showing projections are extreme point of a set in Hilbert space Show that the set $E(H)=\left\{T \in B(H):T \geq0, ||T|| \leq 1\right\}$ for a Hilbert space H, is convex set.
prove that any projection in $B(H)$ is an extreme point of this set.

A point $x$ of a convex set $A$ is an extreme point of $A$ if $$x=ty+(1-t)z, t\in[0,1],z,y\in A$$ then $x=y$ or $x=z$

i have done proving convex part. now if $P$ is projection and $$P=tT_1+(1-t)T_2$$. then since $||P||=1$ this gives $||tT_1+(1-t)T_2||=t||T_1||+(1-t)||T_2||=1$
from this how  can i some show that $t=0$ or $t=1$
any suggestion or any other approach .thanks in advanced.
 A: Let $p=t\,A+(1-t)\,B$, $t\notin\{0,1\}$. Since $A,B$ are positive let $a,b$ be their unique positive roots. Let $x\in \ker(p)$, you have 
$$0=\langle px,x\rangle=t\langle Ax,x\rangle +(1-t)\langle Bx,x\rangle = t \|ax\|^2+(1-t)\|bx\|^2$$
it follows $ax=bx=0$ and thus $A$ and $B$ are both zero on $\ker(p)$. The kernels of $A,B$ cannot be larger than this, for suppose $x\not\in\ker(p)$ and $Ax=0$, by subtracting an element of $\ker(p)$ we can assume $x$ to be in the subspace $p$ projects on. Then
$$\|x\|^2=\|px\|^2 = (1-t)^2\|Bx\|^2≤(1-t)^2\,\|B\|^2\|x\|^2≤(1-t)^2\|x\|^2$$
and $t=0$ must follow.
So $\ker(p)=\ker(A)=\ker(B)$. From self-adjointness follows that their images must be $\mathrm{im}(p)=\ker(p)^\perp$. So by looking at $H'=\mathrm{im}(p)$ we can assume $p=\mathbb1$.
Now we do a spectral theory argument to show that $\sigma(A)=\{1\}$ and thus $A=\mathbb1$. In the finite dimensional case it works like this: Let $x$ be an eigenvector of $A$ to eigenvalue $\lambda<1$. Then
$$x=t\,\lambda x+(1-t)Bx\implies Bx=\frac{1-\lambda t}{1-t}x$$
but $\frac{1-\lambda t}{1-t}>1$ from $\lambda<1$, contradicting $\|B\|≤1$. Now in the infinite dimensional case lets look at a resolution of identity $P_\lambda$ for $A$. If we assume that $A$ has spectrum not equal to $\{1\}$ then we must have some $\lambda<1$ so that $P_{\lambda+\epsilon}-P_\lambda\neq0$ for any $\epsilon$. For $x$ in this corresponding "eigenspace" we have $\|Ax-\lambda x\|≤\epsilon\,\|x\|$, and thus
$$x=t\,\lambda x+ t\,\tilde x+(1-t)Bx\implies Bx=\frac{1-\lambda\,t}{1-t}x-\frac{t}{1-t}\tilde x$$
here $\tilde x$ has norm $≤\epsilon \|x\|$. By taking norms we find $$\|Bx\|≥\frac{1-\lambda t}{1-t}\|x\|-\frac{t}{1-t}\epsilon\|x\|$$
and for some $\epsilon$ this will be greater than $\|x\|$, contradicting $\|B\|≤1$.
