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Define the operator $T: l^2 \rightarrow l^2$ on the canonical orthonormal system $(e_k)_k$ by: $$ Te_k := \frac{e_k}{k} + \frac{e_{k+1}}{k+1}, $$ such that for $a\in l^2$: $$ T((a_i)_i) = (a_1, (a_1+a_2)/2, (a_2+a_3)/3, ...). $$ I have to show that the spectrum contains 0 and the eigenvalues 1/k for $k\in N$. I don't see why the 1/k's are eigenvalues: $$ Tx = T\sum_{i=1}^{\infty}x_ie_i = \sum_{i=1}^{\infty}x_i(e_i/i + e_{i+1}/(i+1)) = \lambda x $$ yields $\forall i$ $$ \lambda e_i = e_i/i + e_{i+1}/(i+1). $$ I don't understand how I should continue here.

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Welcome to Math.SE! You should have equated the coefficients of $e_i$, not the $i$th terms in the sum. To do so, use the index shift: $\sum_{i=1}^\infty x_i e_{i+1}/(i+1)= \sum_{i=2}^\infty x_{i-1} e_{i}/i $. – user53153 Jan 13 '13 at 15:50
Thanks for the welcome Pavel! If I use your hint I find $x_{k-1} = (\lambda k -1 )x_k$. I still don't see how I can deduce the possible values for lambda from this? – Funzies Jan 13 '13 at 16:24
up vote 1 down vote accepted

The problem statement helps us by giving a formula for $T((a_i)_i)$ (not just the images of basis vectors), so let's use it. If $T((a_i)_i)=\lambda (a_i)_i$, then $$a_1=\lambda a_1, \frac{a_1+a_2}{2}=\lambda a_2, \frac{a_2+a_3}{3}=\lambda a_3, \frac{a_3+a_4}{4}=\lambda a_4, \dots$$ The first equation looks the simplest, so begin with it.

If $a_1\ne 0$ then $\lambda=1$. The second equation now says that $a_2=a_1$. The third says that $a_3=a_2/2$, and so on: $a_k=a_{k-1}/(k-1)$. This is your eigenvector for $\lambda=1$.

If $a_1=0$, then the second equation becomes $a_2/2 = \lambda a_2$. Now if $a_2\ne 0$, then $\lambda = 1/2$ and the rest goes similarly to the above. If $a_2=0$, then we look at the next equation...

To organize writing, begin as follows: "suppose $T((a_i)_i)=\lambda (a_i)_i$. Let $k$ be the smallest integer such that $a_k\ne 0$..." and show that $\lambda=1/k$.

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Thank you for this detailed answer 5PM, it is very much appreciated! Maybe you could help me clarify one last thing. I've been given that zero is an element of the spectrum of T. The question remains to which part of the spectrum it belongs. If the closure of Im(-T) is $l^2$, it is a part of the continuous spectrum of T and if not it is part of the residual spectrum of T. The definition of the closure would be in this case (I think): $\forall x \in l^2, \epsilon>0 \thereis y \in Im(-T): d(x,y) < \epsilon$. I have some problem proving/disproving this. – Funzies Jan 14 '13 at 9:00
@Erik Hint: the closure of the range of $T$ is the orthogonal complement of the kernel of $T^*$. – user53153 Jan 14 '13 at 12:27
I have calculated: $T^*e_k = e_k/k + e_{k-1}/{k-1}$. Calculating the kernel of $-T^*$ yields $\sum_i x_i(e_i/i + e_{i-1}/(i-1))=0$ which obviously means that $ker(-T^*) = {0}$. That means that the orthogonal complement is the whole space, thus the closure of Im($-T^*)$ is indeed the whole space. That would make zero an element of the continuous spectrum of T. Is this reasoning correct? – Funzies Jan 14 '13 at 20:04
@Erik Looks good to me. – user53153 Jan 14 '13 at 20:20

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