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?

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    $\begingroup$ Since $(v_i) \mapsto (\sum_j e^{-|i-j|} v_j)$ is bounded $\ell_2 \to \ell_2$, isn't it the case $\|Q - P\| \lesssim \alpha$? $\endgroup$ Jan 10, 2019 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? $\endgroup$ Jan 10, 2019 at 17:01
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    $\begingroup$ @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, 2019 at 17:17
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    $\begingroup$ @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, 2019 at 19:59
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    $\begingroup$ @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. $\endgroup$ Jan 10, 2019 at 21:56

1 Answer 1


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.

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    $\begingroup$ I had not read carefullly the comments, but this is exactly Willie Wong's argument. $\endgroup$ Jan 11, 2019 at 22:06

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