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I need help with this problem:

Let $\{u_{n}\}$ and $\{v_{n}\}$, $(n \in \mathbb{N})$ be two different Hilbert bases in a Hilbert space $H$. Define a linear operator $T$ so that, for $x \in H$: $$T(x) = \sum_{n=1}^{\infty} \lambda(n)u_{n}⟨v_{n} \mid x⟩$$ where $\lambda(n)$ is a fixed sequence of complex numbers with $|\lambda(n)| < 1, \forall n$ . Show that $T$ is bounded, and find its adjoint operator.

So to figure out that it is bounded I guess I could compute the norm somehow?

And to find the adjoint I use the definition that for $x,y \in H$ then:

$$\langle T(x) \mid y\rangle = \langle x \mid T^{*}(y) \rangle$$

where $T^*$ is the adjoint of $T$.

I am quite new to functional analysis and would be grateful is someone have a solution for this.

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  • $\begingroup$ Hint: Assuming $\{ u_{n}\}$ and $\{ v_{n} \}$ is orthonormal basis the boundedness of $T$ is clear. $<Tx,Tx> = \sum |\lambda(n)|^{2}|<v_{n},x>|^{2} \leq \sum |<v_{n},x>|^{2} = ||x||^{2}$. Thus norm of $T < 1$; $\endgroup$ – DBS Jul 26 '13 at 8:03
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We have for $x \in H$ $\def\norm#1{\left\|#1\right\|}\def\sp#1{\left<#1\right>}$ \begin{align*} \norm{Tx}^2 &= \norm{\sum_{n=1}^\infty \lambda_n u_n\sp{v_n, x}}^2\\ &= \sp{\sum_{n=1}^\infty\lambda_n\sp{v_n, x}u_n, \sum_{m=1}^\infty \lambda_m \sp{v_m, x}u_m}\\ &= \sum_{n=1}^\infty\sum_{m=1}^\infty \overline{\lambda_n \sp{v_n, x}}\lambda_m \sp{v_m, x} \underbrace{\sp{u_n, u_m}}_{\delta_{nm}=}\\ &= \sum_{n=1}^\infty |\lambda_n\sp{v_n, x}|^2\\ &\le \sup_n |\lambda_n|\sum_{n=1}^\infty|\sp{v_n, x}|^2\\ &\le \norm x^2 \end{align*} So $T$ is bounded with $\norm T \le 1$, we moreover see that $\norm T = \sup_n |\lambda_n|$.

To compute the adjoint, let $x,y\in H$, we have \begin{align*} \sp{T^*x, y} &= \sp{x, Ty}\\ &= \sp{x, \sum_{n=1}^\infty \lambda_n \sp{v_n, y}u_n}\\ &= \sum_{n=1}^\infty \lambda_n \sp{v_n, y}\sp{x, u_n}\\ &= \sp{\sum_{n=1}^\infty \overline{\lambda_n\sp{x,u_n}}v_n, y}\\ &= \sp{\sum_{n=1}^\infty \overline{\lambda_n}\sp{u_n, x}v_n, y} \end{align*} So for each $x \in H$: $$ T^* x = \sum_{n=1}^\infty \overline{\lambda_n}\sp{u_n, x}v_n. $$

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Assuming that both these basis are orthonormal, we get

$$||T(x)||^{2} \leq \sum|<v_{n},x>|^{2} = ||x||^{2}$$

where the last inequality is due to Parseval's formula. Hence the operator is bounded.

For the adjoint, calculate $<Tx,y>$ and surmise what $T^{*}y$ should be to get the same inner product. Then prove that what you found is actually a linear operator and is the adjoint.

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