$X,Y$ Banach spaces, $F: X \to Y$. Then the operator-equation "$Fx = y$" is well-posed, if for all $y \in Y$

(1) there exists $x \in X: Fx = y$

(2) the solution is unique

(3) the solution continuously depends on y

Q: What exactly is meant by (3)? Is it just continuity in $x$?


1 Answer 1


(3) means that the function $F^{-1}$ is continuous. In other words, fix a solution $Fx = y$. Then, for all $\varepsilon > 0$, there exists $\delta > 0$ such that, for all $y'$ with $||y-y'||<\delta$ (and writing $Fx' = y'$), we have $||x-x'||<\varepsilon$.


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