Let $x,y$ be vectors of some Hilbert space of unit length.

Then we can consider the projection $P_x:=\langle \bullet, x \rangle x$ and similarly $P_y.$

Assume then that we know that $\left\lVert x-y \right\rVert\le \alpha.$

What is the optimal estimate on $\left\lVert P_x-P_y \right\rVert_1$ we can get, where $1$ denote the trace norm.

I am very grateful for any comment/remark on this.

A very pedestrian estimate would be to smuggle in the map $P_{xy}=\langle \bullet,x \rangle y.$ This way we can use the triangle inequality to estimate

$\left\lVert P_x-P_y \right\rVert_1 \le 2 \left\lVert x-y \right\rVert$

which does not seem to be very sophisticated. Is there a better estimate?-I cannot see when my trivial estimate is optimal.

  • 2
    $\begingroup$ Hint: $P_x-P_y$ is self-adjoit, has rank two and trace $0$, so $\|P_x - P_y\|_1 = \sqrt{2} \|P_x-P_y\|_2$. $\endgroup$ – Mikael de la Salle Sep 17 '18 at 9:15
  • $\begingroup$ so $\sqrt{2} \sqrt{\sum_{i,j} \left\lvert \langle x,e_i \rangle \langle x,f_j \rangle- \langle y,e_i \rangle \langle y,f_j \rangle \right\rvert^2}...$ for orthonormal bases $(e_i)$ and $(f_j)$ but how can I estimate this using the norm $\left\lVert x-y \right\rVert$?- Sorry, maybe I fail to see something obvious here. $\endgroup$ – Xing Wang Sep 17 '18 at 9:30
  • $\begingroup$ @MikaeldelaSalle sorry, do you think you could elaborate, please? $\endgroup$ – Xing Wang Sep 17 '18 at 11:33

$\newcommand{\R}{\mathbb{R}} \newcommand{\ep}{\epsilon}$

Without loss of generality (wlog), the space is the two dimensional space spanned by $x,y$. So, wlog the space is $\R^2$, with $x=[1\ 0]^T$ and $y=[a\ b]^T$ for some real $a,b$ such that $a^2+b^2=1$. Then $$\ep:=\|x-y\|=\sqrt{2-2a}$$ and $a=1-\ep^2/2$. Also, identifying $P_x,P_y$ with their matrices in the standard basis of $\R^2$, we have $(P_x-P_y)^2=(1-a^2)I$, where $I$ is the identity matrix. So, $$\|P_x-P_y\|_1=\text{trace}\sqrt{(P_x-P_y)^2}=2\sqrt{1-a^2}=2\ep\sqrt{1-\ep^2/4}.$$ For small $\ep$, this exact expression of the trace norm of $P_x-P_y$ is close to your bound, $2\ep$.


Your Answer

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.