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On the Hilbert space ${\ell }^2(\mathbb{N})$, the left shift deletes the first coordinate. It is the adjoint operator of the unilateral shift operator $S$. The identities $LS=I$ and $SLf=f-f_1{e}_1$ show that it is also the Moore–Penrose inverse of an operator for $S$; both maps have operator norm one.

In a unital Banach algebra, for $\lambda \ne 0$ the existence of $R=(\lambda 1-ab{)}^{-1}$ gives $(\lambda 1-ba{)}^{-1}={\lambda }^{-1}(1+bRa)$, as verified by multiplying on either side. Interchanging $a,b$ proves the converse. Zero may differ: for the unilateral shift operator $S$ on ${\ell }^2({\mathbb{N}}_0)$, $S^{\ast }S=1$ has spectrum $\{1\}$ and $S{S}^{\ast }=1-P_0$ has spectrum $\{0,1\}$.

On ${\ell }_1$, the unilateral shift operator $R$ and left shift operator $L$ satisfy $LR=I$ and $\Vert L\Vert =1$, while $L$ has nontrivial kernel. If $\Vert A\Vert <1$, the Neumann series makes $L(R+A)=I+LA$ invertible. Were $R+A$ invertible, this identity would express $L$ as a product of invertible operators, a contradiction. Thus the open unit ball about $R$ contains no invertible operator.

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.

The retraction of a Hilbert ball onto its sphere exists. In the real Hilbert space ${\ell }^2$, let $S(x_0,x_1,\dots )=(0,x_0,x_1,\dots )$ be the unilateral shift operator, and let $e_0=(1,0,\dots )$. For $x$ in the closed unit ball $\mathbb{B}$, defineThe orthogonality of $e_0$ and $Sx$ gives $\Vert F(x){\Vert }^2=1$. Thus $F$ is a continuous map into the unit sphere. It has no fixed point: $F(x)=x$ would imply $x_{j+1}=x_j$ for every $j\ge 0$. A constant sequence in ${\ell }^2$ must be zero, whereas $F(0)=e_0\ne 0$.

Put $d=x-F(x)$, which is nonzero everywhere, and defineThis is a continuous map on the whole closed unit ball, including zero. To check its range, observe that $⟨F(x),d⟩=⟨F(x),x⟩-1$, soIf $\Vert x\Vert =1$, then $\Vert x-F(x){\Vert }^2=2(1-⟨F(x),x⟩)$, hence $\alpha (x)=1$ and $r(x)=x$. Therefore $r:\mathbb{B}\to \partial \mathbb{B}$ is a retraction.

Geometrically this is the second intersection of the line from $F(x)$ through $x$ with the unit sphere. In factThe construction illustrates why the finite-dimensional hypothesis in the no-retraction theorem matters. The closed unit ball of ${\ell }^2$ is not a compact set in the norm topology.

Set $a_n=\Vert x^n\Vert$ and $D={\inf }_{n\ge 1}{a}_n^{1/n}$. If some $x^m=0$, all later powers vanish, their root norms tend to zero, and $D=0$. Otherwise fix $m\ge 1$ and write $n=qm+s$, $0\le s<m$. Submultiplicativity givesSince $q/n\to 1/m$, it follows that ${\mathrm{limsup}}_n{a}_n^{1/n}\le a_m^{1/m}$. Taking the infimum over $m$ and using $a_n^{1/n}\ge D$ provesThis establishes existence of the limit and its infimum characterization before identifying it with the spectral radius $r(x)={\max }_{\lambda \in \sigma (x)}|\lambda |$.

For $|\lambda |>D$, the root test makes ${\sum }_{n\ge 0}{x}^n/{\lambda }^{n+1}$ converge absolutely. Multiplication of partial sums shows that its limit is $(\lambda 1-x{)}^{-1}$. Thus $r(x)\le D$. For the reverse inequality, take any $\rho >r(x)$ and the circle $|\lambda |=\rho$. The resolvent of an element is holomorphic off the spectrum, andTo justify this identity, first use a radius larger than $\Vert x\Vert$ and integrate the uniformly convergent Neumann series term by term. Then deform to radius $\rho$: the intervening annulus has no spectral points. The Cauchy theorem for these Banach-valued integrals follows by applying bounded linear functionals and using their point separation. Consequently, with $M_{\rho }={\max }_{|\lambda |=\rho }\Vert (\lambda 1-x{)}^{-1}\Vert$,Taking root limits gives $D\le \rho$; let $\rho \downarrow r(x)$. Hence the full spectral radius formula is

For the product comparison, if $\lambda \ne 0$ and $R=(\lambda 1-xy{)}^{-1}$, direct multiplication on both sides verifiesIndeed $(\lambda 1-yx)y=y(\lambda 1-xy)$, and the remaining terms cancel against the inverse equation. Interchanging $x,y$ proves the converse. Thus the nonzero spectra of products in opposite orders agree, soThe possible difference at zero has no effect on the maximum modulus, including when that maximum is zero.

Zero really can differ. In $A=\mathcal{B}({\ell }^2({\mathbb{N}}_0))$, let $S$ be the unilateral shift operator, $S{e}_n=e_{n+1}$, and take $x=S^{\ast }$, $y=S$. Then $xy=1$, whereas $yx=1-P_0$, with $P_0$ the projection onto $e_0$. The latter is zero on $e_0$ and identity on its orthogonal complement, and these blocks also explicitly invert it away from zero and one. Hence

Finally, for $t>r(x)$ the root norms of $(x/t{)}^n$ tend to $r(x)/t<1$, so their norms eventually decay geometrically and the powers converge to zero. Conversely, if the powers converge to zero, choose $N$ with $\Vert (x/t{)}^N\Vert <1$. Homogeneity and the infimum formula implyso $t>r(x)$. The admissible positive $t$ are exactly $(r(x),\infty )$, proving the power-decay characterization of the spectral radiusThis includes quasinilpotent elements with radius zero; no positive endpoint must be attained.

The Fredholm index measures the finite-dimensional obstruction to inverting an operator. For a bounded operator $T:H_1\to H_2$ between Hilbert spaces, the Fredholm conditions are that its range of a bounded linear operator is closed and its kernel and cokernel are finite-dimensional. The orthogonal complement of the range is $\ker T^{\ast }$, soAn invertible linear map has Fredholm index zero; the converse needs an additional finite-dimensional correction, since an operator can have equally large kernel and cokernel.

First construct an inverse up to finite-rank operators. On $(\ker T{)}^{\perp }$, the restriction of $T$ is a bounded bijection onto its closed range, with bounded inverse by the bounded inverse theorem. Extend that inverse by zero on $(\mathrm{ran}T{)}^{\perp }$, obtaining $S:H_2\to H_1$. ThenConversely, suppose $ST-I$ and $TS-I$ are compact operators. On $\ker T$, the first identity expresses the identity as a compact operator, so this kernel is finite-dimensional. If $T$ were not bounded below on $(\ker T{)}^{\perp }$, there would be unit vectors $x_j$ in that complement with $T{x}_j\to 0$. Write $ST=I+K$. Compactness of $K$ gives a subsequence for which $K{x}_j$ converges, hence $x_j=ST{x}_j-K{x}_j$ converges to a unit vector $x$ in $(\ker T{)}^{\perp }$ with $Tx=0$, a contradiction. This lower bound makes the range closed. Applying the analogous argument to $T^{\ast }$, using $S^{\ast }{T}^{\ast }=I+K^{\prime \ast }$, proves that its kernel is finite-dimensional. This proves the Atkinson theorem: a bounded operator is Fredholm exactly when it has an inverse modulo compact operators. For operators on one Hilbert space this says that its image in the Calkin algebra is invertible.

The Fredholm index is stable. Decompose the domain as $N\oplus N^{\perp }$, where $N=\ker T$, and the target as $C\oplus R$, where $C=\ker T^{\ast }$ and $R=\mathrm{ran}T$. A sufficiently small norm perturbation has block matrixMultiplying by invertible triangular block matrices reduces this matrix to the direct sum of $D$ and the Schur complement $a-b{D}^{-1}c:N\to C$. The latter's kernel dimension minus cokernel dimension is $\dim N-\dim C$, whatever its rank. Therefore the Fredholm set is open and the index is locally constant. If $K$ is compact, the same parametrix works modulo compact operators for every $T+tK$, $0\le t\le 1$. Local constancy along this path gives

For Fredholm operators $A:H_1\to H_2$ and $B:H_2\to H_3$, multiplication of parametrices shows that $BA$ is Fredholm. Its index is additive. Here is a direct proof, rather than an assertion of additivity. There is an exact sequence of finite-dimensional spacesThe map from $\ker B$ takes a vector to its class modulo $\mathrm{ran}A$; the next takes the class of $y$ to that of $By$; the last is the quotient map. For example, a class $y+\mathrm{ran}A$ lies in the kernel of the middle map exactly when $By=BAx$ for some $x$, so $y-Ax\in \ker B$. This verifies the only less immediate exactness assertion; the others follow directly from their definitions. Alternating the dimensions givesAlso $\mathrm{ind}{T}^{\ast }=-\mathrm{ind}T$, and indices add under direct sums. For $I+K$ with $K$ compact, the index is zero, so injectivity is equivalent to surjectivity. More generally $Tx=f$ is solvable exactly when $f$ is orthogonal to $\ker T^{\ast }$: these are the finitely many compatibility conditions in the Fredholm alternative.

Toeplitz operators exhibit how this analytic index detects a topological winding. Let $H^2\subset L^2(S^1)$ be the Hardy space of the circle, spanned by $1,z,z^2,\dots$, and let $P$ be its orthogonal projection. For a continuous symbol $g$, define $T_g=P{M}_g{|}_{H^2}$, where $M_g$ is multiplication by $g$. Then $\Vert T_g\Vert \le \Vert g{\Vert }_{\infty }$. The product defect isFor trigonometric polynomials the crossing of negative and nonnegative Fourier modes has finite rank. Uniform approximation of continuous symbols by trigonometric polynomials, together with the displayed norm estimate, therefore makes this defect a compact operator for all continuous $f,g$. If $g$ is nowhere zero, $1/g$ is continuous and $T_{1/g}$ is a two-sided parametrix modulo compact operators. Thus $T_g$ is Fredholm.

The converse is instructive. If $g(\zeta )=0$ at $\zeta \in S^1$, consider the normalized Hardy kernelsTheir norms are one, their individual Fourier coefficients tend to zero as $r\uparrow 1$, and hence they converge weakly to zero. Their squared moduli are the Poisson kernel, so continuity of $g$ givesA Fredholm operator is bounded below off its finite-dimensional kernel. The projection of these weakly vanishing vectors onto that kernel tends to zero, contradicting this lower bound. Thus a continuous-symbol Toeplitz operator is Fredholm exactly when its symbol never vanishes. Applying the same argument to $g-\lambda$ gives its essential spectrum as $g(S^1)$, with essential spectrum defined by failure of the Fredholm property.

For nonvanishing $g$, let $m=\mathrm{wind}(g,0)$. Lifting a continuous argument of $g(e^{it})$ on $0\le t\le 2\pi$ gives an argument change $2\pi m$. After division by $z^m$, its argument has matching endpoints, so there is a continuous function $h:S^1\to \mathbb{C}$ with $g(z)=z^m{e}^{h(z)}$. The nonvanishing-symbol homotopy $g_t=z^m{e}^{th}$ reduces the index to that of $T_{z^m}$. For $m\ge 0$ this is the $m$th power of the unilateral shift operator, whose kernel is zero and whose range has codimension $m$. For $m<0$ it is the $(-m)$th power of the adjoint shift, whose kernel has dimension $-m$ and whose range is all of $H^2$. ConsequentlyThe minus sign comes from counting missing nonnegative Fourier modes. This is a concrete bridge between the Fredholm index, stability under compact perturbations, and the winding number of a loop.

For $\lambda \ne 0$, suppose $R=(\lambda 1-ba{)}^{-1}$ exists. The identities $(\lambda 1-ab)a=a(\lambda 1-ba)$ and $b(\lambda 1-ab)=(\lambda 1-ba)b$ show, by multiplication on either side, thatFor example, multiplying on the left gives $1-{\lambda }^{-1}ab+{\lambda }^{-1}a(\lambda 1-ba)Rb=1$, and the right product gives the same cancellation. Thus invertibility of $\lambda 1-ba$ implies invertibility of $\lambda 1-ab$. Interchanging $a,b$ proves the converse. The nonzero spectra of products in opposite orders are therefore equal:In particular the requested inclusion follows. The exclusion of zero matters: for the unilateral shift operator $S$, $S^{\ast }S=1$ has Banach algebra spectrum $\{1\}$, whereas $S{S}^{\ast }=1-P_0$ has Banach algebra spectrum $\{0,1\}$.

Use the spectral definition of a Positive element of a C-star algebra: $h$ is positive if $h=h^{\ast }$ and ${\sigma }_A(h)\subseteq [0,\infty )$. We use the standard continuous functional calculus for a Hermitian element of a C-star algebra, including its real Banach algebra spectrum and the equality of its norm with its spectral radius. To prove $a^{\ast }a$ positive without assuming this as part of the definition, first establish two elementary cone facts.

If $p,q$ are positive and $s=\Vert p\Vert +\Vert q\Vert$, the continuous functional calculus gives $\Vert \Vert p\Vert 1-p\Vert \le \Vert p\Vert$ and the corresponding bound for $q$, henceSince $p+q$ is self-adjoint, every spectral value $t$ is real and satisfies $|s-t|\le s$, so $t\ge 0$. Thus the positive cone of a C-star algebra is closed under addition. It is proper: if both $h$ and $-h$ are positive, their Banach algebra spectra force $\sigma (h)\subseteq \{0\}$, and the self-adjoint norm equality gives $h=0$. It is also norm closed: if $p_n\ge 0$ converges to $p$, choose $r\ge {\sup }_n\Vert p_n\Vert$; then $\Vert r1-p_n\Vert \le r$ passes to the limit and forces every spectral value of $p$ to be nonnegative.

Now put $h=a^{\ast }a$, and form its negative part $d=\max (-h,0)$ by continuous functional calculus. Let $c=a{d}^{1/2}$. Because $d$ is a function of $h$, the scalar identity $t\max (-t,0)=-\max (-t,0{)}^2$ givesThis has nonpositive Banach algebra spectrum. The previously proved equality of nonzero product Banach algebra spectra shows that $c{c}^{\ast }$ also has nonpositive Banach algebra spectrum; it is self-adjoint, so both $c^{\ast }c$ and $c{c}^{\ast }$ are negative elements. Write $c=u+iv$, where $u=(c+c^{\ast })/2$ and $v=(c-c^{\ast })/(2i)$ are self-adjoint. Direct multiplication givesBy the spectral mapping theorem, the squares $u^2,v^2$ are positive, so their sum is positive by the cone fact above. But the same sum is negative, since both its original summands are negative. Properness forces $c^{\ast }c+c{c}^{\ast }=0$. Then $c^{\ast }c=-c{c}^{\ast }$ is both negative and positive, so $c^{\ast }c=0$, whence $d^2=0$. The C-star identity applied to the self-adjoint $d$ gives $\Vert d{\Vert }^2=\Vert d^2\Vert =0$. Thus $d=0$, provingThis establishes positivity of adjoint products from spectral positivity. It also proves positivity under congruence: for $p\ge 0$ and any $z$, write $p=r^2$ using its functional-calculus square root; then $z^{\ast }pz=(rz{)}^{\ast }(rz)\ge 0$.

For the comparison of squares, we first derive inversion reverses the order of strictly positive elements. If $0<x\le y$ with both positive and invertible, thenThe continuous functional calculus gives $k^{-1}\le 1$, because the scalar function $1-t^{-1}$ is nonnegative on $\sigma (k)\subseteq [1,\infty )$. Congruence then givesNo commutation of $x,y$ was assumed.

Next derive the square-root resolvent integral in a C-star algebra. For $s\ge 0$,For $s>0$, substituting $t=s{u}^2$ reduces the integral to $\sqrt{s}{\pi }^{-1}{\int }_0^{\infty }2/(1+u^2)du=\sqrt{s}$; for $s=0$ it is zero. For a positive element $x$, the norm of $x(t1+x{)}^{-1}$ is at most $\min (1,\Vert x\Vert /t)$. Thus the integral converges in norm, with integrable bounds $t^{-1/2}$ near zero and $\Vert x\Vert t^{-3/2}$ near infinity, uniform on $\sigma (x)$. The continuous functional calculus consequently givesPut $\alpha =a^2$ and $\beta =b^2$. The assumption $b^2-a^2\ge 0$ says $\alpha \le \beta$. For every $t>0$, inversion reverses the order of the strictly positive elements $t1+\alpha$ and $t1+\beta$. Subtracting the two square-root formulas therefore yieldsHere ${\alpha }^{1/2}=a$ and ${\beta }^{1/2}=b$ because $a,b$ themselves are positive. Every truncated integral has a positive integrand, so is positive; norm convergence and closedness of the positive cone of a C-star algebra make its limit positive. This proves order preservation by the positive square root in the required case:It is important that this argument does not factor $b^2-a^2$ as $(b-a)(b+a)$, which would be invalid for noncommuting elements.

For $p=b-a\ge 0$, apply the continuous function $t\mapsto \sqrt{t}$ on its nonnegative compact Banach algebra spectrum. The resulting element $p^{1/2}$ is self-adjoint, has nonnegative Banach algebra spectrum by continuous functional calculus, and satisfies $(p^{1/2}{)}^2=p$. Thus $b-a$ has a positive square root in a C-star algebra.

Finally, if the positive element $a$ is invertible, its compact Banach algebra spectrum is bounded away from zero. Let $\epsilon =\min \sigma (a)>0$; continuous functional calculus gives $a\ge \epsilon 1$. Since $b\ge a$, we also have $b\ge \epsilon 1$, so $\sigma (b)\subseteq [\epsilon ,\infty )$. The reciprocal function is continuous on this Banach algebra spectrum and supplies $b^{-1}$. Therefore $\boxed{a\text{ invertible}⟹b\text{ invertible}}$ under the given positivity assumptions.

For an element $a$ of a complex unital Banach algebra, its spectrum of an element isThe nonzero spectra of products in opposite orders agree:For $\lambda \ne 0$, if $R=(\lambda 1-ab{)}^{-1}$ exists, direct multiplication on both sides verifiesInterchanging $a,b$ proves the converse. Zero may differ. On the Hilbert space ${\ell }^2({\mathbb{N}}_0)$, let $S$ be the unilateral shift operator $S{e}_n=e_{n+1}$ and take $a=S^{\ast }$, $b=S$. Then $ab=S^{\ast }S=I$, whereas $ba=S{S}^{\ast }=I-P_0$, where $P_0$ is projection onto the first coordinate. HenceBoth values for the second operator occur on its two nonzero invariant summands; every other value has an inverse obtained separately on those summands.

We use the spectral definition of a Positive element of a C-star algebra: $h$ is positive when $h=h^{\ast }$ and $\sigma (h)\subseteq [0,\infty )$. The continuous functional calculus, supplied by the Commutative Gelfand--Naimark theorem for the algebra generated by a Normal element of a C-star algebra and its adjoint, sends $F\in C(\sigma (h))$ to $F(h)$. It preserves products, sums, conjugation and the identity, is isometric, and satisfies $\sigma (F(h))=F(\sigma (h))$. For a self-adjoint element the spectrum is real, so real-valued functions produce self-adjoint elements.

For example, $F(t)=\sqrt{t}$ gives the positive square root in a C-star algebra of a Positive element of a C-star algebra. It satisfies $(h^{1/2}{)}^2=h$. Its uniqueness follows from the composition rule: if $d\ge 0$ and $d^2=h$, then the continuous functional calculus in the algebra generated by $d$ gives $\sqrt{d^2}=d$. For positive invertible $h$, the functions $t^{-1}$ and $t^{-1/2}$ give positive inverses and inverse square roots. If $0\le h$ and $r\ge \Vert h\Vert$, the function $r-t$ gives $r1-h\ge 0$; the same calculus givesThese transfer elementary scalar inequalities for functions of one element to the algebra. They do not justify applying a scalar inequality to arbitrary noncommuting elements.

To prove closure under addition without assuming an order theorem, let $a,b\ge 0$, set $\alpha =\Vert a\Vert$, $\beta =\Vert b\Vert$ and $s=\alpha +\beta$. If $s=0$, both elements are zero. Otherwise,The element $a+b$ is self-adjoint, so any $\lambda \in \sigma (a+b)$ is real and satisfies $|s-\lambda |\le s$. Thus $\lambda \ge 0$, provingWe will also use that the positive cone is proper: if $h\ge 0$ and $-h\ge 0$, then $\sigma (h)\subseteq \{0\}$, and the norm formula for self-adjoint elements gives $h=0$.

We next prove $a^{\ast }a\ge 0$ without presupposing that positivity is defined by such products. Put $h=a^{\ast }a$, which is self-adjoint, and form its negative part by the continuous functional calculus:This is positive. Let $c=a{h}_{-}^{1/2}$. Since $h_{-}$ commutes with $h$,The nonzero product-spectrum identity proved above gives $\sigma (c{c}^{\ast })\setminus \{0\}=\sigma (c^{\ast }c)\setminus \{0\}$. As $c{c}^{\ast }$ is self-adjoint, it too has spectrum in $(-\infty ,0]$, so $c{c}^{\ast }\le 0$.

Write $c=u+iv$ with $u=u^{\ast }$ and $v=v^{\ast }$. The continuous functional calculus makes $u^2$ and $v^2$ positive, and the addition result givesBut both summands on the left are nonpositive, so their sum is also nonpositive. Properness of the cone forces $c^{\ast }c+c{c}^{\ast }=0$. Now $c^{\ast }c=-c{c}^{\ast }$ is positive as well as nonpositive, so $c^{\ast }c=0$. The C-star identity gives $c=0$, whence $h_{-}^2=0$ and $h_{-}=0$. Thus $h$ has no negative spectral part:This proves the positivity of adjoint products from spectral positivity and the equivalence with the usual definition by elements of the form $d^{\ast }d$: the reverse direction is the factorization $h=(h^{1/2}{)}^{\ast }{h}^{1/2}$ for $h\ge 0$.

Finally, if $a\ge 0$ and $c\in A$, writeThe preceding result therefore provesIn particular, congruence preserves positivity in a C-star algebra, and $a\le b$ implies $c^{\ast }ac\le c^{\ast }bc$. These facts will justify the noncommutative order manipulations in the numbered parts.

Use an inner product linear in the first variable. For each $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ is represented uniquely by a vector $T^{\ast }y$ under the Riesz representation theorem, so $⟨Tx,y⟩=⟨x,T^{\ast }y⟩$. Uniqueness and the conjugate-linearity in the second argument imply $T^{\ast }(ay+bz)=a{T}^{\ast }y+b{T}^{\ast }z$. FurthermoreThus the adjoint operator exists, is linear and bounded, and is unique because a vector orthogonal to every $x$ is zero.

A normal operator satisfies $T^{\ast }T=T{T}^{\ast }$. The unilateral unilateral shift operator $S(x_1,x_2,\dots )=(0,x_1,x_2,\dots )$ on ${\ell }^2$ is bounded but not normal: $S^{\ast }S=I$, whereas $S{S}^{\ast }=I-P_{e_1}$. For general $T$,Normality implies equality of norms. Conversely equality for every $x$ makes this self-adjoint operator's quadratic form zero; the polarization identity then gives $T^{\ast }T-T{T}^{\ast }=0$.

An approximate eigenvalue $\lambda$ has unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$. It is necessarily in the spectrum, since a bounded inverse would prevent this. Conversely suppose $\lambda$ is not an approximate eigenvalue. Then $A=T-\lambda I$ is bounded below, $\Vert Ax\Vert \ge c\Vert x\Vert$. It is injective and has closed range: a convergent sequence $A{x}_n$ forces $x_n$ to be Cauchy. The operator $A$ is normal, so $\Vert A^{\ast }x\Vert =\Vert Ax\Vert$ and $\ker A^{\ast }=0$. The identity $(\mathrm{ran}A{)}^{\perp }=\ker A^{\ast }$ makes its range dense. Closed and dense means surjective, and its inverse is bounded by $1/c$. Hence $\lambda$ is in the resolvent. This proves $\boxed{\sigma (T)={\sigma }_{\mathrm{ap}}(T)}$.

The given map is the unilateral shift operator. Its adjoint operator is the left shift operator, sinceThe shift is an isometry with trivial kernel and closed range $\{f:f_1=0\}$. Its range has orthogonal complement $\mathrm{span}\{e_1\}$, so every square-summable datum is admissible. Consequently,Indeed $K^{†}K=I$, while $K{K}^{†}f=f-f_1{e}_1$ is the required orthogonal projection. The discarded first component is the residual of the least-squares solution, not an obstruction to the existence of the Moore–Penrose inverse of an operator.

The closed unit ball $B$ in the infinite-dimensional real Hilbert space ${\ell }^2$ admits a retraction onto its unit sphere. Write $S$ for the forward unilateral shift operator and set $F(x)=\sqrt{1-\Vert x{\Vert }^2}{e}_0+Sx$. Then $\Vert F(x)\Vert =1$ and $F(x)\ne x$: equality would force all coordinates of $x$ to coincide, hence vanish, contradicting $F(0)=e_0$.

For $d=x-F(x)$, setThe nonvanishing denominator makes $r$ a continuous map. Expanding its squared norm gives $\Vert r(x)\Vert =1$. On $\Vert x\Vert =1$, the fraction is one, so $r(x)=x$. Thus the finite-dimensional no-retraction theorem does not extend to this closed unit ball.

A right inverse of an element $a$ is an element $b$ with $ab=1$. A left inverse of an algebra element and a right inverse of an algebra element, when both exist, coincide. For the unilateral shift operator $S$, $S^{\ast }S=1$ but $S{S}^{\ast }\ne 1$, exhibiting the distinction in the Banach algebra of bounded operators.

A shift operator translates the components of a function or sequence by a fixed displacement. Bilateral shifts act on doubly infinite sequences; the unilateral shift operator acts on a one-sided sequence space and has distinct forward and backward versions.