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Im doing the following excercise: Ok, so let $(e_n)$ be a orthonormal basis of $l^2$, and fix arbitrary complex numbers $(\lambda_n)$ and define $T:l^2\to l^2 $ as $$T(\sum x_ne_n)=\sum \lambda_nx_ne_n, $$ and let $$D(T) = \{\sum x_ne_n: \sum |\lambda_nx_n|^2 <\infty\}.$$

Clearly $T$ is densely defined since $e_n\in D(T)$ for all $n$. I want to determine the adjoint $T^*$. I thought as follows:

We find out how $T^*$ acts on the basis-vectors $e_j$. Note that $$(T(e_j), e_j) = \lambda_j.$$ Let $T^*e_j = z = \sum z_ne_n$, then using the orthonormality of the $e_n$ $$( e_j, T^*e_j) = \sum_n \overline{z_n}( e_j, e_n) = \overline{z_j}. $$ Also $(T(e_j),e_k) = 0$ if $k\neq j$ which is equivalent with $(e_j,T^*{e_k}) =0$ for $k\neq j$. It follows that for all $\sum x_ne_n \in D(T^*)$ that $$T^*(\sum x_ne_n) = \sum_n\overline{\lambda_n}x_ne_n.$$

Is this correct? Now here's my problem: As hint they say: Finite dimensional subspaces can be helpful for a precise argumentation that the domain of $T^*$ is what you think it is. I don't understand this, because it seems (assuming my $T^*$ is correct) that $D(T^*) = D(T)$. What am i missing?

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  • $\begingroup$ "We find out how $T^\ast$ acts on the basis-vectors $e_j$." If you have a picky grader, you should give an argument why $e_j \in D(T^\ast)$. "Finite dimensional subspaces can be helpful for a precise argumentation that the domain of $T^\ast$ is what you think it is" indicates that you should give a precise argument for $D(T^\ast) = D(T)$. $\endgroup$ – Daniel Fischer Jan 9 '14 at 15:12
  • $\begingroup$ But isn't $D(T)= D(T^*)$ automatic? Since clearly $\sum|\lambda_nx_n|^2 < \infty$ if and only if $\sum|\overline{\lambda_n}x_n|^2 < \infty$. Why does $e_j\in D(T^*)$ help here?. I'm little confused... $\endgroup$ – DinkyDoe Jan 9 '14 at 15:22
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You are taking the second step before the first when you write "We find out how $T^\ast$ acts on the basis-vectors $e_j$"; the first step would be to prove that $e_j \in D(T^\ast)$. That's easily remedied.

And then, you jump from the representation of $T^\ast$ on subspaces of the form $H_n = \operatorname{span} \{ e_j : 0 \leqslant j \leqslant n\}$ that you computed to the conclusion that $T^\ast$ is defined on the largest domain that $$T^\ast_0 = T^\ast\lvert_{\bigcup H_n}$$

can be (naturally) extended to. That needs a justification.

Remember the definition:

$$\begin{align} D(T^\ast) &:= \left\lbrace y\in l^2 : x\mapsto (T x,y) \text{ is continuous in the norm topology}\right\rbrace\\ &= \left\lbrace y \in l^2 : \bigl(\exists z\in l^2\bigr)\bigl(\forall x \in D(T)\bigr) \bigl( (Tx, y) = (x,z)\bigr)\right\rbrace. \end{align}$$

Now it is easy to see that $e_j \in D(T^\ast)$ for all $j$, since $x\mapsto (Tx,e_j) = x\mapsto \lambda_j\cdot x_j$ is continuous in the norm topology, and indeed $T^\ast(e_j) = \overline{\lambda_j}\cdot e_j$. And thus, since $D(T^\ast)$ is a linear subspace, we know that $H = \bigcup H_n$, the space of all finite linear combinations of the $e_j$ is contained in $D(T^\ast)$, and the restriction $T_0^\ast = T^\ast\lvert_H$ is given by

$$T_0^\ast\left(\sum_{k=0}^m y_{j_k}\cdot e_{j_k}\right) = \sum_{k=0}^m \overline{\lambda_{j_k}} y_{j_k}\cdot e_{j_k}.$$

$T_0^\ast$ has a natural extension to $D(T)$, and not beyond, but is it a priori unthinkable that $D(T^\ast)$ is a proper subspace of $D(T)$?

You need to show that indeed $D(T) \subset D(T^\ast)$, and $T^\ast$ is given by the obvious extension from $H$. And then, for completeness, you need to show that $T^\ast$ is not defined by an unobvious extension on a larger subspace. (Both points aren't hard to show, but need to be shown.)

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  • $\begingroup$ Thanks! That's an awesome, precise answer. Thank you kindly. $\endgroup$ – DinkyDoe Jan 9 '14 at 16:46
  • $\begingroup$ I've tried to prove this question here. However I haven't used any finite dimensional subspaces. $\endgroup$ – simon Jan 20 '14 at 17:37

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