same construction to $\varphi (H),\varphi (r_i)$, choosing partners $s_i^{\prime }$ there. The two completed bases have identical alternating Gram matrices and equal remaining dimensions. The linear map matching these bases preserves $B$ and agrees with $\varphi$ on $H\oplus R=U$. This proves the alternating version over an arbitrary field, including characteristic two.

For a quadratic form $Q$, define its polar bilinear form of a quadratic form by $B(x,y)=Q(x+y)-Q(x)-Q(y)$. Its symmetrypreserves $Q$. The zero-dimensional case is immediate. Induct on $\dim U$: choose $Q(t)\ne 0$ and a hyperplane $K\subseteq U\cap t^{\perp }$. Extending $\varphi {|}_K$ first reduces to $\varphi {|}_K=I$. Put $U=K\oplus Fu$, $v=\varphi (u)$, $d=u-v\in S=K^{\perp }$. If $Q(d)\ne 0$, $s_d$ completes the extension. Otherwise $B(u,d)=B(v,d)=0$. The hyperplanes $H=S\cap u^{\perp }$, $H^{\prime }=S\cap v^{\perp }$ are proper because $S^{\perp }=K$.

For $a\in S\setminus (H\cup H^{\prime })$ with $Q(a)\ne 0$, set $c=u-s_a(v)$. Thenand $s_a{s}_c$ fixes $K$ and sends $u$ to $v$. If no such $a$ exists and $|F|>2$, the identityforces $Q(z)=B(a,z)=0$. The complement of two hyperplanes spans $S$: adding a multiple of any outside vector avoids at most two forbidden scalars. Hence $d\perp S$, so $H=H^{\prime }$, forcing $Q{|}_S=0$, contrary to $t\in S$.

Over ${\mathbb{F}}_2$, choose $a$ outside both hyperplanes. In the remaining case $Q(a)=Q(d)=B(a,d)=0$, and $B(a,u)=B(a,v)=1$. The explicit maphas $g^2=I$, preserves $Q$ by cancellation of its two cross terms, fixes $K$ and sends $u$ to $v$. This closes the induction. The characteristic-independent symmetry argument and its binary exceptional step are also treated in Casselman's quadratic-form notes and Elman, Karpenko and Merkurjev, section 8.

In odd characteristic a symmetric bilinear form is recovered by the polarization identity for $Q(x)=B(x,x)/2$, so the quadratic proof also proves its extension theorem. For completeness, the Hermitian form case has a similarly explicit induction. Take $h$ linear in the first variable, with conjugation $a\mapsto \bar{a}$. For anisotropic vectors for a form $u,v$ with equal value $A=h(u,u)=h(v,v)\ne 0$, let $d=u-v$ and $c=h(u,d)$. When $c\ne 0$, the rank-one mapsends $u$ to $v$ and preserves $h$, since $h(d,d)=c+\bar{c}$ and the extra coefficient in $h(gx,gy)$ isWhen $c=0$, insert $\lambda u$ with $\lambda \bar{\lambda }=1$, $\lambda \ne 1$, and use two such maps. Such a scalar exists for a nontrivial involution: $\lambda =a/\bar{a}$ with $a\ne \bar{a}$ works. If the source vector subspace contains an anisotropic vector for a form, align it this way and induct in its nonsingular orthogonal complement for a sesquilinear form.

If instead every source vector is isotropic, the Hermitian polarization identity makes its whole restricted form zero. Choose $e$ in its basis and a partner $f$ pairing to one with $e$ and to zero with the remaining source basis. Make $h(f,f)=0$ by replacing $f$ by $f-te$, where $t+\bar{t}=h(f,f)$. The trace map is onto the fixed field: in odd characteristic divide by two; in characteristic two scale a nonzero trace. Choose $a+\bar{a}\ne 0$. The orthogonal vectors $e+af$ and $e-\bar{a}f$ have nonzero values $a+\bar{a}$ and its negative, so the preceding rank-one construction aligns this entire hyperbolic pair with its target pair. The remaining source basis lies in the pair's orthogonal complement for a sesquilinear form; induction finishes the Hermitian proof.

The hypotheses matter. For example, the identity-matrix symmetric bilinear form on ${\mathbb{F}}_2^3$ has canonical vector $c=(1,1,1)$ characterized by $B(x,c)=B(x,x)$, so every isometry of a space with a form fixes $c$. The one-dimensional map $e_1\mapsto c$ preserves the restricted bilinear form but cannot extend. Thus “all symmetric forms in characteristic two” is not a correct replacement for the quadratic or alternating versions proved above.

The isomorphic pair is $A_8\cong PS{L}_4(2)$; $PS{L}_3(4)$ is different. All three have order $20160$, so order alone does not settle the question; explicitly $|PS{L}_3(4)|=(64-1)(64-4)(64-16)/(3\cdot 3)=20160$. We give a concrete realization and an element-order distinction.

On $W={\bigwedge }^2{\mathbb{F}}_2^4$, the Pfaffian isIt is a nonsingular plus-type quadratic form: the three displayed pairs give a hyperbolic decomposition. Change of basis multiplies the Pfaffian by the determinant, which is always one in $G{L}_4(2)$. Hence the exterior-square realization of PSL4 over F2 embeds $G{L}_4(2)$ into $O_6^{+}(2)$. The kernel of a group homomorphism is trivial: an element acting trivially on the exterior square fixes every decomposable two-space, then every one-space as an intersection of two-spaces, and hence is a scalar; over ${\mathbb{F}}_2$ this scalar is one.

For a second model let $E$ be the even-weight vector subspace of ${\mathbb{F}}_2^8$ and quotient by the all-one vector. The dot product descends to a nondegenerate alternating bilinear form on the six-dimensional quotient, andis well defined, since complementing an even subset changes half its weight from $w/2$ to $4-w/2$. It has $(1+70+1)/2=36$ zero vectors, so is plus type. Coordinate permutations give a faithful $S_8$ action: fixing the quotient fixes each weight-two subset, because its complement has weight six, and fixing all two-subsets forces the identity permutation.

From the independently derived symplectic order and quadratic orbit count below,Thus $O_6^{+}(2)\cong S_8$ and the image of $G{L}_4(2)=S{L}_4(2)=PS{L}_4(2)$ is its index-two subgroup $A_8$. The index-two subgroup is unique: a nontrivial group homomorphism $S_8\to C_2$ sends every transposition to the same nonidentity element and is the sign map.

To separate $PS{L}_3(4)$, observe that $A_8$ contains $(12345)(678)$ of order 15. An odd-order projective element of $PS{L}_3(4)$ lifts to a semisimple linear operator, since its order upstairs divides three times an odd order in characteristic two. The possible eigenspace degree patterns are $1+1+1$, $2+1$ and $3$. In the split case all eigenvalues have order dividing three, so the projective element has no order 15. In the $2+1$ case the eigenvalues are $\lambda ,{\lambda }^4,{\lambda }^{-5}$ with $\lambda \in {\mathbb{F}}_{16}^{\times }$; its fifth power is scalar because ${\lambda }^{15}=1$, so its projective order divides five. In the irreducible case the determinant-one torus has order $4^2+4+1=21$ and quotienting its scalar subgroup of order three gives projective order dividing seven. There is therefore no element of order 15 in $PS{L}_3(4)$.

Several small-parameter coincidences connect the finite classical groups to permutation groups. The projective-line action immediately gives $PS{L}_2(2)\cong S_3$, $PS{L}_2(3)\cong A_4$, $PG{L}_2(3)\cong S_4$ and $PS{L}_2(4)\cong A_5$ by faithful actions of degrees three, four and five and their orders. Also $PS{L}_2(5)\cong A_5$: its involutions have Klein-four centralizers, giving five Sylow subgroups; their conjugation action is faithful and embeds its order-60 simple group into $S_5$. Simplicity here can be checked from class sizes $1,15,20,12,12$: no proper nontrivial normal class union has order dividing 60. The same five-subgroup action extends to $PG{L}_2(5)$, with trivial kernel of a group homomorphism and order 120, giving $S_5$. Indeed its kernel intersects $PS{L}_2(5)$ trivially, so centralizes that normal subgroup. Commuting projectively with both upper and lower elementary unipotents forces a representing matrix to be scalar, so this kernel is trivial.

The less immediate coincidence $PS{L}_2(9)\cong A_6$ can be exhibited on the ten unordered triple partitions used in Question 4. Let $a=(123)$, $b=(456)$, $w=(12)(36)$, assign $\infty$ to $123\mid 456$ and $0$ to $126\mid 345$, and label the other partitions by applying $a^x{b}^y$ to the zero partition, with $t=x+iy\in {\mathbb{F}}_9$ and $i^2=-1$. Inspection of these nine partitions givesTranslations and inversion generate $PS{L}_2(9)$: conjugating the upper unitriangular matrices by inversion gives the lower ones, and elementary elimination generates $S{L}_2(9)$. Their projective action has order $9(9^2-1)/2=360$. These partition permutations come from even permutations of six letters, whose partition action is faithful as proved below; hence they fill $A_6$.

Finally, on the even-weight module of ${\mathbb{F}}_2^6$ modulo its all-one vector, coordinate permutations preserve a nondegenerate alternating bilinear form in dimension four and act faithfully, again by the weight-two-subset argument. Both $S_6$ and $S{p}_4(2)$ have order 720, proving $S{p}_4(2)\cong S_6$, with commutator subgroup $A_6$. These examples illustrate both the need to quotient group centers and the low-rank exceptions to general simplicity. They do not make equal orders sufficient for group isomorphism, as the two order-20160 examples already demonstrate.

Put $C=R/(1-k)$. The hypothesis scales to every residual $r\in F$: if $r\ne 0$, apply it to $r/\Vert r\Vert$ and multiply the resulting vector by $\Vert r\Vert$ to obtain an approximate preimage of norm at most $R\Vert r\Vert$ with error at most $k\Vert r\Vert$. For $r=0$, choose zero.

Start with $r_0=y$. Inductively choose $u_j\in E$ so thatThen $\Vert r_j\Vert \le k^j\Vert y\Vert$ and ${\sum }_j\Vert u_j\Vert \le R\Vert y\Vert /(1-k)$. Completeness of the Banach space $E$ gives $x={\sum }_j{u}_j$. Boundedness of $T$ givesThusThis geometric correction for approximate surjectivity uses no completeness assumption on $F$: every infinite sum was formed in $E$.

To establish the completeness forced by uniformly bounded lifting, let $(y_n)$ be a Cauchy sequence in $F$. Choose a subsequence $(y_{n_j})$ with $\Vert y_{n_{j+1}}-y_{n_j}\Vert \le 2^{-j}$. Lift $y_{n_1}$ to $v_0\in E$ and each difference to $v_j\in E$ with $T{v}_j=y_{n_{j+1}}-y_{n_j}$ and $\Vert v_j\Vert \le C{2}^{-j}$. The series $v_0+{\sum }_j{v}_j$ converges in $E$, and its image under $T$ is the limit of the subsequence. The original Cauchy sequence has that same limit, by the triangle inequality. Hence $F$ is complete.

The open mapping theorem states that a bounded surjective linear map between Banach spaces is open. In particular, a bounded linear bijection has a bounded inverse. This is the theorem the question permits us to state without proof.

For the closed graph theorem, suppose $T:E\to F$ is everywhere-defined and linear with closed graph $G(T)$. The product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$ makes $E\times F$ a Banach space, so its closed subspace $G(T)$ is complete. Projection $\pi :G(T)\to E$, $(x,Tx)\mapsto x$, is a bounded linear bijection. Its inverse is bounded by the open mapping theorem; thereforeIt follows that $T$ is bounded, proving the closed graph theorem rather than assuming it.

For the final clause, $S$ is the separating space of a linear map. It contains zero, and if $x_n\to 0$, $T{x}_n\to y$ and $z_n\to 0$, $T{z}_n\to w$, then $\alpha x_n+\beta z_n\to 0$ and its image tends to $\alpha y+\beta w$. Thus $S$ is a vector subspace. If $y_j\in S$ and $y_j\to y$, choose one vector $w_j$ from a witnessing sequence for $y_j$ withThen $w_j\to 0$ and $T{w}_j\to y$, so $y\in S$: it is closed. If $T$ is continuous, every defining image limit is zero, hence $S=\{0\}$. Conversely, if $S=\{0\}$ and $x_n\to x$, $T{x}_n\to y$, then $x_n-x\to 0$ and $T(x_n-x)\to y-Tx\in S$, so $y=Tx$. The graph is sequentially closed and therefore closed in the normed, metrizable product. The proved closed graph theorem now gives continuity. ConsequentlyThe original PDF places $S$ in $F$; the TeX's first reference to a subset of $E$ is a transcription error.

Choose a best uniform approximation $p_n\in {\mathcal{P}}_n$. It exists because this vector subspace is a finite-dimensional vector space and a minimizing sequence $q_j$ is bounded: $\Vert q_j{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }+\Vert f-q_j{\Vert }_{\infty }$. Finite-dimensional equivalence of norms then supplies a convergent subsequence of coefficients. If $E_n(f)>0$, the Chebyshev alternation theorem gives $n+2$ alternating error extrema. The intermediate value theorem supplies a zero $x_{n,k}$ of $f-p_n$ in each of the $n+1$ intervening open intervals. These interpolation nodes are distinct, and uniqueness of polynomial interpolation givesThe convergence follows from the Weierstrass approximation theorem. If $E_n(f)=0$, choose any $n+1$ distinct nodes: the polynomial interpolant is still $f=p_n$. This constructs interpolation nodes from best uniform approximation; the nodes depend on $f$, and no universal node system for all continuous functions is asserted.

Both candidate sets contain zero. The sum of subspaces is closed under linear combinations because $a(u_1+w_1)+b(u_2+w_2)=(a{u}_1+b{u}_2)+(a{w}_1+b{w}_2)$, with the two terms in $U$ and $W$. The intersection is closed because any linear combination of vectors lying in both subspaces still lies in both. Thus each is a vector subspace.

Take a basis $s_1,\dots ,s_r$ of the intersection, extend it by $u_1,\dots ,u_a$ to a basis of $U$, and by $w_1,\dots ,w_b$ to a basis of $W$. The combined list $s,u,w$ spans $U+W$. To prove independence, a zero combination makes the $u$ combination equal a vector in both $U$ and $W$, so it is a combination of the $s$'s. Independence of the basis of $U$ makes all $u$ coefficients zero; independence of the basis of $W$ then makes all remaining coefficients zero. Hence $\dim (U+W)=r+a+b$, giving the dimension formula for a sum of subspaces

For the concrete kernels, the equations $Ax=0$ giveApplying the two rows of $B$ to this expression gives $4(s-t)$ and $5(s-t)/3$. Thus the intersection has basis $v_0=(0,2,1,1)$. Choose $v_1=(5,-1,3,0)$ in $U$ and $v_2=(-4,-2,1,0)$ in $W$. ThenIndeed $U$ has dimension two because $A$ has rank two, and $v_0,v_1$ are independent. Likewise $W$ has dimension two and $v_0,v_2$ are independent, so the last list spans the sum. Its first three coordinates have determinant $-48$, confirming independence and dimension three.