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Definition of the problem

Let $\mathcal{H}$ be a separable Hilbert space on $J\subset\mathbb{N}$ an index set. Let $\Phi:=\left(\varphi_{j}\right)_{j\in J}\subset\mathcal{H}$ be a frame for $\mathcal{H}$.

I have to prove that the analysis operator $T_{\Phi}:\mathcal{H}\rightarrow\ell_{2}\left(J\right)$ of the frame $\Phi$, defined by $ T_{\Phi}x:=\left(\left\langle x,\varphi_{j}\right\rangle \right)_{j\in J},\quad x\in\mathcal{H}, $ is injective and has a closed range.

Effort to prove closed range

We need to show that $ ran\, T_{\Phi}closed\Leftrightarrow\forall x_{n}\in\mathcal{H}:\,\lim_{n\rightarrow\infty}T_{\Phi}x_{n}\in\ell_{2}\left(J\right). $

Let $x_{n}\in\mathcal{H}$. We have that $ \lim_{n\rightarrow\infty}T_{\Phi}x_{n}=\lim_{n\rightarrow\infty}\left(\left\langle x_{n},\varphi_{j}\right\rangle \right)_{j\in J}\overset{?}{\in}\ell_{2}\left(J\right). $

For this statement to hold, we have a look at $ \sum_{j\in J}\left|\lim_{n\rightarrow\infty}\left\langle x_{n},\varphi_{j}\right\rangle \right|^{2}. $

Using the fact that $\Phi$is a frame for $\mathcal{H}$, $ \sum_{j\in J}\left|\lim_{n\rightarrow\infty}\left\langle x_{n},\varphi_{j}\right\rangle \right|^{2}\overset{???}{=}\sum_{j\in J}\lim_{n\rightarrow\infty}\left|\left\langle x_{n},\varphi_{j}\right\rangle \right|^{2}=\lim_{n\rightarrow\infty}\sum_{j\in J}\left|\left\langle x_{n},\varphi_{j}\right\rangle \right|^{2}\leq\lim_{n\rightarrow\infty}B\left\Vert x_{n}\right\Vert ^{2}. $

My question 1

How could I use that to show that $T_{\Phi}$has a closed range? Would an upper bound help me at all with this?

Effort to show that $T_{\Phi}$is injective

Denote $\left\{ e_{i}:i\in I\right\} $be an orthonormal basis. Let $x,y\in\mathcal{H}$. Assume $T_{\Phi}x=T_{\Phi}y$. We have that $ \sum_{i\in J}\left\langle x,\varphi_{i}\right\rangle e_{i}=\sum_{i\in J}\left\langle y,\varphi_{i}\right\rangle e_{i} $ $ \Leftrightarrow\forall i\in J:\quad\left\langle x,\varphi_{i}\right\rangle =\left\langle y,\varphi_{i}\right\rangle . $

My question 2

Am I allowed to make the assumption on the orthornormal basis? How can I show that such an orthonormal basis exists? How could I go any further showing that it is injective? We do not know if the inner product $\left\langle \cdot,\cdot\right\rangle $ is one-to-one?!

Thank you, Franck!

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    @Norbert The [definition](http://en.wikipedia.org/wiki/Frame_of_a_vector_space) of a frame in a Hilbert space is reasonably established by now.2012-07-01

1 Answers 1

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2) Since $T_\Phi$ is a linear operator, its injectivity amounts to having the trivial kernel $\{0\}$. Use the lower frame bound.

1) To show the range is closed, it suffices prove that $Tx_n\to 0$ implies $x_n\to 0$. Use the lower frame bound for this too.

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    From $T_\Phi x \rightarrow 0$, \epsilon>\left\Vert \sum_{j\in J}\left\langle x,\varphi_{j}\right\rangle e_{j}\right\Vert \overset{{\scriptstyle orthogonality}}{=}\sum_{j\in J}\left\Vert \left\langle x,\varphi_{j}\right\rangle e_{j}\right\Vert =\sum_{j\in J}\left|\left\langle x,\varphi_{j}\right\rangle \right|\left\Vert e_{j}\right\Vert =\sum_{j\in J}\left|\left\langle x,\varphi_{j}\right\rangle \right|. And using the lower frame bound, we show that \epsilon^2 > A\left\Vert x\right\Vert ^{2}. I was missing the orthogonality step.2012-07-02