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Let $C\subset\mathbb C$ be closed. As $\mathbb C$ is separable then so too is the subset $C$. This means that there exists a countable subset $\{\lambda_n:n\in\mathbb N\}\subset C$ dense in $C$.

In the first comment to the question here the claim is made that the operator $A:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$, the action of which is $A(e_n)=\lambda_ne_n$, is bounded. Here, $\{e_n:n\in\mathbb N\}\subset\ell^2(\mathbb N)$ is an orthonormal basis. This is what I want to show and it should be incredibly easy to demonstrate - I am overlooking something terribly obvious here, to my embarrassment.

Here is my working so far. Recall that in any Hilbert space, $\mathcal H$, with an orthonormal basis, any $x\in\mathcal H$ can be uniquely expressed in the form, $$x=\sum_{n=1}^\infty\langle x,e_n\rangle e_n.$$ Now, consider the quantity, $$\begin{align*} \|Ax\|&=\left\|A\left(\sum_{n=1}^\infty\langle x,e_n\rangle e_n\right)\right\|\\ &=\left\|\sum_{n=1}^\infty\langle x,e_n\rangle Ae_n\right\| &&\text{(assuming A linear)}\\ &=\left\|\sum_{n=1}^\infty\langle x,e_n\rangle \lambda_ne_n\right\|\\ &\le\sum_{n=1}^\infty\|\langle x,e_n\rangle \lambda_ne_n\| &&\text{(triangle inequality)}\\ &=\sum_{n=1}^\infty|\langle x,e_n\rangle| |\lambda_n|\|e_n\|\\ &\le\|x\|\sum_{n=1}^\infty |\lambda_n|\|e_n\|^2 &&\text{(Cauchy-Schwarz).}\\ \end{align*}$$ What I thought here to do was to bound $|\lambda_n|$ by $M:=\max_{n\in\mathbb N}\{\lambda_n\}$ so as to obtain, $$\|x\|\sum_{n=1}^\infty |\lambda_n|\|e_n\|^2\le M\|x\|\sum_{n=1}^\infty \|e_n\|^2.$$ But as the norm of each $e_n\in\ell^2(\mathbb N)$ is unity the series we are left with diverges.

As to trying to show that the operator is bounded, what am I overlooking or failing to do correctly?

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  • $\begingroup$ Take $C=\mathbb N$, which is closed. $\mathbb N$ is a countable dense subset of $C=\mathbb N$, while the operator $Ae_n = n \cdot e_n$ is clearly unbounded. $\endgroup$ – lisyarus Jan 22 '19 at 17:01
  • $\begingroup$ I get the idea, but $\overline{\mathbb N}\neq\mathbb C$? $\endgroup$ – Jeremy Jeffrey James Jan 23 '19 at 11:12
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It's not enough to have $C$ closed; you need it bounded, too. That is, you want $C$ compact. If the sequence $\{\lambda_n\}$ is not bounded, your $A$ will not be bounded.

When dealing with series, the triangle inequality is very unlikely to give you sharp inequalities; you've gone too far. What you want here is Parseval's identity. Write $c=\sup_n|\lambda_n|$. Note also that you can only evaluate on finite sums, since a priori you don't know that $A$ is bounded. You have $$ \left\|\sum_{n=1}^M\lambda\langle x,e_n\rangle\,e_n\right\|^2=\sum_{n=1}^M|\lambda_n|^2\,|\langle x,e_n\rangle|^2\leq c^2\,\sum_{n=1}^M \,|\langle x,e_n\rangle|^2\leq c^2\,\|x\|^2. $$ As you can do this for all $M$ and all $x$, you get that $A$ is bounded on a dense subset of $H$, and thus extends to a bounded operator on all of $H$.

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  • $\begingroup$ Taking $A$ inside the sum is assuming that $A$ is bounded, hence one needs to consider finite sums first. Otherwise the reasoning is circular. $\endgroup$ – Klaus Jan 22 '19 at 17:02
  • $\begingroup$ Taking $A$ inside the sum is just down to $A$ being linear, no? $\endgroup$ – Jeremy Jeffrey James Jan 22 '19 at 17:03
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    $\begingroup$ @JeremyJeffreyJames Not if the sum is infinite, i.e. a series. $\endgroup$ – lisyarus Jan 22 '19 at 17:04
  • $\begingroup$ @Klaus: good point. The estimate is the same, though. $\endgroup$ – Martin Argerami Jan 22 '19 at 17:59
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    $\begingroup$ They are the span of an orthonormal basis. $\endgroup$ – Martin Argerami Jan 28 '19 at 11:49
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You want to assume $C$ is compact, not just closed. Then if $C$ is bounded by $M$, for any $x = (x_1, x_2, \ldots) \in \ell^2$, $A x = (\lambda_1 x_1, \lambda_2 x_2, \ldots)$ with $$ \|A x \|^2 = \sum_{n=1}^\infty |\lambda_n|^2 |x_n|^2 \le M^2 \sum_{n=1}^\infty |x_n|^2 = M^2 \|x\|^2 $$

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  • $\begingroup$ I see that the notation we use must coincide with what you have here, namely that if $x = (x_1, x_2, \ldots) \in \ell^2$ then $A x = (\lambda_1 x_1, \lambda_2 x_2, \ldots)$. In the questions I linked above, the given operator is considered with the problem of finding its spectrum in mind. However, I don't see how if $A x = (\lambda_1 x_1, \lambda_2 x_2, \ldots)$, then we can deduce that the whole sequence $(\lambda_n)_{n=1}^\infty$ is in $\sigma(A)$? $\endgroup$ – Jeremy Jeffrey James Jan 25 '19 at 16:35
  • $\begingroup$ Because this does not mirror the usual eigenvalue problem of finding $\lambda\in\mathbb C$ for which $Ax=\lambda x=(\lambda x_1, \lambda x_2, \lambda x_3 ...)$ $\endgroup$ – Jeremy Jeffrey James Jan 25 '19 at 16:36
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    $\begingroup$ $\lambda_n \in \sigma(A)$ because $A e_n = \lambda_n e_n$. $\endgroup$ – Robert Israel Jan 25 '19 at 17:57
  • $\begingroup$ I see, I wasn't thinking correctly. $\endgroup$ – Jeremy Jeffrey James Jan 25 '19 at 18:59

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