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A quotient operator is a surjective bounded linear operator that is an open map. For a linear map, being an open map is equivalent to the existence of $c>0$ such thatThe inclusion gives a neighbourhood of $Tx$ inside the image of every sufficiently small ball about $x$, by translation and scaling. Conversely, an open map sends $B_X(0,1)$ to a neighbourhood of zero. A linear open map onto $Y$ is automatically surjective, since its image contains a ball and is a linear subspace.

The Banach-space adjoint is $T^{\ast }g=g\circ T$. For $g\in Y^{\ast }$, the ball inclusion givesAlso $\Vert T^{\ast }g\Vert \le \Vert T\Vert \Vert g\Vert$. Thus the Banach-space adjoint is an isomorphic embedding, with lower bound $c$.

The real Hahn-Banach separation theorem says that if $C$ is a nonempty open convex set in a real normed vector space and $y\notin C$, there is a nonzero continuous linear functional $g$ such that $g(y)\ge {\sup }_{z\in C}g(z)$.

Put $\delta =1-1/M\in (0,1)$. The prescribed set is open because it is a union of translates of an open ball, and it is convex because both summands are convex. If $\Vert y\Vert <1$ and $y\notin C$, Hahn-Banach separation gives a nonzero $g\in Y^{\ast }$ withThe displayed supremum is a sum of independent suprema; the symmetry of the unit balls identifies them with the respective operator norms. But $g(y)\le \Vert g\Vert \Vert y\Vert <\Vert g\Vert$, a contradiction. Hence $y=Tx+r$ for some $\Vert x\Vert \le 1$ and $\Vert r\Vert <\delta$.

To obtain the strict bound on $x$ required in the approximation hypothesis, choose $t<1$ sufficiently close to $1$. Then $\Vert tx\Vert <1$ andThus the approximation condition holds with exactly $\delta =1-1/M$. When $X$ is a Banach space, the preceding successive approximation for quotient operators proves $T$ is a quotient operator. Combining this with the Banach-space adjoint estimate proves

We use successive approximation for quotient operators. Fix $y$ with $\Vert y\Vert <1$. The hypothesis gives $x_0$ with $\Vert x_0\Vert <1$ and $\Vert y-T{x}_0\Vert <\delta$. Inductively, ifapply the hypothesis to $r_n/{\delta }^n$, obtaining $x_n$ with $\Vert x_n\Vert <{\delta }^n$ and $\Vert r_n-T{x}_n\Vert <{\delta }^{n+1}$. If a residual is zero, take all remaining terms zero.

The series $\sum x_n$ converges in the Banach space $X$ because $\sum \Vert x_n\Vert <\sum {\delta }^n=(1-\delta {)}^{-1}$. Write its sum as $x$. The continuity of $T$ and the residual estimate giveScaling now yieldsThis ball inclusion proves that $T$ is a quotient operator. Completeness of $Y$ was unnecessary: the convergent series is constructed in $X$ and its image has limit $y$.

Let $T:X\to Y$ be a bounded linear operator with $X$ a Banach space. Suppose every $\Vert y\Vert <1$ has an approximation $Tx$ with $\Vert x\Vert <1$ and error less than $\delta \in (0,1)$. Repeatedly approximate the scaled residual to construct $x_n$ with $\Vert x_n\Vert <{\delta }^n$ and residual less than ${\delta }^{n+1}$. Completeness sums the series, and continuity gives $T\sum x_n=y$. Since $\Vert \sum x_n\Vert <(1-\delta {)}^{-1}$, scaling proves $B_Y(0,1-\delta )\subseteq T(B_X(0,1))$. Hence $T$ is a quotient operator, even without assuming completeness of $Y$.