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In this thread Show compactness/ noncompactness of an operator by approximation I came to the conclusion that the operator $$ T\colon\ell^2\to\ell^2, (a_n)_{n\in\mathbb{N}}\mapsto\left(\frac{a_n+a_{n+1}}{2}\right)_{n\in\mathbb{N}} $$ is not compact. Now my task is to show whether this operator is (i) continuous (or not) and (ii) selfadjoint (or not).

(i) Concerning continuity I wrote down $$ \lVert Tx\rVert_{\ell^2}^2=\frac{1}{4}\sum\limits_{n\geq 1}\lvert x_n+x_{n+1}\rvert^2\leq\sum\limits_{n\geq 1}\left(\lvert x_n\rvert+\lvert x_{n+1}\rvert\right)^2\leq\sum\limits_{n\geq 1}(2\lvert x_n\rvert)^2=4\lVert x\rVert_{\ell^2}^2, $$ so $$ \lVert Tx\rVert_{\ell^2}\leq 2\lVert x\rVert_{\ell^2} $$ and so $T$ is continuous, because it is linear and bounded.

(ii) But I do not know exactly how to show if $T$ is selfadjoint. I try to put it more precisely: $\ell^2$ is a Hilbert space with scalar product $$ \langle (x_n),(y_n)\rangle=\sum\limits_{n\geq 1}x_n\overline{y_n}. $$ Therefore I started with $$ \langle Tx,y\rangle=\sum\limits_{n\geq 1}\frac{x_n+x_{n+1}}{2}\overline{y_n}. $$ Now I have to work with this in order to get an expression with is $\langle x,T^*y\rangle$. But I do not know how to continue.


Could you please say me if (i) is correct and help me with (ii)?

With regards

math12

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  • $\begingroup$ You could take the infinite matrix of $T$ with respect to the canonical orthonormal basis. This will show you immediately that $T$ is not self-adjoint. $\endgroup$ – Julien Feb 17 '13 at 14:12
  • $\begingroup$ I am always bad in finding examples or counterexamples, therefore I determined the adjoint operator explicitly. To my calculation, the adjoint operator is here given by $T^*\colon\ell^2\to\ell^2, (a_n)_{n\in\mathbb{N}}\mapsto\left(\frac{a_n+a_{n-1}}{2}\right)$ for $n\geq 2$ and so it is another operator and different from $T$ itself. Therefore $T\neq T^*$ and so $T$ is not selfadjoint. $\endgroup$ – math12 Feb 17 '13 at 14:35
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The operator is not self-adjoint. Consider $x=(1,2,0,\dots,0,\dots)$ and $y=(3,4,0,\dots,0,\dots)$. It is linear and bounded, and you argued correctly.

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  • $\begingroup$ But I would like to determine the adjoint operator nevertheless. And my problem is to do that. $\endgroup$ – math12 Feb 17 '13 at 14:04
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The expression for the adjoint is easy. Notice that $Tx = \frac12 (\operatorname{id}+R)x$, where $R$ is the right shift operator and $\operatorname{id}$ is the identity map on $\ell^2$. It is easy to check that the adjoint of $R$ is the left shift $L$ (that pads with a zero). Hence the adjoint of $T$ is $T^* = \frac12 (\operatorname{id}+R)^* = \frac12(\operatorname{id}^* + R^*) = \frac12(\operatorname{id} + L).$

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This is not an answer for the question but I want to comment on part (i). It seems to me that your estimate $$\lVert Tx\rVert_{\ell^2}\leqslant 2\lVert x\rVert_{\ell^2}$$is too rough. My calculation shows that $$\lVert Tx\rVert_{\ell^2}\leqslant \lVert x\rVert_{\ell^2}$$and it turns out that $\|T\|=1$ since $\|T\| \geqslant \sqrt{1-\frac 3{4n}}$ for any $n$ if you choose the point $(1,\dots,1,0,\dots)$ to test $\|T\|$.

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  • $\begingroup$ Can you show your calculation for $\| Tx\|_{\ell^2}\le \|x\|_{\ell^2}$, please? $\endgroup$ – Giuseppe Negro May 8 '16 at 10:03
  • $\begingroup$ I hope this works: $$\lVert Tx\rVert_{\ell^2}^2=\frac{1}{4}\sum\limits_{n\geq 1}\lvert x_n+x_{n+1}\rvert^2 \leq \frac 12\sum\limits_{n\geq 1}\left(\lvert x_n\rvert ^2 + \lvert x_{n+1}\rvert ^2 \right) \leq\sum\limits_{n\geq 1}\lvert x_n\rvert^2=\lVert x\rVert_{\ell^2}^2.$$ $\endgroup$ – QA Ngô May 8 '16 at 10:06
  • $\begingroup$ @QANgô: I agree with your computation. (Of course this does not answer the question! ) $\endgroup$ – Giuseppe Negro May 8 '16 at 10:08

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