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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 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.