# Analysis operator $T_\Phi$ is injective and has a closed range

### 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!

-
1) What is your defenition of frame 2) In the first part youwas trying to prove that $T_\Phi$ is bounded. –  Norbert Jul 1 '12 at 17:19
@Norbert The definition of a frame in a Hilbert space is reasonably established by now. –  user31373 Jul 1 '12 at 17:47
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.
Thanks, I still have couple of questions. Are $T_{\Phi}x:=\left(\left\langle x,\varphi_{j}\right\rangle \right)_{j\in J}$ and $T_{\Phi}x:=\sum_{j\in J}\left\langle x,\varphi_{j}\right\rangle e_{j}$ equivalent? How could I show that equality? Furthermore, how do we know that such an orthonormal basis exists? For the proof of the $T_{\Phi}$ having a closed range, I obtain from $T_{\Phi}x\rightarrow0$ that $\left|\sum_{j\in J}\left\langle x,\varphi_{j}\right\rangle e_{j}\right|\leq\sum_{j\in J}\left|\left\langle x,\varphi_{j}\right\rangle \right|$, but how can I reach the lower bound? –  FranckStudiesCommEng Jul 1 '12 at 23:33
Every Hilbert space has an orthonormal basis. The space $\ell_2(J)$ has a standard orthonormal basis: namely, sequences $e_k$ in which the $k$th element is $1$ and all others are $0$. Yes, these are two ways to write down the same element of $\ell_2(J)$. You should read something about Hilbert spaces in any basic textbook on functional analysis. –  user31373 Jul 1 '12 at 23:57
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. –  FranckStudiesCommEng Jul 2 '12 at 16:18