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.

  • $\begingroup$ generally, bounded operator means bounded linear operator. For finite dimensional Hilbert spaces, all linear operators are bounded. $\endgroup$ – Ben Grossmann Dec 10 '14 at 16:42
  • $\begingroup$ but not the other way round. right? so, the theorem only holds for "linear" cases. would that be a correct assumption? $\endgroup$ – Arnab Dec 10 '14 at 16:44
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    $\begingroup$ In finite dimensions, every (linear) operator is of finite rank. Hence, the property you stated is trivial. Is that really what you intended? $\endgroup$ – PhoemueX Dec 10 '14 at 17:52
  • $\begingroup$ 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? $\endgroup$ – Arnab Dec 10 '14 at 19:26
  • $\begingroup$ Contexts in which the word "operator" is used to refer to maps that may not be linear are rare. The word is almost always used exclusively to refer to linearity, in situations where it is not so care is often taken to point out that here one is looking at a different meaning of "operator". What is your context? Which maps may be "operators"? $\endgroup$ – s.harp Aug 8 '19 at 22:49

That's a comment but it got a bit too long.

As far as I know, classically, the approximation property deals with bounded linear operators.

But let us assume your point for a minute.

If it is not "bounded linear" then what is you definition of "bounded" operator? E.g., 1) a bounded function, that is a function with bounded range. But a bounded linear operator is not a bounded function unless it is zero, so it'd be inconsistent with the standard definition of AP; or maybe 2) a function, sending bounded sets to bounded sets.

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.

However, it is true that people consider some versions of the AP for non-linear operators. See, e.g., Question 4 in this article by Godefroy and Ozawa: http://www.ams.org/journals/proc/2014-142-05/S0002-9939-2014-11933-2/home.html


I'd wager that the context in which the questioner found this issue is about linear operators, to begin. So, bounded linear operators.

Then, _in_what_sense_ do we want to, or claim to be able to, approximate bounded linear operators by finite-rank (meaning that the image has finite dimension) (linear!!!) operators? One obvious sense is "in operator norm". Ok, but then the claim is not true, except in finite-dimensional Hilbert spaces. In infinite-dimensional Hilbert spaces, approximability by finite-rank operators is "compact"/"completely-continuous". Not every operator is compact, e.g., the identity operator on infinite-dimensional Hilbert spaces.

(In finite-dimensional spaces, _of_course_ everything is approximable by finite-rank operators, because the whole space is finite-dimensional... Maybe you're not asking what you intend...)


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