Almost orthonormal projection and orthonormal projection in Hilbert space

Question

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?

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.