Bounded operators on an Hilbert space

Question

Let $H$ be an hilbert space and consider the set $L(H,H)$ of bounded linear operator $T: X\rightarrow X$. Now we know that this is going to be a Banach space since $X$ itself is a Banach space. Now is this going to be an Hilbert space ? I have tried getting a counterexample to the parallelogram's law but I got nowhere. Also is there possible that there exists an element $z$ with $||z||\leq 1$ such that $||T(z)||=||T||$? Any tips are aprecciated , Thanks in advance.

Answer 1

1. For an example not satisfying the parallelogram law, almost any pair of $2 \times 2$ matrices will do. For an easy one, try $A=\pmatrix{1 & 0\cr 0 & 0\cr}$ and $B =\pmatrix{0 & 0\cr 0 & 1\cr}$ which have $\|A\|=\|B\| = \|A+B\|=\|A-B\|=1$.

2. Depending on the operator $T$, there may or may not be $x$ with $\|x\|=1$ and $\|Tx\|=\|T\|$. An example where there isn't is $H = L^2([0,1])$ with $Tf(x) = x f(x)$.