Equip the continuous dual space $X^{\prime }$ with the operator normContinuity of a linear functional makes this finite; homogeneity, definiteness and the triangle inequality follow directly from the supremum formula.

For completeness of the dual space, let $(f_n)$ be Cauchy in this norm. For each $x\in X$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $(f_n(x))$ is Cauchy in $\mathbb{R}$. Define $f(x)={\lim }_n{f}_n(x)$. Passing to the limit in the linearity identities makes $f$ linear. Since a Cauchy sequence is bounded, ${\sup }_n\Vert f_n\Vert \le C$ and $|f(x)|\le C\Vert x\Vert$, so $f\in X^{\prime }$.

Given $\epsilon >0$, choose $N$ with $\Vert f_n-f_m\Vert \le \epsilon$ for $n,m\ge N$. Let $m\to \infty$ pointwise; then $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ and therefore $\Vert f_n-f\Vert \le \epsilon$. Thus $X^{\prime }$ is a Banach space. This proof in fact works even if $X$ itself is not complete.

For the first linear map, take $(\alpha ,\beta )=(0,e_n)$. Its input norm is one, while $T_1(0,e_n)=(e_n,0)$ has norm $n^2$. Thus $T_1$ is unbounded.

For the second map, since $j^2\ge 1$,Equality holds at $(0,e_1)$, soBoth formulas map $c_{00}\times c_{00}$ into itself because adding two finitely supported sequences preserves finite support.

Let $D$ be the set of finitely supported sequences with rational coordinates. It is countable: it is the union over $N$ of copies of ${\mathbb{Q}}^N$. Given $\alpha \in {\ell }_{1,2}$ and $\epsilon >0$, choose $N$ with ${\sum }_{j>N}{j}^2|{\alpha }_j|<\epsilon /2$. Choose rational numbers $q_j$ for $1\le j\le N$ so that ${\sum }_{j=1}^N{j}^2|q_j-{\alpha }_j|<\epsilon /2$, and put $q_j=0$ thereafter. Then $q\in D$ and $\Vert q-\alpha {\Vert }_{1,2}<\epsilon$. Thus $D$ is countable and dense, so the space is separable.

Define $U:{\ell }_{1,2}\to {\ell }_1$ by $(U\alpha {)}_j=j^2{\alpha }_j$. This is a linear bijection, with inverse $(U^{-1}\gamma {)}_j=j^{-2}{\gamma }_j$, andThus $U$ is an isometric isomorphism of normed spaces onto the absolutely summable sequence space, which is a Banach space. Explicitly, a Cauchy sequence ${\alpha }^{(n)}$ maps to a Cauchy sequence $U{\alpha }^{(n)}\to \gamma \in {\ell }_1$; setting $\alpha =U^{-1}\gamma$ gives convergence in the weighted norm. Hence ${\ell }_{1,2}$ is a Banach space, an instance of weighted summable sequence space isometry.

For $\alpha ,\beta \in {\ell }_{1,2}$ and real scalars $c,d$, the scalar triangle inequality givesThe zero sequence belongs to the space, so it is a real vector space. The weighted sum is nonnegative and homogeneous, and the same inequality with $c=d=1$ proves its triangle inequality. Since every weight $j^2$ is strictly positive, zero weighted sum forces every coordinate to vanish. Therefore $\Vert \cdot {\Vert }_{1,2}$ is a norm.

Since $f(0)=0$, the fundamental theorem of calculus gives $f=T{f}^{\prime }$, with the same oriented integral on the negative half-interval. Set $h(t)=t^2$. The adjoint calculation gives the real-valued functionChoose any value at zero. This function is bounded and hence belongs to $L^2(-\pi ,\pi ]$. Since it is real, the adjoint operator identity yieldsOne can also check this by integration by parts separately on each half-interval: $g^{\prime }=-t^2$, $g(\pm \pi )=0$, and the two possible jump terms at zero vanish because $f(0)=0$.

For the adjoint of an integration operator anchored at zero, use the inner product $⟨f,g⟩=\int f\bar{g}$. On the positive half-interval, reversing the triangular integration order givesOn the negative half-interval the corresponding calculation givesThese exchanges are justified by the Fubini theorem, since on a finite interval $L^2$ functions belong to $L^1$ and the absolute double integral is at most $\Vert f{\Vert }_1\Vert g{\Vert }_1$. ThereforeAdding $Tg$ gives ${\int }_0^{\pi }g$ on the positive half and $-{\int }_{-\pi }^{0}g$ on the negative half, exactly the prescribed $Ug$. Values at zero do not matter in $L^2$. Consequently

The Hilbert projection theorem states that every vector $x$ has a unique decomposition $x=m+n$ relative to a closed linear subspace $M$, with $m\in M$ and $n\in M^{\perp }$. The vector $m$ is the unique nearest point in $M$.

The Riesz representation theorem states that every bounded complex-linear functional $F$ on a Hilbert space has a unique representation $F(x)=⟨x,y⟩$, with $\Vert F\Vert =\Vert y\Vert$, using the convention of linearity in the first argument. If $F=0$, use $y=0$. Otherwise $M=\ker F$ is closed. Project a vector outside $M$ to obtain $z\in M^{\perp }$ with $F(z)\ne 0$, and rescale it so $F(z)=1$. Then $x-F(x)z\in M$, soThis gives the representation. Uniqueness follows because a vector orthogonal to every $x$ is zero. The Cauchy-Schwarz inequality gives $\Vert F\Vert \le \Vert y\Vert$, and evaluation at $y/\Vert y\Vert$ gives equality when $y\ne 0$.

For the adjoint, fix $v$. The functional $u\mapsto ⟨Tu,v⟩$ is bounded with norm at most $\Vert T\Vert \Vert v\Vert$. By the Riesz representation theorem, there is a unique vector $T^{\ast }v$ satisfyingFor scalars $a,b$, conjugate-linearity in the second argument shows that $T^{\ast }(av+bw)=a{T}^{\ast }v+b{T}^{\ast }w$, by uniqueness of the representing vector. The same norm estimate gives $\Vert T^{\ast }v\Vert \le \Vert T\Vert \Vert v\Vert$. Thus $T^{\ast }$ is a unique bounded linear operator, the adjoint operator.

Expand both generating functions and use orthogonality of the Legendre polynomials, with ${\int }_{-1}^1{P}_n{P}_m=2{\delta }_{nm}/(2n+1)$. The allowed exchanges giveHence it depends only on $r=wz$. Explicitly,The series form alone already proves the required dependence.

The generating function gives $P_n(-t)=(-1{)}^n{P}_n(t)$, so the odd function $f$ is orthogonal to every even-degree Legendre polynomial. For $n\ge 1$, integrate the self-adjoint form of the Legendre differential equation from zero to one:Thus ${\int }_0^1{P}_n=P_n^{\prime }(0)/(n(n+1))$. Differentiating the generating function in $t$ and then setting $t=0$ givesCombining these identities with the normalization of $p_n$ yieldsFor the Legendre expansion of the sign function, $\Vert f{\Vert }_2^2=2$, so the Parseval identity givesTherefore the requested sum equals

A complete orthonormal sequence $(e_n)$ has $⟨e_n,e_m⟩={\delta }_{nm}$ and dense linear span; equivalently, the only vector orthogonal to every $e_n$ is zero. The Parseval identity says $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$, with expansion $x={\sum }_n⟨x,e_n⟩e_n$ in the Hilbert norm.

Put $G(t,z)=(1-2zt+z^2{)}^{-1/2}$ and $D=1-2zt+z^2$. Direct differentiation givesSubstitution yields $(1-t^2)G_{tt}-2t{G}_t+z^2{G}_{zz}+2z{G}_z=0$. Extracting the coefficient of $z^n$ in the generating function givesthe Legendre differential equation.

The weighted right shift $T$ is injective, so zero is not an eigenvalue. If $Ta=\lambda a$ with $\lambda \ne 0$, the first coordinate forces $a_1=0$ and subsequent coordinates force every $a_j=0$. Thus its point spectrum is empty.

For $m\ge 1$,since the largest column sum is attained at $j=1$. Therefore the spectral radius formula gives $\nu (T)={\lim }_m(m!{)}^{-1/m}=0$. The limit follows, for example, from $m!\ge (m/2{)}^{m/2}$ for large $m$. Zero is in the spectrum because the output first coordinate is always zero, so $T$ is not onto. HenceThis is a nonzero quasinilpotent element, specifically the factorial weighted right shift.

For the diagonal operator $S$, the equation $Sa=\lambda a$ gives $(j^{-1}-\lambda )a_j=0$ for every $j$. Thus a nonzero eigenvector exists exactly for $\lambda =1/j$, with eigenvector $e_j$; zero is not an eigenvalue because all diagonal entries are nonzero.

If $\lambda$ lies outside $\{0\}\cup \{1/j:j\ge 1\}$, its distance from this closed set is positive. Dividing each coordinate by $j^{-1}-\lambda$ gives a bounded inverse of $S-\lambda I$. Zero lies in the spectrum: $\Vert S{e}_j{\Vert }_1=1/j\to 0$, excluding a bounded inverse. Therefore

First give the general definitions and radius facts. The spectrum of a bounded operator is $\sigma (A)=\{\lambda :A-\lambda I\text{ has no bounded inverse}\}$, and the point spectrum is $p\sigma (A)=\{\lambda :\ker (A-\lambda I)\ne 0\}$. If $|\lambda |>\Vert A\Vert$, the Neumann series inverts $I-A/\lambda$, hence also $A-\lambda I=-\lambda (I-A/\lambda )$. Thus $\sigma (A)$ is contained in the closed disc of radius $\Vert A\Vert$ and is bounded.

The spectral radius is $\nu (A)={\sup }_{\lambda \in \sigma (A)}|\lambda |$. The spectral radius formula isFor the left shift $R$, an eigenvector satisfies $a_{j+1}=\lambda a_j$, so $a_j={\lambda }^{j-1}{a}_1$. A nonzero such sequence is in ${\ell }_1$ exactly when $|\lambda |<1$. Since $\Vert R\Vert =1$, its spectrum is contained in the closed unit disc. It contains the open unit disc as point spectrum and therefore its closure by closedness of the spectrum. Hence

For $1<p<\infty$, truncation makes the ${\ell }^p$ tail arbitrarily small, since ${\sum }_j|a_j{|}^p$ converges. Approximate the finite remaining coordinates by numbers in $\mathbb{Q}+i\mathbb{Q}$. The resulting countable set of finitely supported sequences is dense. ThereforeThe same construction also works for $p=1$, though that endpoint is not needed here.

Map each bounded sequence $a=(a_n)$ to the continuous function that takes value $a_n$ at the integer $n\ge 1$ and is affine on every interval $[n,n+1]$. Convexity of absolute value gives $|F_a(t)|\le {\sup }_n|a_n|$, and evaluation at integers gives equality of suprema. Thus $a\mapsto F_a$ is an isometric embedding of the l-infinity sequence space into $C^b([1,\infty ))$.

The former is not separable by the allowed fact. If the latter were separable, the subspace result just proved would make its isometric copy of ${\ell }_{\infty }$ separable, a contradiction. Hence

A normed vector space is a real or complex vector space $X$ with a norm satisfying positivity, definiteness, absolute homogeneity and the triangle inequality. It is separable if it has a countable dense subset for the norm distance.

First prove the unlabelled general assertions. If $(x_n)$ is dense in $X$, choose $y_{n,m}\in Y\cap B(x_n,1/m)$ whenever that intersection is nonempty. These choices form a countable subset of $Y$. For any $y\in Y$ and $\epsilon >0$, choose $m$ with $2/m<\epsilon$ and $n$ with $\Vert x_n-y\Vert <1/m$. The corresponding intersection is nonempty and $\Vert y-y_{n,m}\Vert <2/m<\epsilon$. Thus every subspace of a separable normed vector space is separable.

If a countable $S$ has dense linear span, take all finite linear combinations of its elements with rational coefficients in the real case, or coefficients in $\mathbb{Q}+i\mathbb{Q}$ in the complex case. This set is countable and dense in the linear span, by approximating the finitely many coefficients of each combination. Its closure is therefore $X$, proving separability. Conversely, a countable dense set already has dense linear span.

For $c_0$, use all finitely supported sequences with coordinates in $\mathbb{Q}+i\mathbb{Q}$. Truncate a sequence tending to zero to control its supremum tail, then approximate the finite initial coordinates. This produces a countable dense subset, so

Define $f^{♯}(s)=\overline{f(-s)}$. Interchanging the integrals and substituting variables givesMoreover Cauchy-Schwarz in the inner integral givesTaking the supremum over unit $g_2$ proves $\Vert T_f{g}_1{\Vert }_2\le \Vert f{\Vert }_1\Vert g_1{\Vert }_2$, so $T_f$ is bounded with norm at most $\Vert f{\Vert }_1$, and the adjoint identity is justified.

It remains to ensure necessity is a condition on the kernel itself. If $T_h=0$ for some $h\in L^1$, convolve $h$ with a smooth compactly supported approximate identity ${\eta }_{\epsilon }\in L^1\cap L^2$. Since $T_h{\eta }_{\epsilon }=0$, all these convolutions vanish. But $h\ast {\eta }_{\epsilon }\to h$ in $L^1$, by continuity of translations in $L^1$, so $h=0$ almost everywhere. Thus the map from kernels to convolution operators is injective. Applying this to $h=f-f^{♯}$ yields the exact criterionFor real-valued kernels this reduces to evenness; real-valuedness by itself is insufficient.

Write $\lambda =a+ib$ with $b\ne 0$. Since $aI-T$ is self-adjoint, $⟨(aI-T)v,v⟩$ is real. The cross term in the squared norm therefore vanishes:This proves the equality and lower bound, and makes both $S=\lambda I-T$ and $S^{\ast }=\overline{\lambda }I-T$ injective.

If $S{v}_n$ converges, the lower bound shows $v_n$ is Cauchy. Completeness gives $v_n\to v$ and boundedness gives $S{v}_n\to Sv$, so the range is closed. Also $(\mathrm{ran}S{)}^{\perp }=\ker S^{\ast }=0$, so the range is dense. Being both closed and dense, it equals $V$. Hence $S$ is bijective, and its lower bound gives

Complete the square to obtain, initially for real $y,z$,Compare coefficients of $y^m{z}^n$ on the two sides. The left coefficient is $(m!n!{)}^{-1}\int e^{-t^2}{H}_m(t)H_n(t)dt$, and the right is zero unless $m=n$, in which case it is $\sqrt{\pi }{2}^n/n!$. The polynomials are real on the real line, so these are the complex $L^2$ inner products of the ${\varphi }_n$. ThusPolynomial times Gaussian decay also directly shows each ${\varphi }_n$ lies in $L^2$.