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Compactness gives a stronger statement than the one requested. Since $T(B_H)$ has compact closure, finitely many open balls of radius $\delta$ cover it. Choose their centres from $T(B_H)$ and let $E$ be their finite-dimensional linear span. ThenFor the orthogonal projection $P_E$, best approximation in a Hilbert space givesIn particular this holds for every vector $e_n$ in the given orthonormal basis.

The operator $P_{E}T$ has image in $E$ and hence has finite rank, whileThus every compact operator is an operator norm limit of finite-rank operators. This is the finite-rank approximation theorem for compact operators on a Hilbert space.

Write $A=T-\lambda I$. We first prove that $\mathrm{im}A$ is dense. Otherwise some nonzero $y$ lies in its orthogonal complement, soBy part (b), choose finite-dimensional spaces $E_j$ containing $y$ such that $S_j=P_{E_j}T\to T$ in operator norm. Sincethe operator $S_j^{\ast }-\overline{\lambda }I$ is not injective. Therefore $S_j-\lambda I$ is not surjective; the contrapositive of part (c) makes $\lambda$ an eigenvalue of $S_j$. Choose unit vectors $x_j$ with $S_j{x}_j=\lambda x_j$. ThenCompactness gives a subsequence for which $T{x}_j$ converges. Since $\lambda \ne 0$, the displayed relation makes $x_j$ converge to a unit vector $x$, and continuity gives $Tx=\lambda x$, contradicting the hypothesis. Thus $\mathrm{im}A$ is dense.

Next suppose $A$ were not bounded below, or equivalently that its lower norm were zero. There would be unit vectors $x_j$ with $A{x}_j\to 0$, henceAgain a convergent subsequence of $T{x}_j$ forces the corresponding $x_j$ to converge to a unit vector $x$ satisfying $Ax=0$, another contradiction. Therefore some $c>0$ satisfies

This lower bound makes $\mathrm{im}A$ closed: if $A{x}_j$ converges, then $x_j$ is Cauchy and its limit maps to the same limit. Since the image is both dense and closed, it is all of $H$. We have proved the Fredholm alternative for a compact operator: every nonzero spectral value of a compact operator is an eigenvalue.

For $y\in D(A)$, the semigroup property givesHence $U(t)y\in D(A)$ andThe generator domain is therefore an invariant subspace, and the operator norm bound givesThusMoreover, strong continuity applied separately to $y$ and $Ay$ givesTherefore the restrictions form the semigroup restricted to its generator domain. Its derivative at zero exists in the graph norm exactly when $y\in D(A)$ and $Ay\in D(A)$, namely when $y\in D(A^2)$, and then the derivative is $Ay$. Hence its generator is

A solution operator for a nonautonomous evolution equation is an evolution family $U(t,s)$ satisfyingand, on a suitable common domain $\mathcal{D}$,

One applicable nonautonomous generation theorem is the following. Suppose $\mathcal{D}$ is a dense linear subspace of $H$, each $A(t)$ has domain $\mathcal{D}$, the family is a stable family of semigroup generators with constants $M,0$, and $t\mapsto A(t)$ is continuously differentiable as a map from $[0,T]$ to $\mathcal{B}(\mathcal{D},H)$, where $\mathcal{D}$ carries one of the uniformly equivalent graph norms. Then there is a unique evolution family such that:

For the uniform partition $t_j=s+j(t-s)/N$, the frozen-generator product approximation isAs $N\to \infty$, $U_N(t,s)u\to U(t,s)u$ in the norm of $H$ for every $u\in H$, uniformly for $(t,s)$ in the compact time triangle $0\le s\le t\le T$. This is convergence in the strong operator topology, rather than convergence in the operator norm.

It remains to verify the second differential equation. The evolution family law gives, for $h>0$,Divide by $h$. Since $u\in \mathcal{D}$,while strong continuity gives $U(t,s+h)A(s)u\to U(t,s)A(s)u$. ThereforeThe left derivative follows in the same way, so $s\mapsto U(t,s)u$ is differentiable.

Write the local Hamiltonian as $H={\sum }_{j=1}^m{h}_j$. Since its terms commute, their matrix exponentials factor exactly:Each factor acts on at most two qubits and can be compiled over a fixed universal quantum gate set to operator norm error at most $\epsilon /m$. The telescoping bound for products of operators then bounds the total error by the sum of the factor errors, at most $\epsilon$. Because $m$ is polynomial in $n$ and the Solovay--Kitaev theorem gives gate count polynomial in $\log (m/\epsilon )$ for each fixed-dimensional factor, this is an efficient commuting local Hamiltonian simulation. Finally, the eigenvalue equation $H|\Psi ⟩=\lambda |\Psi ⟩$ impliesso $|\Psi ⟩$ remains an eigenstate and its eigenvalue is $e^{-i\lambda t}$.

Let $(T_m)$ be a Cauchy sequence in the operator norm. For each $v\in V$,so $(T_{m}v)$ is a Cauchy sequence. Completeness of the Banach space $V$ permits the definition
$Tv={\lim }_m{T}_{m}v$. Pointwise passage to the limit proves that $T$ is a linear. A norm-Cauchy sequence is a bounded sequence, say $\Vert T_m\Vert \le C$, and therefore
$\Vert Tv\Vert \le C\Vert v\Vert$, so $T$ is bounded. Finally, letting $m\to \infty$ in
$\Vert (T_n-T_m)v\Vert \le \epsilon \Vert v\Vert$ gives
$\Vert T_n-T\Vert \le \epsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal{B}(V,V)$ is Banach.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Frechet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

The continuous dual space of a normed vector space $V$ iswith the operator norm

Let $(f_n)$ be a Cauchy sequence in this norm. For each $v\in V$,so $(f_n(v))$ is Cauchy in the scalar field. Define $f(v)={\lim }_n{f}_n(v)$. Passing to limits in the linearity identities shows that $f$ is linear.

A norm-Cauchy sequence is bounded, say $\Vert f_n\Vert \le M$. Hence $|f(v)|\le M\Vert v\Vert$, so $f\in V^{\ast }$. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ givesfor every $v$, and therefore $\Vert f_n-f\Vert \le \epsilon$. Thus $f_n\to f$ in operator norm, proving completeness of the dual space: $V^{\ast }$ is Banach.

Differentiate the eigenvalue equationat $t=0$, where $A^{\prime }(0)=E$. Suppressing the argument $0$ givesLeft-multiplication by the transpose of the unit left eigenvector $u$ and use of $u^{T}A=\lambda u^T$ cancel the terms involving $v^{\prime }$:Because the eigenvalue is simple, $u^{T}v\ne 0$, and hence the first-order perturbation of a simple eigenvalue isThe definition of the operator norm, followed by the Cauchy-Schwarz inequality, givesTherefore

Write the interpolating spline asThe interpolation equations are $A_{\mathbf{x}}c=f_{\mathbf{x}}$, where $(f_{\mathbf{x}}{)}_i=f(x_i)$, soThe B-splines are nonnegative and form a partition of unity on the spline interval. ConsequentlyTaking the supremum over $t$ and then over $\Vert f{\Vert }_{\infty }\le 1$ proves the operator norm bound

The spectral norm of an operator $O$ is the induced operator normEvery Pauli X gate $X_i$ is unitary, so $\Vert X_i\Vert =1$. The triangle inequality for the norm therefore givesIn fact equality holds, as the product of $+1$ eigenvectors of the $X_i$ has eigenvalue $n$ under $J_X$.

Letwhere the two signs describe the Heisenberg antiferromagnet and Heisenberg ferromagnet. A standard Lieb-Robinson bound is obtained by iterating the Heisenberg picture equation and bounding nested commutators by operator norms. Its constants depend on the interaction only through quantities such asChanging $J$ to $-J$ changes neither the supports nor the norms $\Vert \Phi (X)\Vert$. Every term in the nested-commutator estimate acquires at most an irrelevant sign before its absolute value is taken. Therefore both chains obey exactly the same estimatewith the same $C,\mu$, and Lieb-Robinson velocity $v_{\mathrm{LR}}$. No unitary equivalence of the two Hamiltonians is required.

Start with the sequence from part ii. For each $n$, the definition of the operator norm gives $x_n\in B_X$ withBecause $g_n\stackrel{w^{\ast }}{\to }f$, pass recursively to a subsequence so far out that its $n$th member also satisfies

Part b supplies a weakly Cauchy subsequence $(x_{k_j})$. PutThen $(y_j)$ is weakly null, $h_j\stackrel{w^{\ast }}{\to }f$, and the two preceding estimates giveSince $f(y_j)\to 0$, discard finitely many terms to obtain $|h_j(y_j)|>\epsilon /16$ for every remaining $j$. Relabelling proves the claim.

The operator norm givesUsing part i and the first inequality for the adjoints gives

A linear estimator in nonparametric regression at $x_0$ has the form ${\hat{m}}_n(x_0)={\sum }_i{W}_i(x_0)Y_i$, where the weights may depend on the design, $x_0$, $p$, $h$, and $K$, but not on the responses. Putand define the local polynomial Gram matrixWhen ${\lambda }_0>0$, the weighted least squares normal equations have the unique solutionTherefore ${\hat{m}}_n(x_0)={\sum }_i{W}_i{Y}_i$, where the effective kernel weight is

The identityshows that these weights exactly reproduce at $x_0$ every multivariate polynomial of total degree at most $p$. This is the required polynomial reproduction property of local polynomial regression.

Only grid points with $\Vert x_i-x_0{\Vert }_{\infty }\le h$ have nonzero weight. There are at most $(2n_{1}h+1{)}^d\le (4n_{1}h{)}^d=4^{d}n{h}^d$ such points because $n_{1}h\ge 1/2$. On this support,so the operator norm bound $\Vert B_n^{-1}{\Vert }_{\mathrm{op}}={\lambda }_0^{-1}$ givesBecause the errors are independent random variables with the stated variance bounds,Thus ${\gamma }_1=d$.

Let $T_{x_0}$ be the Multivariate Taylor polynomial of $m$ at $x_0$ through total degree ${\beta }_0$. The assumed Hölder continuity of the derivatives andgive the Taylor remainder boundPolynomial reproduction cancels $T_{x_0}$ in the bias. The same support count and weight bound giveConsequentlyso ${\gamma }_2=\beta$.

A positive contraction is a positive operator whose operator norm is at most one. Its spectrum lies in $[0,1]$, so the spectral theorem for normal operators gives $M^2\le M$. Positive contractions are the effects in a POVM and occur in the pure-state gentle measurement bound.

For a nonnegative bounded weak subsolution and $\beta >1$, test with ${\eta }^2(u+\epsilon {)}^{\beta -1}$. Uniform ellipticity and Cauchy-Schwarz inequality give $\int |Du{|}^2{u}^{\beta -2}{\eta }^2\le 4{\Lambda }_{\ast }^2/[{\lambda }^2(\beta -1{)}^2]\int u^{\beta }|D\eta {|}^2$, with ${\Lambda }_{\ast }$ an operator norm bound for the coefficient matrix. The Sobolev chain rule and $\epsilon \downarrow 0$ justify fractional powers and zero values. An entrywise bound $\Lambda$ yields ${\Lambda }_{\ast }\le n\Lambda$.

Every density operator $\sigma$ with positive partial transpose has squared quantum fidelity at most $1/d$ with a fixed maximally entangled state on $d\otimes d$. The partial transpose of a maximally entangled projector and the trace-adjoint identity for partial transpose express that squared overlap as $\mathrm{Tr}(S{\sigma }^{T_B})/d$. Since ${\sigma }^{T_B}$ is positive of trace one and the swap operator has operator norm one, trace duality gives the bound. A matching product state attains equality.

For an $m$-quasirandom group, uniform expectations, and the normalized convolution on a finite group, a scalar mean-zero function $f$ satisfiesThe Fourier analysis on a finite group proof bounds each nontrivial matrix component of $f$ in operator norm using its weighted Hilbert-Schmidt norm. For subsets of subset density values $a,b,c$, the error in their normalized product count is at most $\sqrt{a(1-a)b(1-b)c(1-c)/m}$. In particular $abc>1/m$ guarantees a solution of $xy=z$ in the three subsets.

Independent verifier circuits on $r$ witness registers induce a threshold acceptance effect built from tensor products of $M$ and $I-M$, where $M$ is the quantum verifier acceptance operator. In a product eigenbasis of $M$, the effect's eigenvalues are threshold probabilities for independent Bernoulli random variables with parameters given by the selected eigenvalues of $M$. Bounding every parameter on a NO instance bounds the effect's operator norm, even if the actual submitted witness is entangled. Independence is used only for the spectral calculation, not assumed for the witness.