# Finite rank approximation of bounded operators on Hilbert space

Let H be a (finite dimensional) Hilbert space. The approximation property states that every bounded operator from H to itself can be approximated by a sequence of finite rank operators.

My question is - does the above statement implicitly assume LINEAR operators? Or, "bounded" alone suffices?

I understand things might play out differently between finite and infinite dimensional cases. I'm particularly interested in operators over finite dimensional spaces.

• generally, bounded operator means bounded linear operator. For finite dimensional Hilbert spaces, all linear operators are bounded. – Omnomnomnom Dec 10 '14 at 16:42
• but not the other way round. right? so, the theorem only holds for "linear" cases. would that be a correct assumption? – Arnab Dec 10 '14 at 16:44
• In finite dimensions, every (linear) operator is of finite rank. Hence, the property you stated is trivial. Is that really what you intended? – PhoemueX Dec 10 '14 at 17:52
• Not really! The approximation property, as stated in many places "seems" to indicate that the "only" requirement is that the operator is bounded (no reference to linearity). In that case, I wonder if this holds for non-linear cases as well - i.e., if I have a bounded non-linear operator, does that admit a similar approximation? or the "linearity" requirement is implicit? – Arnab Dec 10 '14 at 19:26

If 2), you may want to approximate it by linear finite-rank operators. But even in the 1-dimensional case you cannot do it (e.g., try to approximate $x^2$). Alternatively, you may want to approximate it by (non-linear) functions with finite-dimensional range, which is, of course, of no interest in a finite-dimensional space.