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Definition of the problem

Let $\mathcal{H}$ be a separable Hilbert space and $J\subset\mathbb{N}$ an index set. Let $\Phi:=\left(\varphi_{j}\right)_{j\in J}\subset\mathcal{H}$ be a frame for $\mathcal{H}$, i.e. $\exists A,B>0\,\forall x\in\mathcal{H},\, A\left\Vert x\right\Vert ^{2}\leq\sum_{j\in J}\left|\left\langle x,\varphi_{j}\right\rangle \right|^{2}\leq B\left\Vert x\right\Vert ^{2}$.

Show that the closed linear span $cls\left\{ \varphi_{j}:j\in J\right\} :=\overline{span\left\{ \varphi_{j}:j\in J\right\} }$ coincides with $\mathcal{H}$. Furthermore, show that $\left(\left\Vert \varphi_{j}\right\Vert \right)_{j\in J}$ is always bounded.

My efforts

For the first part, I have tried to prove that the synthesis operator $T_{\Phi}^{\star}:\ell_{2}\left(J\right)\rightarrow\mathcal{H}$ is surjective. We know that the analysis operator $T_{\Phi}:\mathcal{H}\rightarrow\ell_{2}\left(J\right)$ of $\Phi$ is injective. Since the synthesis operator is the adjoint operator of the analysis operator, we have that $$ (\star)\ \forall x\in\mathcal{H},\, y\in\ell_{2}\left(J\right):\quad\left\langle T_{\Phi}x,y\right\rangle _{\ell_{2}\left(J\right)}=\left\langle x,T_{\Phi}^{\star}y\right\rangle _{\mathcal{H}}. $$

From this, I derive that $T_{\Phi}^{\star}$ has the form $T_{\Phi}^{\star}\left(c_{j}\right)=\sum_{j\in J}c_{j}\varphi_{j},\,\left(c_{j}\right)\in\ell_{2}\left(J\right)$.

Since $\left(\star\right)$ must hold, we have that $\left(\star\star\right)\ \ker T_{\Phi}=\left(ran\, T_{\Phi}^{\star}\right)^{\perp}.$ And since $T_{\Phi}$ is injective, we have that its kernel is the trivial kernel, i.e. $\ker T_{\Phi}=\left\{ 0\right\} $. From this point, I do not see how to use this to prove $\left(\star\star\right)$.

After having shown that the synthesis operator $T_{\Phi}^{\star}:\ell_{2}\left(J\right)\rightarrow\mathcal{H}$ is surjective, and has the form $T_{\Phi}^{\star}\left(c_{j}\right)=\sum_{j\in J}c_{j}\varphi_{j},\,\left(c_{j}\right)\in\ell_{2}\left(J\right)$, we know that all points in codomain $\mathcal{H}$ are covered by $T_{\Phi}^{\star}$. It seems to me that the synthesis operator has kind of the form of a linear combination of the frame-vectors of $\Phi$.

My question

How can I complete the proof of the surjectivity of the synthesis operator, and how can I properly interpert that to show that the closed linear span of $\Phi$ coincide with $\mathcal{H}$.

Thank you, Franck!

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1 Answer

up vote 1 down vote accepted

If a closed subspace $V\subset \mathcal H$ does not coincide with $\mathcal H$, then there exists a nonzero vector $x$ that is orthogonal to $V$. What happens with the lower frame bound when you plug in such $x$?

To show that $\|\varphi_j\|$ are uniformly bounded, plug $x=\varphi_j$ into the upper frame bound.

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Thanks for the hint! $x\in\mathcal{H}$ such that $x\in\left(span\left\{ \varphi_{j}:j\in J\right\} \right)^{\perp} \Rightarrow\forall j\in J:\,\left\langle x,\varphi_{j}\right\rangle =0$. Thus, the lower frame-bound must be zero. Hence, $\left\{ \varphi_{j}:j\in J\right\} $ cannot be a frame. Here is the contradiction. For the uniformely boundedness: I plug $x=\varphi_j$ in, and I obtain $$\sum_{j\in J}\left|\left\langle \varphi_{n},\varphi_{j}\right\rangle \right|^{2}\leq B\left\Vert \varphi_{n}\right\Vert ^{2}.$$ I am stuck again. Do we know anything about $<\varphi_n,\varphi_j>$? –  FranckStudiesCommEng Jul 2 '12 at 22:21
    
@FranckStudiesCommEng We know something when $n=j$. –  user31373 Jul 2 '12 at 22:24
    
Since $<\cdot,\cdot>$ is an inner-product, it is positive-definite, hence $<\varphi_n,\varphi_n>\geq 0$. Besides that, I tried using Cauchy-Schwarz, but it did not help. What are you precisely thinking about? –  FranckStudiesCommEng Jul 2 '12 at 23:00
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All terms of the sum are nonnegative. Ignoring all of them except the one with $j=n$, we get $|\langle \varphi_n,\varphi_n\rangle|^2\le B\|\varphi_n\|^2$. Do you know the relation of the norm to the inner product of a vector with itself? –  user31373 Jul 2 '12 at 23:21
    
$$\left\Vert \varphi_{n}\right\Vert ^{4}=\left|\left\langle \varphi_{n},\varphi_{n}\right\rangle \right|^{2}\leq\sum_{j\in J}\left|\left\langle \varphi_{n},\varphi_{j}\right\rangle \right|^{2}\leq B\left\Vert \varphi_{n}\right\Vert ^{2} $$ $$ \Rightarrow\left\Vert \varphi_{n}\right\Vert ^{2}\leq B. $$ Thank you again for your great help, I really appreciate that you do not give me directly the answer but that you lead me on the way to the answer! Best regards, Franck! –  FranckStudiesCommEng Jul 2 '12 at 23:28
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