I'm reading a book about Time-Frequency Analysis and I have a question regarding the property weather or not a set is a frame for a Hilbert space:

Let $H$ be a Hilbert space and $E= (e_j)_{j \in J} \subset H$ a subset of elements in $H$ ($J$ countable). We define the associated frame operator $S$ via

$$ S:H \to H, \ \ Sf=\sum_{j \in J} \langle f,e_j\rangle e_j $$

Now assume that $S$ is a bounded operator then it is well known that $S$ is a positive operator. My question is: If we further assume that $S$ is invertible, can we conclude that an estimate of the form

$$ A\|f\|^2 \le \langle Sf,f \rangle \leq B \|f\|^2 $$

holds for $0 < A \leq B ?$


If we assume that $S$ is a bounded (this is true, for instance, if $(e_j)_{j\in J}$ is a Bessel sequence) and invertible linear operator on $H$ , with $B:=\|S\|$, then by the inverse mapping theorem $S^{-1}$ is also bounded with $A^{-1}:=\|S^{-1}\|$, and we have $$A\|f\|\leq \|Sf\|\leq B\|f\|,\qquad B^{-1}\|f\|\leq \|S^{-1}f\|\leq A^{-1}\|f\|,\qquad \forall f\in H $$ with $A\leq B$.

Thus $|\left \langle Sf,f\right\rangle|\leq \|S\|\|f\|^2\leq B\|f\|^2$ for all $f$, and we have obtained the upper bound in your estimate.

For the lower bound, use the generalized Cauchy-Schwarz inequality (see e.g. here), valid for any positive self-adjoint operator $T$ on $H$: $$|\left \langle Tu,v\right\rangle|^2\leq \left\langle Tu,u\right\rangle \left \langle Tv,v\right\rangle,\qquad u,v\in H $$ Let $T=S$ (since $S$ is positive), $u=S^{-1}f$, $v=f$. Then the above inequality becomes $$\|f\|^4\leq \left \langle f,S^{-1}f\right\rangle \left\langle Sf,f\right\rangle \leq A^{-1}\|f\|^2\left\langle Sf,f\right\rangle$$ and hence $$A\|f\|^2\leq \left\langle Sf,f\right\rangle $$

  • $\begingroup$ Great, thank you! $\endgroup$ – Muzi Jun 23 '18 at 11:38

The answer is yes.

Any invertible map $T : H \to H$ is bounded from above and below. Namely, $T$ is bounded so $\|Tf\| \le \|T\|\|f\|$, and also $T^{-1}$ is bounded so $\|f\| = \|T^{-1}Tf\| \le \|T^{-1}\|\|Tf\|$. Hence:

$$\frac1{\left\|T^{-1}\right\|}\|f\|\le \|Tf\| \le \|T\|\|f\|, \quad\forall f \in H$$

Recall that a positive operator $S \ge 0$ has a unique positive square root $S^{1/2} \ge 0$. Since $S$ is invertible, $S^{1/2}$ is also invertible, with the inverse being $\left(S^{1/2}\right)^{-1} = \left(S^{-1}\right)^{1/2}$. Also we have $\left\|S^{1/2}\right\| = \|S\|^{1/2}$.

Now notice:

$$\langle Sf, f\rangle = \langle S^{1/2}f, S^{1/2}f\rangle = \left\|S^{1/2}f\right\|^2, \quad\forall f \in H$$

Using the above relation with $T =S^{1/2}$ we obtain:

$$\frac1{\left\|S^{-1}\right\|}\left\|f\right\|^2 =\frac1{\left\|S^{-1/2}\right\|^2}\left\|f\right\|^2 \le \left\|S^{1/2}f\right\|^2 \le \left\|S^{1/2}\right\|^2\|f\| = \|S\|\|f\|^2, \quad\forall f \in H$$


$$\frac1{\left\|S^{-1}\right\|}\left\|f\right\|^2 \le \langle Sf,f\rangle \le \|S\|\|f\|^2, \quad\forall f \in H$$

Furthermore, we see that $A = \frac1{\left\|S^{-1}\right\|}$ and $B = \|S\|$ are the optimal constants for this frame.


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