Compact operators in H is the only closed non-trivial ideal.

Question

Let $H$ denote a separable Hilbert space. There is a well-known fact, that compact operators $K(H)$ forms unique proper and closed ideal in $B(H)$ - (space of bounded operator equipped with operator norm). Is there any elementary proof which is not using any spectral theorem explicitly?

Answer 1

As all separable, infinite-dimensional Hilbert spaces are isomorphic, without loss of generality we may suppose that $H=\ell_2$.

Suppose that $T\in B(H)$ is not compact. This means that there exists a sequence $(f_n)_{n=1}^\infty$ in the unit ball of $H$ such that $\inf_n \|Tf_n\|>0$. As $H$ is reflexive, we may pass to a weakly convergent subsequence $(f_{n_k})_{k=1}^\infty$. Let $f$ be the weak limit and set $g_k = f_{n_k} - f$ ($k\in \mathbb{N}$); certainly this is a weakly null sequence and $\|g_k\|\leqslant 2$. As $(g_k)_{k=1}^\infty$ fails to converge to $0$ in norm, without loss of generality $\inf_k \|Tg_k\|>0$. As finitely supported vectors in $\ell_2$ are dense in $\ell_2$, we may assume that $g_k$ and $Tg_k$ ($k\in \mathbb{N}$) are finitely supported and $(g_k)_{k=1}^\infty$ is still weakly null and $\inf_k \|Tg_k\|>0$.

Now comes the key observation. We may pass to a further subsequence convergent of $(g_k)_{k=1}^\infty$ so that the supports of $g_k$ are pairwise disjoint. Otherwise, there is $n_0$ such that for infinitely many $k$, $g_k(n_0)=c\neq 0$ for some $c$. This is not possible as $(g_k)_{k=1}^\infty$ is weakly null, so $ \langle g_k, e_{n_0}\rangle \to 0$. Applying this reasoning once again we may infer that the vectors $Tg_k$ are disjointly supported too.

In this case, for any finitely supported sequence $(a_k)_{k=1}^\infty$ of scalars we have

$$\|\sum_{k=1}^\infty a_k Tg_k\|^2 = \sum_{k=1}^\infty \|a_k Tg_k\|^2\geqslant \delta^2 \sum_{k=1}^\infty|a_k|^2,$$ where $\delta = \inf_k\|Tg_k\|>0$. Moreover

$$ \sum_{k=1}^\infty \|a_k Tg_k\|^2\leqslant 2^2\|T^2\| \sum_{k=1}^\infty|a_k|^2.$$

This means that the image of $T$ contains a subspace isomorphic to the $\ell_2$; namely $N=\overline{{\rm span}}\{Tg_k\colon k\in \mathbb{N}\}$. Thus, $T$ restricted to $M=\overline{{\rm span}}\{g_k\colon k\in \mathbb{N}\}$ is an isomorphism. Let $P_M$ be the orthogonal projection onto $M$. Let $S$ be any bounded extension to $H$ of the inverse of $T|_M$. Also, let $U\colon H\to M$ be any isomorphism and let $V$ be any bounded extenstion to $H$ of $U^{-1}$. Then $$I_H = VSTP_MU$$ is in the ideal generated by $T$.