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?

New contributor
D. Driggs is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • 1
    $\begingroup$ Since $(v_i) \mapsto (\sum_j e^{-|i-j|} v_j)$ is bounded $\ell_2 \to \ell_2$, isn't it the case $\|Q - P\| \lesssim \alpha$? $\endgroup$ – Willie Wong Jan 10 at 16:59
  • $\begingroup$ 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? $\endgroup$ – Willie Wong Jan 10 at 17:01
  • 1
    $\begingroup$ @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? $\endgroup$ – D. Driggs Jan 10 at 17:17
  • 1
    $\begingroup$ @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? $\endgroup$ – D. Driggs Jan 10 at 19:59
  • 2
    $\begingroup$ @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. $\endgroup$ – Willie Wong Jan 10 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.

  • 1
    $\begingroup$ I had not read carefullly the comments, but this is exactly Willie Wong's argument. $\endgroup$ – Mikael de la Salle Jan 11 at 22:06

Your Answer

D. Driggs is a new contributor. Be nice, and check out our Code of Conduct.

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.