4
$\begingroup$

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 $0 < A \leq B ?$

$\endgroup$
1
3
$\begingroup$

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 $$A\|f\|^2\leq \left\langle Sf,f\right\rangle $$

$\endgroup$
1
  • $\begingroup$ Great, thank you! $\endgroup$ – Muzi Jun 23 '18 at 11:38
2
$\begingroup$

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:

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

Therefore

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.