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.



Optimal estimate in trace norm
$\newcommand{\R}{\mathbb{R}}
\newcommand{\ep}{\epsilon}$


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, $2\ep$. 


