In $X=\ell^p$, $p\in[1,\infty]$ we consider: $$ T(x_1,x_2,x_3,\ldots)=(0,x_1,0,x_3,\ldots) $$ Prove that $T$ isn't a compact operator and that $T^2$ is a compact operator. I think I solved the second point of the exercise because $T^2 x= 0$ always considering $T^2=T(T(x))$ (but I'm not sure this is $T^2$...). Anyway I'd like to find a sequence $\{x_n\}$ such that $x_n\rightharpoonup x$ but $Tx_n\nrightarrow Tx$.
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2$\begingroup$ Second part's ok. For the first, you're not taking the correct approach. You need to find a bounded sequence $(x_n)$ such that $(Tx_n)$ has no convergent subsequence. Think of the action of $T$ on the unit vectors. $\endgroup$– David MitraFeb 4, 2014 at 10:47
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$\begingroup$ The approach of finding a weakly convergent series of $\ell^p$-series which doesn't converge in norm only works for reflexive banach spaces - but your range for $p$ is to large to allow that. In particular: What does $\ell^0$ mean in this context? $\endgroup$– RolandFeb 4, 2014 at 10:56
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$\begingroup$ Sorry, I wrote wrong! $\endgroup$– rusca91Feb 4, 2014 at 10:57
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$\begingroup$ That is my idea but I can't think about a single sequence $x_n$ which can make a proper counterexample, even thinking about unit vectors. $\endgroup$– rusca91Feb 4, 2014 at 11:33
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1$\begingroup$ Take the sequence of "odd" unit vectors $(e_1,e_3,\ldots)$. They are mapped by $T$ to the sequence of "even" unit vectors $(e_2,e_4,\ldots)$. The former sequence is bounded in norm, the latter sequence has no subsequence converging in norm. (Here, $e_i$ is the sequence of reals that has value $1$ in the $i$th coordinate and value $0$ elsewhere.) $\endgroup$– David MitraFeb 4, 2014 at 11:38
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4 Answers
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Consider the subspace $U=\{(x_1,0,x_3,0,\dots)\}\subset\ell^p$. The restriction $T_{|_U}$ is the right shift operator on $U$. But we know that this is not a compact operator (e.g. looking at the spectrum).
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Take $\boldsymbol{x}_n=\boldsymbol{e}_{2n+1},\,n\in\mathbb N$ - where $\boldsymbol{e}_{n}$is the element with all zeros except at position $n$, where it has an $1$. Then $\boldsymbol{x}_{n}$ is a bounded sequence in $\ell^p$, as $\|\boldsymbol{x}_{n}\|_p=1$, and $$T\boldsymbol{x}_n=T\boldsymbol{e}_{2n+1}=\boldsymbol{e}_{2n+2}$$ does not have a converging subsequence, since $$ \|\boldsymbol{e}_{m}-\boldsymbol{e}_{n}\|=2^{1/p}, $$ wherever $m\ne n$.
answered Feb 4, 2014 at 11:39
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Take the action of $T$ two distinct unit vectors $e_i$ (by vector I mean infinite dimensional) which is obviously bounded. Then use the equivalent formulation of a compact operator in terms of the properties of $(Tx_n)$.
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Define the operator : $$Tx= \sum_{n=1}^\infty \left(\sum_{k=0}^\infty a_{nk} x_n \right) e_n \qquad\text{with}\qquad a_{nk}=\frac{1}{n+2k}.$$ Show that T is compact. That is if $x^j \in l^2$ is a bounded sequence. Then there exists a subsequence such that $Tx^{kj}$ converges.
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$\begingroup$ I don't know how begin because the sum of $a_{nk}$ diverge... $\endgroup$ Dec 1, 2019 at 14:10