# Is Gram-Schmidt on a separable Hilbert space operator norm continuous?

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 $$f_i=\frac{Ae_i-\sum_{j Define a new operator $$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.

• If $A=0$, then how is $GS(A)$ defined? Is $GS(0)$ the identity operator? – Skeeve Feb 28 '19 at 8:08
• Does "Linear embedding" mean injective bounded linear operator with closed range"? – Pietro Majer Feb 28 '19 at 8:15
• @Pietro: I did not want to include "closed range" in the definition of embedding. However, it follows from Nik's answer that I better should have: Put $Ae_i=e_{2i}$. Then the linear homotopy joining $A$ and the identity runs through injective bounded linear operators with non-closed range. If there were a continuous retraction as in my question, then $A$ and the identity would be homotopic in $\mathcal I$. However, $ind(A)=\infty$ whereas $ind(id)=0$. – Benedikt Hunger Feb 28 '19 at 10:26
• @Skeeve: $0$ is not an embedding. – Benedikt Hunger Feb 28 '19 at 10:26

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.
• You may be right, but once you know it for unitaries it follows easily for isometries. Given two isometries $V_1$, $V_2$ whose ranges have codimension $n$, there is a unitary $U$ such that $V_2 = UV_1$. A path from $U$ to $I$ in the unitaries then yields a path from $V_1$ to $V_2$ in the isometries with codimension $n$. – Nik Weaver Feb 28 '19 at 14:02
• Look, if ${\rm codim}({\rm ran}(V_1)) > {\rm codim}({\rm ran}(V_2))$ then there must be a unit vector in ${\rm ran}(V_2)$ that is orthogonal to ${\rm ran}(V_1)$. So $\|V_1-V_2\| \geq \sqrt{2}$. – Nik Weaver Feb 28 '19 at 15:50
• Yes, since the linear homotopy $\gamma$ joining the isometry $A\colon e_i\mapsto e_{2i}$ to the identity is injective (and bounded) on every stage. But since $A$ has index $-\infty$ and $id$ has index $0$, they cannot be connected in $\mathcal I$. Now if $\phi\colon\mathcal E\to\mathcal I$ were a retraction then $\phi\circ\gamma$ would connect $A$ and $id$ in $\mathcal I$. – Benedikt Hunger Feb 28 '19 at 18:21