Are all bounded operators a compact operator?

Question

Let $H$ be a Hilbert space. I was trying for a bounded operator which is not compact operator.

My attempt:- I got this example from a textbook "Hilbert Space Operators in Quantum Physics" by Jiří Blank, Pavel Exner, Miloslav Havlíček. Why does unit operator infinite dimensional subspace of $H$ is bounded but not a compact operator.

I was confused whether the author mean for identity operator or a unitary operator. We can see that either identity operator or unitary operator is bounded. How do I prove that it is not compact?

Answer 1

Take the identity operator on the unit sphere in

$l_2\iff (\forall x)\left(x=\{x_n\}_{n=1}^{\infty}, x_i\in R\;, \forall i, \; ||x||^2=\sum_{i=1}^\infty{x_i^2}<\infty\right)$

now this operator is bounded on the unit sphere.

now take the sequence

$x_n=\delta_{in}$ i.e the sequence of zeros with 1 at the nth location.

It is clear that $\{x_n\}\subseteq S$ yet it is not Cauchy.

$||x_n-x_m||=\sqrt{2}, n\neq m$

so any subsequence you take will not converge.

the unit sphere is not compact.

and so the identity operator is not compact.