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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.

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

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, $2\ep$.

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