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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 $Sy = Ty = 0$.

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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 $B_+$.

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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 –  no identity Nov 16 '12 at 21:59
Thank you very much –  math Nov 16 '12 at 22:38

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