Write $q(x)={\sup }_k|{\sum }_{n=1}^k{x}_n|$. This is indeed a norm: the triangle inequality and homogeneity follow from the scalar absolute value, and vanishing of every partial sum forces every coordinate to vanish. It is finite on ${\ell }^1$ because $q(x)\le \Vert x{\Vert }_1$.

Let $a_n=(-1{)}^{n+1}/n$ and let $a^{(N)}$ be its first $N$ entries followed by zeros. The alternating-series bound gives, for $M>N$,Thus these truncations are Cauchy for $q$. But $|x_n|\le 2q(x)$ for every coordinate, since $x_n$ is the difference of two consecutive partial sums. Any $q$-limit would consequently have coordinates $a_n$. This sequence is not in ${\ell }^1$. Hence

Let $e_j$ be the coordinate vectors and define finitely supported vectorsEach $v_k$ belongs to ${\ell }^1$, and $\Vert v_k{\Vert }_{\infty }=2^{-k}$. Hence ${\sum }_k{v}_k$ is absolutely convergent for the supremum norm. Its partial sums converge uniformly in ${\ell }^{\infty }$ to the sequence $v$ whose entries on the $k$th block all equal $2^{-k}$. But every block contributes one to its ${\ell }^1$ sum:Thus $v\notin {\ell }^1$. Any supremum-norm limit in ${\ell }^1$ would have the same coordinates as $v$, an impossibility. The partial sums are Cauchy and have no limit in the space. Therefore $({\ell }^1,\Vert \cdot {\Vert }_{\infty })$ is not Banach, and the displayed series is the required absolutely convergent but nonconvergent series in that space.

Show that $c$ is a closed linear subspace of ${\ell }^{\infty }$. Suppose $a^{(j)}\in c$ and $\Vert a^{(j)}-a{\Vert }_{\infty }\to 0$. Given $\epsilon >0$, choose $j$ with uniform error below $\epsilon /3$. Since $a^{(j)}$ converges, choose $N$ such that $|a_m^{(j)}-a_n^{(j)}|<\epsilon /3$ for $m,n\ge N$. ThenThus $(a_n)$ is a real Cauchy sequence and converges, so $a\in c$. A closed linear subspace of a Banach space is complete in the inherited norm. Therefore $\boxed{(c,\Vert \cdot {\Vert }_{\infty })\text{ is Banach}}$.

Use the inner product convention $⟨f,h⟩={\int }_{-\pi }^{\pi }f\bar{h}$. For $h=1_{[-\pi /2,\pi /2]}$, integration givesThe nonzero even coefficients vanish, while $\sin ((2n+1)\pi /2)=(-1{)}^n$. The preceding polarized Parseval identity therefore yieldsAbsolute convergence follows from square summability of the Fourier coefficients and of $((2n+1{)}^{-1}{)}_{n\in \mathbb{Z}}$.

For the convolution identity, fix any real $t$ and define $h_t(s)=\overline{g(t-s)}$. Periodicity makes $h_t\in L^2(-\pi ,\pi )$ with $\Vert h_t{\Vert }_2=\Vert g{\Vert }_2$. A change of variable and integration over one full period giveApplying polarized Parseval identity to $f,h_t$ gives, for every fixed $t$,The integral exists by the Cauchy-Schwarz inequality. The same inequality gives ${\sum }_n|⟨f,e_n⟩⟨g,e_n⟩|\le \Vert f{\Vert }_2\Vert g{\Vert }_2$, so the displayed series converges absolutely and uniformly in $t$. In particular the equality holds at every $t$, without requiring pointwise convergence of either original Fourier series. This is Fourier convolution of two periodic L2 functions.

Commutativity and equality of squares giveThus $S-T$ vanishes on $\mathrm{Ran}(S+T)$, and by continuity on its closure. The preceding kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Since $S+T$ is self-adjoint, part (i) and the Hilbert projection theorem give the orthogonal direct sumTherefore $S-T$ vanishes on all of $X$, provingThis proves commuting positive operators with equal squares are equal without assuming a square-root uniqueness theorem.

By the Cauchy-Schwarz inequality for a positive operator, for every $y$,Taking $y=Tx$ gives $Tx=0$, and $(S+T)x=0$ then gives $Sx=0$. ThusThe reverse inclusion is immediate. This is the kernel of a sum of positive operators identity.

For $x\in \ker (S+T)$,Both terms are nonnegative because $S,T$ are positive operators. Their sum is zero, so each vanishes. In particular,

Set $B(x,y)=⟨Tx,y⟩$. It is a positive semidefinite Hermitian form. Write $A=B(x,x)$, $C=B(y,y)$ and $c=B(x,y)$. For every complex $t$,If $C>0$, choosing $t=-c/C$ gives $A-|c{|}^2/C\ge 0$. If $C=0$ and $c\ne 0$, choosing $t=-sc$ for large positive $s$ makes the expression negative, a contradiction. Thus $c=0$ in that case. In all cases,This is the Cauchy-Schwarz inequality for a positive operator.

For a positive operator, $⟨Tx,x⟩\ge 0$. We show directly that every nonreal or negative real $\lambda$ lies in the resolvent set.

If $\mathrm{Im}\lambda \ne 0$, the Cauchy-Schwarz inequality and reality of $⟨Tx,x⟩$ giveThus $T-\lambda I$ is bounded below, hence injective with closed range: a convergent sequence of outputs has Cauchy preimages, whose limit maps to that output. Its adjoint is $T-\bar{\lambda }I$, which has trivial kernel by the same estimate. The identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$, proved by the same inner-product calculation as in (i), makes its range dense. Being closed and dense, it is all of $X$, and the lower bound makes the inverse bounded.

If $\lambda <0$ is real, positivity gives $⟨(T-\lambda I)x,x⟩\ge (-\lambda )\Vert x{\Vert }^2$, so the same argument applies with lower bound $-\lambda$. Here the operator is self-adjoint, so its range is dense because its own kernel is zero. Hence

Self-adjointness and conjugate symmetry of the inner product giveA complex number equal to its conjugate is real, so $⟨Tx,x⟩\in \mathbb{R}$ for every $x$.

A bounded linear operator on a complex Hilbert space is self-adjoint if $T=T^{\ast }$, equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y$.

If $Tx=0$, then $⟨x,Ty⟩=⟨Tx,y⟩=0$ for all $y$, so $x\in (\mathrm{Ran}T{)}^{\perp }$. Conversely, membership in that orthogonal complement gives $⟨Tx,y⟩=0$ for every $y$, and choosing $y=Tx$ yields $Tx=0$. Therefore

The unilateral shift operator $R$ is not onto, since its outputs have first coordinate zero. The left shift operator $L$ is not injective, since $L{e}_1=0$. Both have operator norm one, and $LR=I$. ConsequentlyIf $\Vert A\Vert <1$, then $\Vert LA\Vert <1$, so the Neumann series makes $I+LA$ invertible. If $R+A$ were invertible too, the identity above would givea product of invertible operators, contradicting the nontrivial kernel of $L$. ThusThis is norm stability of noninvertibility for the unilateral right shift.

For the multiplication operator, $\Vert Tf{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }\Vert f{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. Taking $f=1$ gives equality, so $\Vert T\Vert =1$.

An eigenfunction obeys $(g(t)-\lambda )f(t)=0$ at every $t$. If $\lambda \in [0,1)$, its level set is the single point $t=\lambda$, and a continuous function supported at a single point is zero. If $\lambda \notin [0,1]$, that level set is empty. For $\lambda =1$, every continuous function vanishing on $[0,1]$ is an eigenfunction, and nonzero examples exist, such as $f(t)=\max \{t-1,0\}$. Thus ${\sigma }_p(T)=\{1\}$.

If $\lambda \notin [0,1]$, multiplication by $(g-\lambda {)}^{-1}$ is a bounded inverse of $T-\lambda I$. If $\lambda \in [0,1]$, choose $t_0$ with $g(t_0)=\lambda$. Every function in the range of $T-\lambda I$ vanishes at $t_0$, so the range omits the constant function one. Hence it is not onto. Therefore

If $|\lambda |>\Vert T\Vert$, write $T-\lambda I=-\lambda (I-T/\lambda )$. The Neumann series applies to $T/\lambda$, so $\lambda$ is outside the spectrum of a bounded operator. Thus $\sigma (T)\subseteq \{\lambda :|\lambda |\le \Vert T\Vert \}$ and $r(T)\le \Vert T\Vert$.

The polynomial spectral mapping theorem gives $\sigma (T^m)=\{{\lambda }^m:\lambda \in \sigma (T)\}$, and henceTaking $m$th roots and then the limit, whose existence is allowed, provesOnly this inequality is required; the reverse inequality is the additional content of the full spectral radius formula.

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded inverse}\}$. The point spectrum consists of those $\lambda$ for which $\ker (T-\lambda I)\ne 0$, namely the eigenvalues. The spectral radius is $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$.

If $\Vert T\Vert <1$, the Neumann series $S={\sum }_{n=0}^{\infty }(-T{)}^n$ converges in the Banach space of bounded linear operators, since its term norms are bounded by a convergent geometric series. Multiplying the finite sums on either side by $I+T$ gives $I-(-T{)}^{N+1}$, which tends in operator norm to $I$. Therefore

Put $\rho =r-\sqrt{r^2-1}$. Since $r>1$, $0<\rho <1$ and ${\rho }^2-2r\rho +1=0$. The nonzero square-summable sequence ${\beta }_j={\rho }^{j-1}$ defines a continuous linear functional on ${\ell }_2$ byThe Cauchy-Schwarz inequality proves continuity. For every $k$,Hence $f$ vanishes on the closed linear span $M$, but $f(e_1)=1$. ThereforeMore precisely, any vector in the orthogonal complement of $M$ satisfies the recurrence ${\beta }_{k+2}-2r{\beta }_{k+1}+{\beta }_k=0$. Its roots are $r\pm \sqrt{r^2-1}$, one less than one and the other greater than one. Square summability eliminates the growing solution, so $M^{\perp }$ is exactly the span of $({\rho }^{j-1})$. Thus $M=\ker f$ is a closed hyperplane of codimension one. This is annihilator of a second-difference span in l2.

If a bilinear map is continuous, continuity at $(0,0)$ gives $\delta >0$ such that $\Vert F(x,y)\Vert \le 1$ whenever $\Vert x\Vert ,\Vert y\Vert <\delta$. For nonzero $x,y$, apply this to $\delta x/(2\Vert x\Vert )$ and $\delta y/(2\Vert y\Vert )$. Bilinearity givesIf either input is zero, the inequality follows from bilinearity directly.

Conversely, suppose the product bound holds with $C\ge 0$. At a fixed point $(x,y)$,Its norm is at most $C(\Vert h\Vert \Vert y\Vert +\Vert x\Vert \Vert k\Vert +\Vert h\Vert \Vert k\Vert )$, which tends to zero as $(h,k)\to (0,0)$. Thus the product bound is equivalent to joint continuity. This is boundedness criterion for a bilinear map.