Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Some one help me to solve this problem i put some hints Show that every projection $P \in \mathcal{B}(H)$ is and extreme point in the convex set
$$B_+ = \{T \in \mathcal{B}(H) : T \geq 0, \Vert T\Vert \leq 1\}$$ Hints: If $P = \lambda S + ( 1 – \lambda) T$ with $S$ and $T$ in $B_+$ , then $x = Sx = Tx$ for every $x \in P(H)$. Furthermore $\langle Ty,y\rangle = 0$ for every $y \in P(H)^\perp$ and because of the positivity and the Cauchy Schwarz inequality this implies that $Sy = Ty = 0$.

share|cite|improve this question
up vote 1 down vote accepted

Since $S,T\in B_+$, then $\Vert S\Vert\leq 1$ and $\Vert T\Vert\leq 1$. Hence $$ \Vert Sx\Vert\leq \Vert x\Vert\qquad \Vert Tx\Vert\leq \Vert x\Vert$$ for all $x\in H$. Assume there exist $x_0\in P(H)$ such that $Sx_0\neq Tx_0$, then $$ |\langle Sx_0,Tx_0\rangle|<\Vert Sx_0\Vert\Vert Tx_0\Vert\leq \Vert x\Vert^2 $$ Since $x_0\in P(H)$, then $x_0=Px_0$ so $$ \begin{align} \Vert x_0\Vert^2&=\Vert Px_0\Vert^2=\Vert\lambda Sx_0+(1-\lambda) Tx_0\Vert^2\\ &=\lambda^2\Vert Sx_0\Vert^2+2\lambda(1-\lambda)\mathrm{Re}\langle Sx_0,Tx_0\rangle+(1-\lambda)^2\Vert Tx_0\Vert^2\\ &\leq\lambda^2\Vert x_0\Vert^2+2\lambda(1-\lambda)|\langle Sx_0,Tx_0\rangle|+(1-\lambda)^2\Vert x_0\Vert^2\\ &<\lambda^2\Vert x_0\Vert^2+2\lambda(1-\lambda)\Vert x_0\Vert^2+(1-\lambda)^2\Vert x_0\Vert^2\\ &=\Vert x_0\Vert^2 \end{align} $$ Thus $\Vert x_0\Vert^2<\Vert x_0\Vert^2$, contradiction. Therefore, for all $x\in P(H)$ we have $$ Sx=Tx\\ x=Px=\lambda Sx+(1-\lambda)Tx=\lambda Sx+(1-\lambda)Sx=Sx=Tx $$ Since $T\in B_+$ we have $T\geq 0$ and in particular for all $y\in P(H)^\perp$ holds $\langle Ty,y\rangle\geq 0$. Assume there exist $y_0\in P(H)^\perp$ such that $\langle Ty_0,y_0\rangle>0$. Since $y_0\in P(H)^\perp$ we have $$ 0=Py_0=\lambda Sy_0+(1-\lambda)T y_0 $$ hence $S y_0=-\lambda^{-1}(1-\lambda)t y_0$. Since $\lambda\in(0,1)$ then $$ \langle S y_0,y_0\rangle=-\lambda^{-1}(1-\lambda)\langle Ty_0, y_0\rangle<0 $$ Hence $S\not{\geq}0$ and $S\notin B_+$. Contradiction, so $$ \langle Ty,y\rangle=0 $$ for all $y\in P(H)^\perp$. Let $y_1,y_2\in P(H)^\perp$, then $$ \langle T(y_1+y_2),y_1+y_2\rangle=0\\ \langle T(y_1+iy_2),y_1+iy_2\rangle=0 $$ Since $T\in B_+$, we have $T=T^*$. Using this fact previous two equations can be simplified to $$ \langle Ty_1,y_1\rangle+2\mathrm{Re}\langle Ty_1,y_2\rangle+\langle Ty_2,y_2\rangle=0\\ \langle Ty_1,y_1\rangle+2\mathrm{Im}\langle Ty_1,y_2\rangle+\langle Ty_2,y_2\rangle=0 $$ Since $y_1,y_2\in P(H)^\perp$, then $$ \langle Ty_1,y_1\rangle=\langle Ty_2,y_2\rangle=0 $$ so we get $\mathrm{Re}\langle Ty_1,y_2\rangle=\mathrm{Im}\langle Ty_1,y_2\rangle=0$ which is equivalent to $$ \langle Ty_1,y_2\rangle=0 $$ Since $y_1\in P(H)^\perp$, then for all $x\in P(H)$ we have $\langle y_1,x\rangle=0$. Since $T=T^*$ we get $\langle Ty_1,x\rangle=\langle y_1,T^*x\rangle=\langle y_1,Tx\rangle=\langle y_1,x\rangle=0$ for all $x\in P(H)$. This means that $Ty_1\in P(H)^\perp$. Hence we may consider case $y_2=Ty_1$, then $$ \Vert Ty_1\Vert^2=\langle Ty_1,Ty_1\rangle=\langle Ty_1,y_2\rangle=0 $$ so $Ty_1=0$. Thus for all $y_1\in P(H)^\perp$ we have $Ty_1=0$.

Now take arbitrary $z\in H$, and consider representation $z=Pz+(1_H-P)z$. Since $Pz\in P(H)$ and $(1_H-P)z\in P(H)^\perp$ we have $$ Tz=T(Pz)+T((1_H-P)z)=T(Pz)=Pz $$ Since $z\in H$ is arbitrary $T=P$. Similar argument shows that $S=P$. Since $S=T=P$ we see that $P$ is an extreme point of $B_+$.

share|cite|improve this answer
ok could u please little bit expand this things with cauchy inequality and poditivity – math Nov 16 '12 at 17:05
could u please elaborate your hints so i can put it in beautiful proof – math Nov 16 '12 at 17:14
@motu See edits to my answer – Norbert Nov 16 '12 at 21:59
Thank you very much – math Nov 16 '12 at 22:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.