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Let $H_1, H_2$ be complex Hilbert Spaces. Suppose that $\{e_n\}_{n=1}^\infty$ is an orthonormal basis for $H_1$, that $\{f_n\}_{n=1}^\infty$ is an orthonormal basis for $H_2$ and that $\{\lambda_n\}_{n=1}^\infty$ is a bounded sequence in C. For $x \in H_1$ define $$Tx=\sum_{n=1}^\infty \lambda_n (x,e_n)f_n$$

a) Show that the series right hand side of the above equation converges in $H_2$.

b) Show $T$ is bouneded.

c) Compute $||T||$.

d) Give an explicit expression for $T^*:H_2 \to H_1$

I have really no idea how to start this.. can anyone help?

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  • $\begingroup$ What properties of orthonormal bases do you know? Anything connected to the name "Bessel", perhaps? $\endgroup$ – PhoemueX Dec 29 '16 at 15:53
  • $\begingroup$ Yeah I know the Bessel Inequality! :) $\endgroup$ – Yuhe Dec 29 '16 at 15:55
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a) It is enough to show that $lim_n \|\sum_{i\geq n}\lambda_i(x,e_i)f_i\|=0$. Write $A=sup_n|\lambda_n|$. We have $\|\sum_{i\geq n}(x,e_i)f_i\|\leq \sum_{i\geq n}|\lambda_i| |(x,e_i)|\leq A\sum_{i\geq n}|(x,e_i)|$. The result follows from the fact that $lim_n\sum_{i\geq n}|(x,e_i)|=0$, since $x=\sum_i(x,e_n)e_n$ thus the sequence $\sum_{i=1}^n(x,e_n)e_n$ converges towards $x$ and thus is a Cauchy sequence.

b) $\|T(x)\|\leq A\sum_{i\geq i}|(x,e_n)|=A\|x\|$ since $(e_i)$ is orthonormal.

c) The proof of $b$ shows that the norm of $T$ is bounded $A$. Let's shows that it is $A$.For every $c>0$, there exists $n$ such that $|A|-|\lambda_n|<c$. We have $T(e_n)=\lambda_nf_n$ and $\|T(e_n)\|=|\lambda_n|$. This implies that $Sup_{\|x\|=1}\|T(x)\|=A$.

d) $(T(e_n),f_m)=(e_n,T^*(f_m))=\lambda_n(f_n,f_m)$. This implies that if $n\neq m$, $(e_n,T^*(f_m))=0$ and if $n=m$, $(e_n,T^*(f_n))=\lambda_n$. We deduce that $T^*(f_n)=\bar\lambda_ne_n$.

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  • $\begingroup$ Thank you a lot! What is happening with the $f_i$ in part a)? $\endgroup$ – Yuhe Dec 29 '16 at 16:07
  • $\begingroup$ $(f_i)$ is an orthonormal base, so $\|\sum_{i\geq n}(x,e_i)f_i\|=\sum_{i\geq n}|(x,e_i)|$. $\endgroup$ – Tsemo Aristide Dec 29 '16 at 16:10

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