I`m having a homework question that goes like this: X is a Hilbert space, a complete inner product space, show that B(X) is not a Hilbert space.
I`m quite stuck and I would love to understand this problem!
My attempt so far:
Let $T \in B(X)$ be a linear operator, $\exists M > 0$ such that $||Tx|| \leq M||x||$, $\forall x \in X$. ( $T: X \rightarrow X$)
$T_n$ is a Cauchy sequence in $B(X)$ such that $T_n \rightarrow T$.
I think I have to show that $B(X)$ is not complete under the norm $||x||=\sqrt{(x,x)}$, so show that $B(X)$ is open. I can also show that the parallelogram law isnt satisfied, but I dont see how.
Proof: $\forall \epsilon > 0 $, $\exists N \in \mathbb{N}$ such that $||T_n-T|| < \epsilon $. Also we have $||T_nx-Tx|| \leq ||T_n-T||||x|| < \epsilon||x||$ $\forall, x \in X$.
I dont know how to get further, please help!