2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $T$.

$\endgroup$
  • $\begingroup$ I can not understand your argument about disjoint supports for weakly null sequence $g_{k}$, but anyways, in fact, more simple fact of "almost orthogonality" are needed for your first estimate, which is obvious. Thank you very much for your help! $\endgroup$ – Byobe Feb 16 '17 at 10:06
  • $\begingroup$ @Rick_Student, this is the principle of small perturbations: books.google.co.uk/… $\endgroup$ – Tomek Kania Feb 16 '17 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.