Prove that $d(T_1,T_2)=\|T_1-T_2\|$ is a metric on $V$

Question

Let $V$ be the vector space of all bounded linear operators on the Hilbert space $H$. Note that the operator norm is written like $\| \cdot \|$.

Prove that $$d(T_1,T_2)=\|T_1-T_2\|$$ is a metric on $V$.

1) if $d(T_1,T_2)=0 \Rightarrow \|T_1-T_2\|=0$ then it must $T_1-T_2=0 \Rightarrow T_1=T_2$

Anyone can help me with this problem... Thanks in advance for your time.

Answer 1

That $\|T_1-T_2\|=\|T_2-T_1\|$ holds, should be clear. Hence $d(T_1,T_2)=d(T_2,T_1)$.

If we denote the norm on $H$ by $\|\cdot\|_H$, then we get for $A,B \in V$ and $x \in H$:

$\|(A+B)x\|_H=\|Ax+Bx\|_H \le (\|A\|+\|B\|)\|x\|_H$, thus

$\|A+B\| \le \|A\|+\|B\|$.

Use this result to show that

$\|T_1-T_2\|=\|T_1-T_3+T_3-T_2\| \le \|T_1-T_3\|+\|T_3-T_2\|$.

Can you proceed?