a question about contractions on Hilbert spaces

Question

Let $\cal{H}$ be a Hilbert space, $T_1,T_2\in\cal{B(H)}$,

1. $\|T_1(h_1)+T_2(h_2)\|^2\leq\|h_1\|^2+\|h_2\|^2$ for all $h_1,h_2\in\cal{H}$.

2. $T_1T^\ast_1+T_2T^\ast_2\leq I$.

Then can we verify that 1 holds if and only if 2 holds ?

Answer 1

Consider the operator $T:\mathcal H\oplus \mathcal H\to\mathcal H$ defined by $$T(h_1\oplus h_2):= T_1(h_1)+T_2(h_2)\, . $$

Then (1) says exactly that $\Vert T\Vert\leq 1$. Since $\Vert T\Vert=\Vert T^*\Vert$, this is equivalent to the condition $\Vert T^*\Vert\leq 1$, which is again equivalent to $$TT^*\leq I\, .$$

Now, you will easily verify that $T^*:\mathcal H\to \mathcal H\oplus \mathcal H$ is given by the formula $$T^*(u)=T_1^*(u)\oplus T_2^*(u)\, . $$

So we have $T(T^*(u))=T_1T_1^*(u)+T_2T_2^*(u)$, i.e. $$TT^*=T_1T_1^*+T_2T_2^*\, .$$

Hence, (1) is equivalent to (2).

Answer 2

As Paul remarked above- this is not true. Take $h_1$ and $h_2$ to be two orthogonal vectors in $\mathcal{H}$, and $T_1=T_2=I$. Then, your condition 1) holds, but condition 2) doesn't.