# Almost orthonormal projection and orthonormal projection in Hilbert space

Let $$(e_i)_i$$ be a family of vectors in a Hilbert space being almost orthonormal but not quite, i.e.

$$\langle e_i, e_j \rangle \approx \delta_{i,j} + \alpha e^{-\vert i-j \vert}$$ and $$\alpha$$ is a small positive number.

It is tempting then to say that

$$Q:=\sum_{i=1}^{\infty} \langle \bullet, e_i \rangle e_i$$

is almost the orthogonal projection onto the closed span of the $$e_i$$ which we denote by $$P.$$

However, this is not quite right, as the basis is not fully orthonormal, so is there an estimate on the operator norm

$$\Vert Q-P \Vert ?$$

It seems that an error $$\alpha \lesssim \Vert Q-P \Vert$$ is unavoidable (just by testing both against one of the $$e_i$$.

But is there also a bound from above?

• Since $(v_i) \mapsto (\sum_j e^{-|i-j|} v_j)$ is bounded $\ell_2 \to \ell_2$, isn't it the case $\|Q - P\| \lesssim \alpha$? Jan 10, 2019 at 16:59
• Just to be clear, do you in fact intend the almost orthonormal condition to be $|\langle e_i, e_j\rangle - \delta_{ij} | \leq \alpha e^{-|i-j|}$, or do you mean something that is weaker? Jan 10, 2019 at 17:01
• @WillieWong no I mean what you write. And the boundedness you state is also correct. However I do not see how you conclude the bound you write. Would you mind turning this with a few more details into an answer? Jan 10, 2019 at 17:17
• @WillieWong but why do you hesitate to just state your derive your result in an answer if it is that obvious?-Maybe there is a nice trick we are missing? Jan 10, 2019 at 19:59
• @D.Driggs: quick sketch of my thoughts. The orthogonal complement to the span doesn't matter as it is in the kernel of both $Q$ and $P$. For small $\alpha$ your vectors $e_i$ are linearly independent, so if $v = \sum v_i e_i$ then the unique representation of $Pv$ is $\sum v_i e_i$. So $(Q-P)v \approx \sum \alpha e^{-|i-j|} v_i e_j$, Using boundedness you get $\|(Q-P)v\| \lesssim \alpha \|v\|$, noting that $\|(v_i)\|_{\ell_2} \approx \|v\|$. // To your question: Yemon's much better at this than I; if he hesitates, usually I made a mistake somewhere. Jan 10, 2019 at 21:56

Indeed, there is also a bound $$\|P-Q\| \lesssim \alpha$$.
There are certainly many other more involved and elegant proofs, but here is a very basic one. As noticed in the comments, we can assume that the $$e_i$$ span a dense subspace, in which case we have to prove that $$\|Q-Id\|\lesssim \alpha$$. In other words that for every $$x = \sum_i \lambda_i e_i$$ in the linear span, we have $$\|Qx-x\| \lesssim \alpha \|x\|$$.
The right-hand side is equal to the square root of $$\sum_{i,j} \lambda_i \overline{\lambda_j} \langle e_i,e_j\rangle$$, so by the assumption and Cauchy-Schwarz we get $$|\|x\|^2 - \sum_i |\lambda_i|^2 |\leq \alpha \sum_{k \in \mathbb Z} e^{-|k|} \sum_i |\lambda_i \lambda_{i+k}| \leq \alpha \sum_{k \in \mathbb Z} e^{-|k|} \sum_i |\lambda_i|^2.$$
So for $$\alpha$$ small enough we have $$\|x\| \lesssim (\sum_i |\lambda_i|^2)^{\frac 1 2} \lesssim \|x\|.$$
Similarly, $$Qx-x= \sum_{i,j} \lambda_i (\langle e_i,e_j\rangle e_j - \delta_{i,j}) e_j$$, so by the preceding inequality for $$Qx-x$$, $$\|Qx-x\| \simeq (\sum_{j} |\sum_i \lambda_i(\langle e_i,e_j\rangle - \delta_{i,j})|^2)^{\frac 1 2} \lesssim \alpha (\sum_{j,s,t} |\lambda_s \lambda_{t}| e^{-|s-j|-|t-j|})^{\frac 1 2}.$$ Bound $$\sum_j e^{-|s-j|-|t-j|} \lesssim e^{-|s-t|/2}$$. You obtain $$\|Q_x-x\|\lesssim \alpha (\sum_{s,t} |\lambda_s\lambda_t| e^{-|s-t|/2})^{\frac 1 2} \lesssim \alpha (\sum_i |\lambda_i|^2)^{\frac 1 2}.$$ This is $$\lesssim \alpha \|x\|$$ by the first computation.