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For $m>n$, the $L^1$ difference is ${\int }_n^m{x}^{-1}dx=\log (m/n)$. Choosing $m=2n$ makes it $\log 2$, so the sequence is not Cauchy in $L^1$. Its squared $L^2$ difference is ${\int }_n^m{x}^{-2}dx\le 1/n$, which tends uniformly to zero for $m>n$. Hence it is Cauchy in $L^2$, with norm limit $x^{-1}$ on $(1,\infty )$.

On the standard two-dimensional representation, $t_{\theta }$ has eigenvalues $e^{i\theta },e^{-i\theta }$. Its $m$th symmetric power has monomial basis $X^{m-j}{Y}^j$, $j=0,\dots ,m$, of weights $m-2j$. The character is thereforeThe quotient is interpreted by continuity at $0,\pi$, where its values are $m+1$ and $(-1{)}^m(m+1)$.

Apply the Weyl integration formula for SU2 to the character product:The last equality follows directly by product-to-sum integration. A finite-dimensional representation of a compact group decomposes into irreducibles, and the squared character norm is the sum of squared multiplicities. Since this norm is one, there is exactly one constituent with multiplicity one. Thus every symmetric power is irreducible, and the different powers are pairwise inequivalent.

If a linear map satisfies $\Vert Tx\Vert \le M\Vert x\Vert$, then $\Vert Tx-T{x}^{\prime }\Vert \le M\Vert x-x^{\prime }\Vert$, so it is continuous. Conversely continuity at zero gives $\delta >0$ such that $\Vert x\Vert <\delta$ implies $\Vert Tx\Vert <1$. Apply this to $\delta x/(2\Vert x\Vert )$ for $x\ne 0$ to obtain $\Vert Tx\Vert \le 2\Vert x\Vert /\delta$. Hence a linear map is continuous exactly when it is bounded.

For the separately continuous bilinear map, put $T_x(y)=F(x,y)$ for $\Vert x\Vert \le 1$. Every $T_x:Y\to Z$ is a bounded linear map by the first result. For each fixed $y$, continuity in the other variable gives $\Vert F(x,y)\Vert \le C_y\Vert x\Vert$, so ${\sup }_{\Vert x\Vert \le 1}\Vert T_{x}y\Vert <\infty$. The Uniform boundedness principle therefore gives ${\sup }_{\Vert x\Vert \le 1}\Vert T_x\Vert =M<\infty$, and bilinearity then yields

To spell out the uniform-boundedness step, define the closed subsets $E_m=\{y:{\sup }_{\Vert x\Vert \le 1}\Vert T_{x}y\Vert \le m\}$ of the Banach space $Y$. They cover $Y$. The Baire category theorem gives an $E_m$ containing a ball $B(y_0,r)$. For $\Vert h\Vert <r$, both $y_0+h$ and $y_0$ lie in $E_m$, so ${\sup }_x\Vert T_{x}h\Vert \le 2m$. Scaling $h=(r/2)y$ for $\Vert y\Vert \le 1$ bounds all the operator norms by $4m/r$. This proves the claimed uniform estimate without presuming joint continuity.

For the modified Hebbian learning rule,It increases below $s_{\ast }={\alpha }^{-1/2}$ and decreases above $s_{\ast }$, without crossing that equilibrium. More precisely, for a nonzero initial norm,Under persistent excitation, meaning the integral tends to infinity, this givesWithout persistent excitation the limiting norm need not reach this value; zero weights remain zero. The quartic norm stabilization of Hebbian learning bounds the norm and supplies negative feedback at large amplitudes.

Set $s=|w{|}^2$. The exact Hebbian learning rule givesThus the norm never decreases, and increases whenever the output is nonzero. Because learning is slow relative to input fluctuations, its averaged equation is $\tau \dot{w}=Cw$, where $C=⟨x{x}^T⟩$ is the input second-moment matrix. It is a covariance matrix only if the inputs have zero mean. Diagonalizing the positive semidefinite matrix $C$ gives components $w_j(t)=w_j(0)e^{{\lambda }_{j}t/\tau }$. Hence weights grow without bound whenever the initial vector has a component in an excited positive-eigenvalue direction. A vector entirely in $\ker C$, or identically zero input, is an important stationary exception.

Unmodified Hebbian learning reinforces correlated input and output but has no mechanism limiting synaptic strength. The second rule adds activity-dependent negative feedback and stabilizes the norm at an input-scale-independent value, provided the input keeps exciting the output. The third has a cubic weight penalty independent of the instantaneous output; its equilibrium norm instead depends on the strength of the input second moment.

For the third averaged rule, an equilibrium along eigenvector $i$ has radial linearized eigenvalue $-2{\lambda }_i/\tau$ and transverse eigenvalues $({\lambda }_j-{\lambda }_i)/\tau$. Thus a positive simple largest eigenvalue gives stable equilibria in its two opposite directions, whereas lower eigen-directions are unstable to larger components. With generic initial weights, it learns the dominant correlation direction, linking the rule to principal component analysis. These are idealized normalization and feature-extraction mechanisms, not claims that biological synapses implement these exact equations.

We first justify the maximum in the proposed supremum norm. If a null sequence $x$ is nonzero, choose an index $j$ with $|x_j|>0$. All sufficiently late terms have absolute value smaller than $|x_j|/2$, since $x_i\to 0$. Hence the largest absolute value occurs among a finite initial set of terms, and is finite. For the zero sequence the maximum is zero.

The norm axioms now follow directly. The maximum is nonnegative and vanishes precisely when every coordinate vanishes. For a scalar $a$, $\Vert ax{\Vert }_{\infty }=|a|\Vert x{\Vert }_{\infty }$. For two null sequences, coordinatewise addition remains a null sequence andTaking the maximum proves the triangle inequality, so this really is a norm on $V$.

Let $e_n$ have a single $1$ in position $n$. If $m\ne n$, then $\Vert e_m-e_n{\Vert }_{\infty }=1$. A convergent sequence in a normed vector space must be a Cauchy sequence, by the triangle inequality. These distances violate the Cauchy condition with, for example, $\epsilon =1/2$. Thus $(e_n)$ does not converge in this norm, even though each fixed coordinate converges to zero.

First scale the stated unit-ball approximation. For every nonzero residual $r\in F$, apply the hypothesis to $r/\Vert r\Vert$ and multiply the resulting vector by $\Vert r\Vert$. This gives $v\in E$ withFor zero residual use $v=0$. Start with $r_0=y$, choose $v_j$ by this rule for $r_j$, and set $r_{j+1}=r_j-T{v}_j$. Inductively,The Banach space $E$ therefore contains the sum $x={\sum }_{j\ge 0}{v}_j$, withThe partial sums satisfy $T{\sum }_{j=0}^{n-1}{v}_j=y-r_n$. Boundedness of $T$ gives convergence of the left side to $Tx$, while $r_n\to 0$ gives $Tx=y$. Thus $T$ is surjective with the stated lifting bound. This geometric correction for approximate surjectivity used completeness of $E$, not any unproved completeness of $F$.

Put $C=R/(1-k)$. Given a Cauchy sequence $(y_n)$ in $F$, choose a subsequence $(y_{n_j})$ such that $\Vert y_{n_{j+1}}-y_{n_j}\Vert \le 2^{-j}$ for $j\ge 1$. Lift each difference to $u_j\in E$ with $T{u}_j=y_{n_{j+1}}-y_{n_j}$ and $\Vert u_j\Vert \le C{2}^{-j}$. Lift $y_{n_1}$ to $u_0$. The sum $u=u_0+{\sum }_{j\ge 1}{u}_j$ exists in $E$, and its image is the limit of the subsequence. A Cauchy sequence with a convergent subsequence converges to the same limit: use the triangle inequality between an arbitrary late term, a later subsequence term, and the limit. This proves $F$ is complete, the completeness forced by uniformly bounded lifting principle.

The open mapping theorem states that a bounded surjective linear operator between Banach spaces maps open sets to open sets. In particular a bounded linear bijection between Banach spaces has a bounded inverse. To deduce the closed graph theorem, let $S:E_1\to F_1$ be an everywhere-defined linear map between Banach spaces with closed graph. Its graph is a closed subspace of the Banach product $E_1\times F_1$, with norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$, so it is itself Banach. The projectionis a bounded linear bijection. The open mapping theorem makes its inverse bounded. Composing that inverse with the bounded second projection proves that $S$ is bounded. This is the required closed graph conclusion, rather than a continuity assumption on $S$.

Finally consider the identity map $I:(C(X),\Vert \cdot \Vert )\to (C(X),\Vert \cdot {\Vert }_{\infty })$. If $f_n\to f$ in the new norm and $f_n\to g$ uniformly, each assumed continuous point evaluation functional gives $f_n(x)\to f(x)$, while uniform convergence gives $f_n(x)\to g(x)$. Hence $f=g$. Since both spaces are metric, this sequential argument proves the graph is closed. Both norms are complete, so the closed graph theorem makes $I$ bounded. Its inverse is bounded by the open mapping theorem. Consequently there are constants $c,C^{\prime }>0$ such thatThis proves Banach norm rigidity from continuous point evaluations. Pointwise continuity alone would not give a common bound on all evaluations; completeness supplies that through the closed graph argument.

Work with a nonzero complex unital Banach algebra, so $1\ne 0$. The spectrum of an element isFor $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in norm and is a two-sided inverse, by multiplying its partial sums and letting the remainder tend to zero. Thus the spectrum lies in $|\lambda |\le \Vert x\Vert$. The assumed openness of the invertible group makes its complement closed, so the spectrum is compact.

To prove nonemptiness, suppose every $\lambda$ has an inverse $R(\lambda )=(\lambda 1-x{)}^{-1}$. Locally,so the resolvent is holomorphic. At infinity its Neumann series gives $\Vert R(\lambda )\Vert =O(|\lambda {|}^{-1})$. For any bounded linear functional $\ell \in A^{\ast }$, the scalar entire function $\ell (R(\lambda ))$ is bounded: it is bounded on compact disks by continuity, and tends to zero outside them. The Liouville theorem makes it constant, and its limit makes that constant zero. The Hahn-Banach theorem separates points of the normed space $A$, so $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Hence the spectrum is nonempty and compact.

On the algebra of entire functions, setIt is finite, homogeneous and satisfies the triangle inequality and submultiplicativity. If it vanishes, the identity theorem makes $f$ identically zero, so it is an algebra norm. However, no algebra norm on the entire-function algebra can be complete. The coordinate function $Z(z)=z$ satisfies $\sigma (Z)=\mathbb{C}$ algebraically: $Z-\lambda$ vanishes at $z=\lambda$ and has no entire multiplicative inverse. A complete algebra norm would make this a Banach algebra, contradicting the proved boundedness of its spectrum. This is the entire function algebra admits no complete algebra norm obstruction; it applies to every proposed complete algebra norm, not only the displayed one.

For all continuous functions on $\mathbb{C}$, choose continuous cutoffsTheir supports are disjoint and escape every compact set. Thus $f(z)={\sum }_{n\ge 1}n{h}_n(z)$ is a locally finite sum and is continuous. Let $g_n(z)=\max (0,1-4|z-3n|)$. This nonzero continuous function is supported where $h_n=1$ and all the other cutoffs vanish. Hence $f{g}_n=n{g}_n$. Any algebra norm would implyso $n\le \Vert f\Vert$ for every positive integer, impossible. Therefore the algebra of all continuous functions on $\mathbb{C}$ admits no algebra norm, even an incomplete one. This proves the continuous functions on the complex plane admit no algebra norm obstruction directly.

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

Identify ${\mathbb{R}}^4$ with the quaternions, and use the identification of unit quaternions with $SU(2)$ proved in the root Solution. The actionpreserves the Euclidean norm by multiplicativity of quaternion norms. It is a homomorphism $SU(2)\times SU(2)\to SO(4)$: its domain is connected, so the orthogonal image has determinant $+1$.

An element in the kernel fixes $v=1$, giving $q_L=q_R=q$. Fixing every other $v$ means $q$ commutes with all quaternions, so $q$ is real; since it has unit norm, $q=\pm 1$. Hence the kernel is the diagonal subgroup $\{(1,1),(-1,-1)\}$.

The differential acts as $v\mapsto av-vb$ for imaginary quaternions $a,b$. If it vanishes, $v=1$ gives $a=b$, and commutation with all $v$ makes $a$ real and imaginary, hence zero. The differential is injective, and both Lie algebras have dimension six. The image is therefore an open subgroup of connected $SO(4)$, hence all of $SO(4)$. Thuswhere ${\mathbb{Z}}_2$ acts diagonally, not separately on the two factors.