Let $T:H\to H$ be defined by $Tx=\sum_{n=1}^\infty \lambda_n \langle x,\varphi_n \rangle \varphi_n$ where $\{\varphi_n\}_{n=1}^\infty$ is an orthogonal sequence and $\{\lambda_n\}_{n=1}^\infty$ is a sequence of numbers.

If I write $T^2(x)=(T\circ T)(x)$ as $\sum_{n=1}^\infty \mu_n \langle x,\varphi_n \rangle \varphi_n$ and $(a_mT^m+a_{m-1}T^{m-1}+\dotsm+a_1T^1)$ as $\sum_{n=1}^\infty \mu_n \langle x,\varphi_n \rangle \varphi_n$ can I say $\{\mu_n\}_n$ is a bounded sequence? How can I show it for this case?


$$T^2x=\sum_n\lambda_n\langle Tx,\phi_n\rangle\phi_n=\sum_n\lambda_n\left\langle \sum_m \lambda_m\langle x,\phi_m\rangle\phi_m,\phi_n\right\rangle \phi_n\\ =\sum_n\lambda_n\sum_m\lambda_m\langle x,\phi_m\rangle\langle\phi_m,\phi_n\rangle \phi_n=\sum_n\lambda_n\sum_m\lambda_m\langle x,\phi_m\rangle||\phi_n||^2\delta_{m,n} \phi_n\\ =\sum_n\lambda_n\lambda_n\langle x,\phi_n\rangle||\phi_n||^2\phi_n$$

So if you write $T^2(x)$ as $$\sum_n \mu_n\langle x,\phi_n \rangle\phi_n$$ we have $\mu_n=||\phi_n||^2\lambda_n^2$. And similarly $\mu_n$ for your $2$nd assumption can be calculated. In case $\{\phi_n\}$ is an orthonormal sequence for the $2$nd case we have $$ \mu_n=\sum_j a_j\lambda_n^{j} $$

  • $\begingroup$ I lost you at $\sum_n\lambda_n\sum_m\lambda_m\langle x,\phi_m\rangle\langle\phi_m,\phi_n\rangle \phi_n=\sum_n\lambda_n\sum_m\lambda_m\langle x,\phi_m\rangle||\phi_n||^2\delta_{m,n} \phi_n\\$. Can you kindly expand on how you got to the norm? also what is $\delta_{m,n}$ and how does $\sum_m\lambda_m\langle x,\phi_m\rangle \delta_{m,n} $ converges to $\lambda_n\langle x,\phi_n \rangle$? $\endgroup$ – havakok Jun 6 '16 at 19:39
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    $\begingroup$ $\langle x,x\rangle=||x||^2$ by definition of norm and $\delta_{m,n}$ is defined as $0$ when $m\ne n$ and $1$ when $m=n$. We get to $\delta_{m,n}$ using the fact that $\{\phi_{n}\}$ is orthogonal i.e. for $m\neq n $ we have $\langle\phi_m,\phi_n\rangle=0$. If you don't want $\delta$ you can just skip to the next line it comes straight from the line before $\delta$ by means of orthogonality. $\endgroup$ – Jack's wasted life Jun 6 '16 at 19:42
  • $\begingroup$ I can see that $\mu_n$ is bounded by $sup(\| \phi_n\|^2\lambda_n \lambda_n)$ for the first case. How can i show whether or not $\sum_j a_j \lambda_n^{j-1}$ is bounded? Note that I have edited the definition of $T^2x$ from $\sum_n \mu_n \lambda_n \langle x,\phi_n \rangle\phi_n$ to $\sum_n \mu_n\langle x,\phi_n \rangle\phi_n$ $\endgroup$ – havakok Jun 10 '16 at 18:10
  • $\begingroup$ If $|\lambda_n|<M$ we have $|\mu_n|<\sum_{j=1}^m|a_j|M^j$. The RHS is a finite sum and therefore bounded. $\endgroup$ – Jack's wasted life Jun 10 '16 at 18:16
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    $\begingroup$ But the sum in RHS is over $j$ not over $n$ and $j$ ranges from $1$ to $m$ not from $1$ to $\infty$. $\endgroup$ – Jack's wasted life Jun 10 '16 at 18:24

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