Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $l_2^{+}$ be the Hilbert space of all square summable sequences $\{x_n\}, n \in \mathbb{N}$ under some definition of inner-product $\langle,\rangle_l$. Define $B[l_2^{+}]$ as the set of all bounded linear transformation from $l_2^{+}$ to $l_2^{+}$. Finally, let $A[l_2^{+}]$ be the set of all bounded transformations from $l_2^{+}$ to $l_2^{+}$. Note that $B[l_2^{+}] \subset A[l_2^{+}]$ and that $A[l_2^{+}]$ contains transformations that are not necessarily linear (the superposition property does not hold).

Define the inner product on $A[l_2^{+}]$:

$$\langle T_1,T_2\rangle = \sup_{x \in l_2^{+} - \{0\}} \frac{\langle T_1x,T_2x \rangle_l}{\langle x,x \rangle_l}\text{, for all }T_1, T_2 \in A[l_2^{+}]$$

Question: is $B[l_2^{+}]$ dense on $A[l_2^{+}]$?

share|cite|improve this question
A hlibert space is still a vector space...what do you mean by a operator which is not linear? – s.b Sep 14 '12 at 19:23
This is not the core of the question. Anyways, a nonlinear operator is an operator which does not have the superposition property. – user38397 Sep 14 '12 at 19:57
Your "inner product" is not an inner product. – Robert Israel Sep 14 '12 at 20:03
Presumably you are asking whether for any bounded nonlinear operator $T$ and $\epsilon > 0$ there is a linear operator $S$ such that $\langle T-S, T-S \rangle \le \epsilon$, i.e. that $\|(T - S) x \|^2 \le \epsilon \|x\|^2$ for all $x$. That is clearly not the case: e.g. you can find $T$ where $T x = T(-x) = x$ for some $x \ne 0$. – Robert Israel Sep 14 '12 at 20:13
@Robert I think what you wrote is actually an answer. I'll add a comment for the OP: it's a good idea to test conjectures about Hilbert spaces on finite-dimensional examples. For example, can we approximate $f(x)=|x|$ by linear functions (in one dimension)? Of course not. – user31373 Sep 15 '12 at 19:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.