Let $\mathcal H$ be a separable Hilbert space, with inner product $\langle\cdot,\cdot\rangle$, and with orthonormal basis $(e\_i)\_{i\in\mathbb N}$. Consider a continuous linear embedding $A\colon\mathcal H\to\mathcal H$. Then one can apply the Gram-Schmidt process to the (linearly independent) vectors $Ae\_i$. That is, we define recursively
$$
f\_i=\frac{Ae\_i-\sum\_{j<i}\langle Ae\_i,f\_j\rangle f\_j}{\|Ae\_i-\sum\_{j<i}\langle Ae\_i,f\_j\rangle\|}.
$$
Define a new operator $GS(A)$ by $Ae\_i=f\_i$. Since the $f\_i$ are an orthonormal family of vectors, $GS(A)$ will be an isometric embedding $\mathcal H\to\mathcal H$.


Let $\mathcal E$ be the space of all continuous linear embeddings $\mathcal H\to\mathcal H$, and $\mathcal I$ the space of all isometric embeddings $\mathcal H\to\mathcal H$. Then the above construction defines a map $GS\colon\mathcal E\to\mathcal I$.


Is this map $GS$ continuous with respect to the operator norm topologies on both domain and target? If the answer is no, what is the largest subspace of $\mathcal E$ such that the restricted map $GS$ is continuous? (obviously $GS|\_{\mathcal I}$ is the identity) Also if the answer is no: Is there a continuous alternative, i.e. a continuous retraction $\mathcal E\to\mathcal I$?


If it were, then the in particular the $f\_i$ would depend continuously on $A$ *uniformly in $i$*, and I do not even see if this is true.



Is Gram-Schmidt on a separable Hilbert space operator norm continuous?
The answer to the main question is no. Working on $l^2$, let $A$ be the operator $A: e\_n \mapsto \frac{1}{n}e\_n$ and for each $i$ let $A\_i$ be $A$ followed by the unitary $U\_i$ that switches $e\_i$ and $e\_{i+1}$ and fixes the other standard basis vectors. Then $A\_i \to A$ in norm but $(U\_i)$ does not converge in norm.


To the second question, there is no "largest" subspace on which the map is continuous, however for each $N$ its restriction to the set of operators for which $A^{-1}: {\rm ran}(A) \to H$ is bounded, with norm at most $N$, is continuous.


To the third question, I'm pretty sure there is a continuous retraction, but this is infinite dimensional topology and I wouldn't know where to find this result. You can treat each Fredholm index separately since the isometries with index $n$, for $n=0,1,\ldots,\infty$, are the connected components of the set of all isometries.


