Randomized continuous frames in time-frequency analysis

Abstract

Recently, a Monte Carlo approach was proposed for processing highly redundant continuous frames. In this paper, we present and analyze applications of this new theory. The computational complexity of the Monte Carlo method relies on the continuous frame being so-called linear volume discretizable (LVD). The LVD property means that the number of samples in the coefficient space required by the Monte Carlo method is proportional to the resolution of the discrete signal. We show in this paper that the continuous wavelet transform (CWT) and the localizing time-frequency transform (LTFT) are LVD. The LTFT is a time-frequency representation based on a 3D time-frequency space with a richer class of time-frequency atoms than classical time-frequency transforms like the short time Fourier transform (STFT) and the CWT. Our analysis proves that performing signal processing with the LTFT has the same asymptotic complexity as signal processing with the STFT and CWT (based on FFT), even though the coefficient space of the LTFT is higher dimensional.

Appendices

Appendix 1: Pseudo inverse of the analysis operator

For an injective linear operator with close range $B:\mathcal {V}\rightarrow \mathcal {W}$ between the Hilbert spaces $\mathcal {V}$ and $\mathcal {W}$, we define the pseudo inverse [26, Section 2.1.2]

$$B^{+}:\mathcal{W}\rightarrow \mathcal{V}, \quad B^{+} = \big(B\big|_{\mathcal{V}\rightarrow B\mathcal{V}}\big)^{-1}R_{B\mathcal{V}},$$

where $R_{B\mathcal {V}}:\mathcal {W}\rightarrow B\mathcal {V}$ is the surjective operator given by the orthogonal projection from $\mathcal {W}$ to $B\mathcal {V}$ and restriction of the image space to the range $B\mathcal {V}$, and $B\big |_{\mathcal {V}\rightarrow B\mathcal {V}}$ is the restriction of the image space of B to its range $B\mathcal {V}$, in which it is invertible. Note that $R_{B\mathcal {V}}^{*}$ is the operator that takes a vector from $B\mathcal {V}$ and canonically embeds it in $\mathcal {W}$, and $P_{B\mathcal {V}}=R_{B\mathcal {V}}^{*}R_{B\mathcal {V}}:\mathcal {W}\rightarrow \mathcal {W}$ is the orthogonal projection upon $B\mathcal {V}$. Note that $V_{f}^{+}$ exists. Indeed, by the frame inequality (7), V_f is bounded both from above and below, so it must be injective with closed range [2, Chapter 2]. In the following, we collect basic properties from frame analysis (see, e.g., [42], [9, Section 5.6] and [26, Section 2.1.2])

Lemma 25

Let $f:G\rightarrow {\mathscr{H}}$ be a continuous frame with frame bounds A, B. Then the following properties hold. Let $s\in {\mathscr{H}}$.

• 1. The operator $V_{f}^{*}$ is the pseudo inverse of $V_{f}^{+*}$, and $V_{f}^{+*}[{\mathscr{H}}]=V_{f}[{\mathscr{H}}]$.

• 2. $V_{f}^{+*}V_{f}^{*}=V_{f}V_{f}^{+}=P_{V_{f}[{\mathscr{H}}]}$.

• 3. $S_{f}^{-1}=(V_{f}^{*}V_{f})^{-1}= V_{f}^{+}V_{f}^{+*}$.

• 4. $V_{f}^{+*}[s]=V_{S_{f}^{-1}f}[s]=V_{f}[S_{f}^{-1}s]$.

• 5. $V_{f}^{+}=S_{f}^{-1}V_{f}^{*}$.

• 6. $\left \|V_{f}^{+}\right \|{~}_{2}\geq A^{-1/2}$.

Proof

We prove 5, and note that the rest are basic properties of dual frames and pseudo inverse. See, e.g., [42] and [9, Section 5.6] for dual frames, and [26, Section 2.1.2] for pseudo inverse.

$$S_{f}^{-1}V_{f}^{*} = V_{f}^{+}V_{f}^{+*}V_{f}^{*} = V_{f}^{+}P_{V_{f}[\mathcal{H}]} =V_{f}^{+}. $$

□

Appendix 2: Proofs

1.1 2.1 Proofs of Section 4

The mapping (x, ω, τ)↦f_{x, ω, τ} is continuous as a mapping $[\tau _{1},\tau _{2}]\times \mathbb {R}^{2}\rightarrow L^{2}(\mathbb {R})$, since $\mathcal {T}(x),{\mathscr{M}}(\omega ),\mathcal {D}(\gamma )$ are continuous, in x, ω, and γ respectively, in the strong topology [22, Section 9.2]. Hence, $V_{f}[s]:G\rightarrow \mathbb {C}$ is a continuous function for every $s\in L^{2}(\mathbb {R})$, and thus measurable. To show that f is continuous frame, it is left to show the existence of frame bounds $0<A\leq B<\infty$ satisfying (7). Equivalently we show that V_f is injective and

$$ \left\|V_{f}\right\| \leq B^{1/2}, \quad \left\|V_{f}^{+}\right\| \leq A^{-1/2} $$

(56)

where $V_{f}^{+}:L^{2}(G)\rightarrow L^{2}(\mathbb {R})$ is the pseudo inverse of V_f as defined in Section 3.1.

We start by deriving a formula for the LTFT atoms $\hat {f}_{x,\omega ,\tau }$ in the frequency domain.

Lemma 26

LTFT atoms take the following form in the frequency domain.

$$ \begin{array}{@{}rcl@{}} \hat{f}_{x,\omega,\tau}(z) & = &\left\{ \begin{array}{ccc} \mathcal{M}(-x) \mathcal{T}(\omega) \mathcal{D}(a_{\tau}/\tau) \hat{f}(z) & \text{if} & \left|\omega\right|<a_{\tau} \\ \mathcal{M}(-x) \mathcal{T}(\omega) \mathcal{D}(\omega/\tau) \hat{f}(z) & \text{if} & a_{\tau}\leq\left|\omega\right|\leq b_{\tau} \\ \mathcal{M}(-x) \mathcal{T}(\omega) \mathcal{D}(b_{\tau}/\tau)\hat{f}(z) & \text{if} & b_{\tau}<\left|\omega\right| \end{array} \right. \\ & =& \left\{ \begin{array}{ccc} \sqrt{\frac{\tau}{a_{\tau}}}\hat{f}\big(\frac{\tau}{a_{\tau}}(z-\omega)\big)e^{-2\pi i x z} & \text{if} & \left|\omega\right|<a_{\tau} \\ \sqrt{\frac{\tau}{\omega}}\hat{f}\big(\frac{\tau}{\omega}(z-\omega)\big)e^{-2\pi i x z} & \text{if} & a_{\tau}\leq\left|\omega\right|\leq b_{\tau} \\ \sqrt{\frac{\tau}{b_{\tau}}}\hat{f}\big(\frac{\tau}{b_{\tau}}(z-\omega)\big)e^{-2\pi i x z} & \text{if} & b_{\tau}<\left|\omega\right|. \end{array} \right. \end{array} $$

(57)

For f_τ(t) = τ^− 1/2f(τ^− 1t)e^2πit we have

$$ \quad \hat{f}_{\tau}(z) =\tau^{1/2}\hat{f}\big(\tau (z-1)\big), $$

(58)

and another formula in case $a_{\tau }\leq \left |\omega \right |\leq b_{\tau }$ is

$$\hat{f}_{x,\omega,\tau}(z) = \omega^{-1/2}\hat{f}_{\tau}(\omega^{-1}z)e^{-2\pi i x z}.$$

Proof

Equation (57) is a direct result of Lemma 4 and (30).

We can write $f_{\tau } = {\mathscr{M}}(1) \mathcal {D}(\tau )h$. Another formula in case $a_{\tau }\leq \left |{\omega }\right |\leq b_{\tau }$ is

$$f_{x,\omega,\tau} = \mathcal{T}(x)\mathcal{D}(\omega^{-1})\hat{f}_{\tau},$$

so

$$\hat{f}_{x,\omega,\tau} = \mathcal{M}(-x)\mathcal{D}(\omega)\hat{f}_{\tau}.$$

We can also write

$$\hat{f}_{x,\omega,\tau} =\mathcal{M}(-x)\mathcal{D}(\omega) \mathcal{T}(1) \mathcal{D}(\tau^{-1})\hat{f}.$$

□

For convenience, we repeat here Definition 13. The sub-frame filter kernels $\hat {S}_f^{\text {low}}, \hat {S}_f^{\text {mid}}$ and $\hat {S}_f^{\text {high}}$ are the functions $\mathbb {R}\rightarrow \mathbb {C}$ defined by

$$ \hat{S}_f^{\text{band}}(z) = {\int}_{{\tau}_1}^{{\tau}_2} {\int}_{\mathcal{J}^{\text{band}}_{\tau}} \left|{\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}}\right|^2 d\omega d{\tau}. $$

(59)

The frame filter kernel $\hat {S}_f:\mathbb {R}\rightarrow \mathbb {C}$ is defined by

$$ \hat{S}_f = \hat{S}_f^{\text{low}}+\hat{S}_f^{\text{mid}}+\hat{S}_f^{\text{high}}. $$

(60)

Let band ∈{low,mid,high}. For convenience, we recall (34)

$$ \hat{f}^{\text{band}}(\tau,\omega; z-\omega) = \left\{ \begin{array}{ccc} \sqrt{\frac{\tau}{a_{\tau}}}\hat{f}\big(\frac{\tau}{a_{\tau}}(z-\omega)\big) & \text{if} & \left|{\omega}\right|<a_{\tau} \\ \sqrt{\frac{\tau}{\omega}}\hat{f}\big(\frac{\tau}{\omega}(z-\omega)\big) & \text{if} & a_{\tau}\leq\left|\omega\right|\leq b_{\tau} \\ \sqrt{\frac{\tau}{b_{\tau}}}\hat{f}\big(\frac{\tau}{b_{\tau}}(z-\omega)\big) & \text{if} & b_{\tau}\leq\left|\omega\right|. \end{array} \right. $$

(61)

Thus, by (57),

$$\hat{f}_{x,\omega,\tau}(z) = \hat{f}^{\text{band}}(\tau,\omega; z-\omega)e^{-2\pi i x z},$$

for the band corresponding to ω.

Proof of Theorem 12

Denote by $V_f^{\text {band}}[s]$ the restriction of V_f[s] to (x, ω, τ) satisfying $\omega \in \mathcal {J}_{\tau }^{\text {band}}$. We offer the following informal computation.

$$ \begin{array}{@{}rcl@{}} && \left\|V_{f}^{\text{band}}[s]\right\|{~}_{2}^{2} \\ && = {\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{\mathbb{R}} \left|V_{f}[s](x,\omega,{\tau}) \right|{~}^{2} dx d\omega d{\tau}\\ && = {\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{\mathbb{R}} {\int}_{y\in\mathbb{R}} \hat{s}(y)\overline{\hat{f}_{x,\omega,{\tau}}(y)} dy {\int}_{z\in\mathbb{R}} \overline{\hat{s}(z)} \hat{f}_{x,\omega,{\tau}}(z) dz dx d\omega d{\tau} \\ && = {\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{\mathbb{R}} {\int}_{y\in\mathbb{R}} \hat{s}(y)\overline{\hat{f}^{\text{band}}(\tau,\omega; y-\omega)e^{-2\pi i x y}} dy \\ && \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad {\int}_{z\in\mathbb{R}} \overline{\hat{s}(z)}\hat{f}^{\text{band}}(\tau,\omega; z-\omega)e^{-2\pi i x z} dzdx d\omega d{\tau} \\ && = {\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{z\in\mathbb{R}}{\int}_{y\in\mathbb{R}}\Big({\int}_{\mathbb{R}} e^{-2\pi i x (z-y)} dx\Big) \hat{s}(y)\overline{\hat{f}^{\text{band}}(\tau,\omega; y-\omega)} \\ && \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \overline{\hat{s}(z)}\hat{f}^{\text{band}}(\tau,\omega; z-\omega) dy dy d\omega d{\tau} \\ && = {\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{z\in\mathbb{R}} {\int}_{y\in\mathbb{R}} \delta(z-y) \hat{s}(y)\overline{\hat{f}^{\text{band}}(\tau,\omega; y-\omega)} \overline{\hat{s}(z)} \hat{f}^{\text{band}}(\tau,\omega; z-\omega) dy dz d\omega d{\tau} \\ && = {\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{z\in\mathbb{R}} \hat{s}(z)\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)} \overline{\hat{s}(z)} \hat{f}^{\text{band}}(\tau,\omega; z-\omega) dz d \omega d{\tau} \end{array} $$

$$ = {\int}_{z\in\mathbb{R}}\left|\hat{s}(z)\right|{~}^{2} {\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{J}^{\text{band}}_{\tau}} \left|\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}\right|{~}^{2} d\omega d{\tau}dz . $$

(62)

Here, δ is the delta functional, and the informal computation with the delta functional can be formulated appropriately similarly to the usual Calderón’s reproducing formula in continuous wavelet analysis (see, e.g., [12, Proposition 2.4.1 and 2.4.1], [8, Theorem 1], and [11, Theorem 2.5]).

Now note that by the fact the three $\mathcal {J}^{\text {band}}_{\tau }$ domains are disjoint, so

$$\left\|V_{f}[s]\right\|{~}_{2}^{2} = \left\|V_{f}^{\text{low}}[s]\right\|{~}_{2}^{2}+\left\|V_{f}^{\text{mid}}[s]\right\|{~}_{2}^{2}+\left\|V_{f}^{\text{high}}[s]\right\|{~}_{2}^{2}.$$

Thus, by (62) and (60),

$$\left\|V_{f}[s]\right\|{~}_{2}^{2} = {\int}_{z\in\mathbb{R}}\left|\hat{s}(z)\right|{~}^{2}\hat{S}_{f}(z) dz. $$

Our goal now is to show that $\hat {S}_{f}(z)$ is bounded from below by some A > 0 and from above by some B > 0 for every $z\in \mathbb {R}$. The constants A, B are the frame bounds. In the following, we construct implicit upper and lower bounds for $\hat {S}_{f}(z)$, without any effort to make these bound realistic estimates of $\left \|V_{f}\right \|{~}^{2}$ and $\left \|V_{f}^{+}\right \|{~}^{2}$. The goal is to prove that f is a continuous frame, rather than to obtain good frame bounds. In Section 4.2, we explain separately that numerically estimating $\hat {S}_{f}$ and $\hat {S}_{f}^{-1}$ give good estimates for the frame bounds.

Next, we show that there is some A > 0 such that for every $z\in \mathbb {R} \hat {S}_{f}(z)\geq A$. For simplicity, we consider the case where a_τ = a and b_τ = b are constants. The general case is shown similarly with the appropriate modifications. Let z ≥ 0, and note that the case z ≤ 0 is shown symmetrically. By the fact that f is a non-negative function,

$$ \hat{f}(0) = \left\|f\right\|{~}_{1}>0. $$

(63)

Since f is compactly supported, $\hat {f}$ is smooth, so there is some ν > 0 such that for every z ∈ (−ν, ν)

$$ \hat{f}(z) \geq \frac{1}{2}\left\|f\right\|{~}_{1}=:C_{0}. $$

(64)

We now distinguish between three cases: $z\in [0,a], z\in \left .\left (a,b\right .\right ]$, and $z\in (b,\infty )$.

In case z ∈ [0, a],

$$ \hat{f}^{\text{band}}(\tau,\omega; z-\omega) = \sqrt{\frac{\tau}{a}}\hat{f}\big(\frac{\tau}{a}(z-\omega)\big). $$

(65)

By lugging (64) in (65), for any ω satisfying

$$\omega \in (z-\nu\frac{a}{\tau_{2}}, z+\nu\frac{a}{\tau_{2}})$$

we have

$$ \left|\hat{f}^{\text{band}}(\tau,\omega; z-\omega)\right| \geq \sqrt{\frac{\tau_{1}}{a}}C_{0}. $$

(66)

Let $\mathcal {I}_{z}$ denote the interval $(z-\nu \frac {a}{\tau _{2}}, z+\nu \frac {a}{\tau _{2}})\cap (-a,a)$, and note that the length of $\mathcal {I}_{z}$ is bounded from below by

$$ \mu(\mathcal{I}_{z})\geq \nu\frac{a}{\tau_{2}}, $$

(67)

where μ is the standard Lebesgue measure or $\mathbb {R}$. Thus, by the fact that the integrand of (35) is non-negative, the fact that μ_τ([τ₁, τ_s]) = 1, and by (66) and (67),

$$\hat{S}_{f}^{\text{band}}(z) = {\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{J}^{\text{band}}_{\tau}} \left|\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}\right|{~}^{2} d\omega d{\tau} $$

$$ \geq {\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{I}_{z}} \left|\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}\right|{~}^{2} d\omega d{\tau} \geq \nu\frac{a}{\tau_{2}}\frac{\tau_{1}}{a}{C_{0}^{2}} = C_{1}. $$

(68)

If $z\in \left .\left (a,b\right .\right ]$,

$$\hat{f}^{\text{band}}(\tau,\omega; z-\omega) = \sqrt{\frac{\tau}{\omega}}\hat{f}\big(\frac{\tau}{\omega}(z-\omega)\big). $$

By (64), for any ω satisfying

$$\omega \in (z-\nu\frac{a}{\tau_{2}}, z+\nu\frac{a}{\tau_{2}})$$

we have

$$ \left|\hat{f}^{\text{band}}(\tau,\omega; z-\omega)\right| \geq \sqrt{\frac{\tau_{1}}{b}}C_{0}. $$

(69)

Let $\mathcal {I}_{z}$ denote the interval $(z-\nu \frac {a}{\tau _{2}}, z+\nu \frac {a}{\tau _{2}})\cap (a,b)$, and note that the length of $\mathcal {I}_{z}$ is bounded from below by

$$ \mu(\mathcal{I}_{z})\geq \nu\frac{a}{\tau_{2}}. $$

(70)

Thus, by the fact that the integrand of (35) is non-negative, by μ_τ([τ₁, τ_s]) = 1, (69) and (70),

$$\hat{S}_{f}^{\text{band}}(z) = {\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{J}_{\tau}} \left|\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}\right|{~}^{2} d\omega d{\tau} $$

$$ \geq {\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{I}_{z}} \left|\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}\right|{~}^{2} d\omega d{\tau} \geq \nu\frac{a}{\tau_{2}}\frac{\tau_{1}}{b}{C_{0}^{2}} = C_{2}. $$

(71)

Last, if $z\in (b,\infty )$,

$$\hat{f}^{\text{band}}(\tau,\omega; z-\omega) = \sqrt{\frac{\tau}{b}}\hat{f}\big(\frac{\tau}{b}(z-\omega)\big). $$

By (64), for any ω satisfying

$$\omega \in (z-\nu\frac{b}{\tau_{2}}, z+\nu\frac{b}{\tau_{2}})$$

we have

$$ \left|\hat{f}^{\text{band}}(\tau,\omega; z-\omega)\right| \geq \sqrt{\frac{\tau_{1}}{b}}C_{0}. $$

(72)

Let $\mathcal {I}_{z}$ denote the interval $(z-\nu \frac {b}{\tau _{2}}, z+\nu \frac {b}{\tau _{2}})\cap (b,\infty )$, and note that the length of $\mathcal {I}_{z}$ is bounded from below by

$$ \mu(\mathcal{I}_{z})\geq \nu\frac{b}{\tau_{2}}. $$

(73)

Thus, by the fact that the integrand of (35) is non-negative, by μ_τ([τ₁, τ_s]) = 1, (72) and (73),

$$\hat{S}_{f}^{\text{band}}(z) = {\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{J}_{\tau}} \left|\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}\right|{~}^{2} d\omega d{\tau} $$

$$ \geq {\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{I}_{z}} \left|\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}\right|{~}^{2} d\omega d{\tau} \geq \nu\frac{b}{\tau_{2}}\frac{\tau_{1}}{b}{C_{0}^{2}} = C_{3}. $$

(74)

By taking $A=\min \limits \{C_{1},C_{2},C_{3}\}$, for every z ≥ 0

$$\hat{S}_{f}^{\text{band}}(z) \geq A,$$

and thus

$$\left\|V_{f}[s]\right\|{~}_{2}^{2}\geq A\left\|s\right\|{~}_{2}^{2}.$$

Next, we bound $\left \|V_{f}\right \|{~}^{2}$ from above. Note that

$$\left\|V_{f}\right\| =\left\|V_{f^{\text{low}}}\right\|+\left\|V_{f^{\text{mid}}}\right\|+\left\|V_{f^{\text{high}}}\right\|,$$

where $V_{f^{\text {band}}}$ now denotes V_f restricted to (x, ω, τ) with $\omega \in \mathcal {J}^{\text {band}}_{\tau }$, for any band ∈{low, mid, high}. The systems f^low and f^high are both STFT systems restricted in phase space to a sub-domain of frequencies, and integrated along τ ∈ (τ₁, τ₂). By extending f^low and f^high to the whole frequency axis $\mathbb {R}$, we increase $\left \|{V_{f^{\text {low}}}}\right \|$ and $\left \|{V_{f^{\text {high}}}}\right \|$ to the frame bound of the STFT which is 1. This shows that

$$\left\|V_{f^{\text{low}}}\right\|, \ \left\|V_{f^{\text{high}}}\right\| \leq 1.$$

It is left to bound $\left \|V_{f^{\text {mid}}}\right \|{~}^{2}$ from above. Note that f^mid cannot be extended to a CWT frame, since this system is not based on an admissible wavelet.

Recall the pseudo mother wavelet defined in (31)

$$f_{\tau}(t) = \tau^{-1/2}f(\tau^{-1}t)e^{2\pi i t} .$$

In the following, we use the bound

$$ \left\|\hat{f}_{\tau}\right\|{~}_{\infty} \leq \left\|f_{\tau}\right\|{~}_{1} = \tau^{1/2}\int \left|f(\tau^{-1}t)\right| \tau^{-1}dt = \tau^{1/2}\left\|h\right\|{~}_{1} \leq \tau_{2}^{1/2}\left\|h\right\|{~}_{1} \leq \tau_{2}^{1/2}\left\|{h}\right\|{~}_{2} = C_{0}. $$

(75)

By (62)

$$\left\|V_{f^{\text{mid}}}\right\|{~}^{2}= {\int}_{z} \left|\hat{s}(z)\right|{~}^{2}\hat{S}_{f}^{\text{mid}}(z) dz,$$

and by Lemma 26,

$$ \hat{S}_{f}^{\text{mid}}(z)={\int}_{\tau_{1}}^{\tau_{2}}{\int}_{\omega\in[-b,-a]\cap[a,b]}\omega^{-1}\left|\hat{f}_{\tau}(\omega^{-1}z)\right|{~}^{2}d\omega d\tau. $$

(76)

Let us change variable in (76) and consider the interval ω ∈ [a, b]. By ω^− 1z = y, we have ω = zy^− 1, dω = −zy^− 2dy, and

$$\omega=a \Leftrightarrow y=a^{-1}z , \quad \omega=b \Leftrightarrow y=b^{-1}z.$$

Thus

$$\hat{S}_{f}^{\text{mid}}(z)={\int}_{\tau_{1}}^{\tau_{2}}{\int}_{b^{-1}z}^{a^{-1}z}yz^{-1}\left|\hat{f}_{\tau}(y)\right|{~}^{2}z y^{-2}dy d\tau ={\int}_{\tau_{1}}^{\tau_{2}}{\int}_{b^{-1}z}^{a^{-1}z}\left|{\hat{f}_{\tau}(y)}\right|{~}^{2} y^{-1}dy d\tau.$$

Therefore, by (75),

$$ \begin{array}{@{}rcl@{}} \hat{S}_{f}^{\text{mid}}(z) & =&{\int}_{\tau_{1}}^{\tau_{2}}{\int}_{b^{-1}z}^{a^{-1}z}\left|\hat{f}_{\tau}(y)\right|{~}^{2} y^{-1}dy d\tau \leq {\int}_{\tau_{1}}^{\tau_{2}}C_{0} {\int}_{b^{-1}z}^{a^{-1}z} y^{-1}dy d\tau \\ & =& {\int}_{\tau_{1}}^{\tau_{2}}C_{0} \ln\Big(\frac{b}{a}\Big) d\tau =(\tau_{2}-\tau_{1})C_{0} \ln\Big(\frac{b}{a}\Big)=C^{\prime}. \end{array} $$

To conclude,

$$ \left\|V_{f}\right\| \leq \left\|V_{f^{\text{low}}}\right\|+\left\|V_{f^{\text{mid}}}\right\|+\left\|V_{f^{\text{high}}}\right\| \leq 2+\sqrt{C^{\prime}} =: \sqrt{B}. $$

□

Next, we prove Proposition 14, that states that for any $s\in L^{2}(\mathbb {R})$, the frame operator of the LTFT is given in the frequency domain by

$$\mathcal{F} [S_{f} s] = \hat{S}_{f}^{\text{low}} \hat{s} +\hat{S}_{f}^{\text{mid}} \hat{s} +\hat{S}_{f}^{\text{high}} \hat{s}.$$

Proof of Proposition 14

For any band ∈{low,mid,high}, define the operator $S_{f}^{\text {band}}$ by

$$ S_{f}^{\text{band}} s ={\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{J}^{\text{band}}_{\tau}}{\int}_{\mathbb{R}} V_{f}[s](x,\omega,\tau) dx d\omega d\tau,$$

and observe that

$$ S_{f}=S_{f}^{\text{low}}+S_{f}^{\text{mid}}+S_{f}^{\text{high}}.$$

We show that for any band ∈{low,mid,high}, and any $s\in L^{2}(\mathbb {R})$,

$$ \mathcal{F} S_{f}^{\text{band}} s (z)= \hat{S}_{f}^{\text{band}}(z)\hat{s}(z). $$

(77)

We offer the following informal computation, analogues to the proof of Theorem 12.

$$ \begin{array}{@{}rcl@{}} \mathcal{F} S_{f}^{\text{band}} s (z) & = &{\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{\mathbb{R}} V_{f}[s](x,\omega,{\tau}) \hat{f}_{x,\omega,{\tau}}(z) dx d\omega d{\tau}\\ & =& {\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{\mathbb{R}} {\int}_{\mathbb{R}} \hat{s}(y)\overline{\hat{f}_{x,\omega,{\tau}}(y)} dy \hat{f}_{x,\omega,{\tau}}(z) dx d\omega d{\tau} \\ & = &{\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{\mathbb{R}} {\int}_{\mathbb{R}} \hat{s}(y)\overline{\hat{f}^{\text{band}}(\tau,\omega; y-\omega)e^{-2\pi i x y}} dy\\ && \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \hat{f}^{\text{band}}(\tau,\omega; z-\omega)e^{-2\pi i x z} dx d\omega d{\tau} \\ & =& {\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{\mathbb{R}} \Big({\int}_{\mathbb{R}} e^{-2\pi i x (z-y)} dx\Big) \\ & &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \hat{s}(y)\overline{\hat{f}^{\text{band}}(\tau,\omega; y-\omega)} \hat{f}^{\text{band}}(\tau,\omega; z-\omega) dy d\omega d{\tau} \\ & = &{\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} {\int}_{\mathbb{R}} \delta(z-y) \hat{s}(y)\overline{\hat{f}^{\text{band}}(\tau,\omega; y-\omega)} \hat{f}^{\text{band}}(\tau,\omega; z-\omega) dy d\omega d{\tau} \\ & = &{\int}_{{\tau}_{1}}^{{\tau}_{2}}{\int}_{\mathcal{J}^{\text{band}}_{\tau}} \hat{s}(z)\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)} \hat{f}^{\text{band}}(\tau,\omega; z-\omega) d\omega d{\tau}\\ & = &\hat{s}(z) {\int}_{{\tau}_{1}}^{{\tau}_{2}} {\int}_{\mathcal{J}^{\text{band}}_{\tau}} \left|\overline{\hat{f}^{\text{band}}(\tau,\omega; z-\omega)}\right|{~}^{2} d\omega d{\tau} . \end{array} $$

Here, δ is the delta functional, and the formal computation with the delta functional can be formulated appropriately as explained in the proof of Theorem 12. □

1.2 2.2 Proofs of Section 5.1

Recall that the sequence of spaces $\{V_{M}\cap \mathcal {R}_{C}\}_{M\in \mathbb {N}}$, from Definition 15 and (40), is the discretization on which we prove the LVD property of the CWT. Recall that ξ is supported in (− 1/2, 1/2), non-negative, and continuously differentiable, and obtains zero only outside (− 1/2, 1/2). Recall that the class $\mathcal {R}_{C}$ is the set of all signals q ∈ L²[− 1/2, 1/2] such that

$$ \left\|\xi^{-1}q\right\|{~}_{\infty}< C\left\|q\right\|{~}_{\infty} $$

(78)

and

$$ \quad \left\|q\right\|{~}_{\infty} < C\left\|q\right\|{~}_{2}. $$

(79)

Moreover, recall that

$$ V_{M}= \text{span}\{e^{2 \pi i m x}\xi(x)\}_{m=-M}^{M}. $$

(80)

Next, we prove Proposition 17, which states that $\{V_{M}\cap \mathcal {R}_{C}\}_{M\in \mathbb {N}}$ is a discretization of $\mathcal {R}_{C}$.

Proof of Proposition 17

Let δ. Given $q\in \mathcal {R}_{C}$, we can approximate q by a smooth function $p\in L^{2}(-1/2,1/2)\cap L^{\infty }(-1/2,1/2)$ that vanishes in a neighborhood of − 1/2 and 1/2, up to the small errors

$$ \left\|q-p\right\|{~}_{2}<\delta, \quad \left|\ \left\|q\right\|{~}_{\infty}-\left\|p\right\|{~}_{\infty} \right| < \delta, \quad \left|\ \left\|\xi^{-1}q\right\|{~}_{\infty}-\left\|\xi^{-1}p\right\|{~}_{\infty} \right| < \delta. $$

(81)

Since p and ξ^− 1p have continuously differentiable periodic extensions, their Fourier series converge to p and ξ^− 1p respectively in both L²(− 1/2, 1/2) and $L^{\infty }(-1/2,1/2)$ [48, Section 4.4]. We denote by v_M the truncation of the Fourier series of ξ^− 1p up to the frequency M, multiplied by ξ. Namely,

$$v_{M}(x)= \sum\limits_{m=-M}^{M}\left\langle \xi^{-1}p , e^{2\pi i m(\cdot)} \right\rangle\xi(x)e^{2\pi i mx}.$$

There is thus a large enough M, such that v_M ∈ V_M satisfies

$$ \left\|v_{M}-p\right\|{~}_{2}<\delta, \quad \left\|v_{M}-p\right\|{~}_{\infty} < \delta, \quad \left\|\xi^{-1}v_{M}-\xi^{-1}p\right\|{~}_{\infty} < \delta. $$

(82)

Given any $\delta ^{\prime }>0$, by choosing δ small enough, and M large enough, (81) and (82) guarantee that $v_{M}\in V_{M}\cap \mathcal {R}_{C}$ and $\left \|v_{M}-q\right \|{~}_{2}<\delta ^{\prime }$. □

We prove Proposition 18 in a sequence of claims. Consider the phase space domain G_M of (41). Recall that [−S, S] is the support of the mother wavelet f. Thus, [−S/ω, Sω] is the support of the dilated wavelet ω^1/2f(ωz). Since the signal q is supported in time in [− 1/2, 1/2], V_f[q](x, ω) is zero for any x∉(− 1/2 − S/ω, 1/2 + S/ω). As a result, restricting V_f[q] to the phase space domain

$$ G_{M}^{\prime} = \big\{(x,\omega)\ \big|\ W^{-1}M^{-1} < \omega <WM \big\} $$

(83)

is equivalent to restricting V_f[q] to G_M. We thus consider without loss of generality the domain $G_{M}^{\prime }$ instead of G_M in this section.

We define the following inner product that corresponds to enveloping by ξ.

Definition 27 (Weighted Lebesgue space)

For any two measurable $q,p:[-1/2,1/2]\rightarrow \mathbb {C}$

$$\left\langle q,p\right\rangle_{\xi } =\left\langle\xi^{-1}q,\xi^{-1}p\right\rangle = {\int}_{-1/2}^{1/2} \frac{1}{\xi^{2}(x)}q(x)\overline{p(x)}dx$$

where $\left \langle \xi ^{-1}q,\xi ^{-1}p\right \rangle$ is the L²[− 1/2, 1/2] inner product. Denote

$$\left\|q\right\|{~}_{\xi }= \sqrt{\left\langle q,p\right\rangle_{\xi }}.$$

Denote by L_2;ξ(− 1/2, 1/2) the Hilbert space of signals with $\left \|q\right \|{~}_{\xi }<\infty$.

The following lemma characterizes the behavior of admissible wavelets about the zero frequency.

Lemma 28

Let $f\in L_{2}(\mathbb {R})$ be an admissible wavelet supported in [−S, S]. Then

$$\left|\hat{f}(z)\right| \leq 2\pi S\left\|f\right\|{~}_{1} \left|z\right| \leq 2^{3/2}\pi S^{3/2}\left\|f\right\|{~}_{2}\left|z\right|.$$

Proof

First note that by the fact that f is supported in (−S, S), it is also in $L^{1}(\mathbb {R})$ by the Cauchy–Schwarz inequality, with

$$\left\|f\right\|{~}_{1} \leq \sqrt{2S}\left\|f\right\|{~}_{2}.$$

By the wavelet admissibility condition $\hat {f}(0)=0$. Moreover, since f is compactly supported, $\hat {f}$ is smooth. Thus, for z > 0,

$$\hat{f}(z) = {{\int}_{0}^{z}} \hat{f}^{\prime}(y)dy.$$

As a result

$$ \left|\hat{f}(z)\right| \leq {{\int}_{0}^{z}} \left|\hat{f}^{\prime}(y)\right|dy \leq \left\|\hat{f}^{\prime}\right\|{~}_{\infty}z \leq \left\|2\pi yf(y)\right\|{~}_{1}z \leq 2S\pi\left\|f\right\|{~}_{1}z \leq 2^{3/2}\pi S^{3/2}\left\|{f}\right\|{~}_{2}\left|{z}\right|, $$

where $\left \|\hat {f}^{\prime }\right \|{~}_{\infty } \leq \left \|2\pi yf(y)\right \|{~}_{1}$ since $(\mathcal {F}\hat {f}^{\prime })(y)=2\pi i yf(y)$, and

$$\left\|\hat{f}^{\prime}\right\|{~}_{\infty} = \sup_{\omega\in\mathbb{R}}\left|{\int}_{\mathbb{R}} 2\pi i yf(y) e^{-2\pi i \omega y} dy\right| \leq \sup_{\omega\in\mathbb{R}}{\int}_{\mathbb{R}} 2\pi \left|yf(y)\right| dy = \left\|2\pi yf(y)\right\|{~}_{1}.$$

For z < 0 the proof is similar. □

Note that the basis $\{e^{2\pi i m x}\xi (x)\}_{m=-M}^{M}$ of V_M is an orthonormal system in L_2;ξ(− 1/2, 1/2), since $\{e^{2\pi i m x}\}_{m=-M}^{M}$ are orthonormal in L₂(− 1/2, 1/2). By Parseval’s identity, for q ∈ V_M satisfying

$$q(x) = \sum\limits_{m=-M}^{M} c_{n} e^{2\pi i m x}\xi(x),$$

we have

$$\left\|q\right\|{~}_{\xi} = \sqrt{ \sum\limits_{m=-M}^{M} \left|c_{m}\right|{~}^{2}} = \left\|\{c_{m}\}_{m=-M}^{M}\right\|{~}_{2}.$$

We prove Proposition 18 by embedding the signal class $\mathcal {R}_{C}$ in a richer space, and proving linear volume discretization for the richer space. Consider the signal space $\mathcal {S}_{E}$ of signals q ∈ L²(− 1/2, 1/2) satisfying

$$ \left\|q\right\|{~}_{\xi} \leq E\left\|q\right\|{~}_{2} $$

(84)

for some fixed E > 0.

Lemma 29

$\mathcal {R}_{C}\subset \mathcal {S}_{E}$ for any E ≥ C².

Proof

Let $q=\xi s \in \mathcal {R}_{C}$. By Cauchy-Schwartz inequality and by (37) and (38)

$$\left\|q\right\|{~}_{\xi } = \left\|s\right\|{~}_{2} \leq \left\|s\right\|{~}_{\infty}= \left\|\xi^{-1}q\right\|{~}_{\infty} \leq C \left\|q\right\|{~}_{\infty} \leq C^{2} \left\|q\right\|{~}_{2}$$

so

$$\left\|q\right\|{~}_{\xi } \leq C^{2} \left\|q\right\|{~}_{2} \leq E \left\|q\right\|{~}_{2}.$$

□

Proposition 30

Under the above construction, with ξ satisfying (44) with k > 2, we have μ(G_M) ≤ 5WM, and for every $q_{M}\in V_{M}\cap \mathcal {S}_{E}$ we have

$$ \frac{\left\|I-\psi_{M} V_{f}[q_{M}]\right\|{~}_{2}}{\left\|V_{f}[q_{M}]\right\|{~}_{2}} = O(W^{-1})+o_{M}(1), $$

(85)

where the O notation O(W^− 1) is with respect to $W\rightarrow \infty$, and o_M(1) is a function that goes to zero as $M\rightarrow \infty$.

Next, we prove Proposition 18, which we copy here for the convenience of the reader.

Proposition 18 Consider a smooth enough ξ in the sense

$$ \hat{\xi}(z) \leq \left\{ \begin{array}{ccc} D & , & \left|z\right|\leq 1 \\ D z^{-k} & , & \left|z\right|>1 \end{array} \right. $$

(86)

for some k > 2 and D > 0, and the corresponding discrete spaces $\{V_{M}\}_{M\in \mathbb {N}}$ of (40). The continuous wavelet transform with a compactly supported mother wavelet $f\in L_{2}(\mathbb {R})$ is linear volume discretizable with respect to the class $\mathcal {R}_{C}$ and the discretization $\{V_{M}\cap \mathcal {R}_{C}\}_{M\in \mathbb {N}}$, with the envelopes ψ_M defined by (43) for large enough W that depends only on 𝜖 of Definition 6.

By Lemma 29, Proposition 18 is now a corollary of Proposition 30, where for every 𝜖 we choose W and M₀ large enough, so that for every M > M₀

$$ \frac{\left\|I-\psi_{M} V_{f}[q_{M}]\right\|{~}_{2}}{\left\|V_{f}[q_{M}]\right\|{~}_{2}} < \epsilon. $$

(87)

Proof of Proposition 30

Since $\{e^{2\pi i m x}\xi (x)\}_{m=-M}^{M}$ is an orthonormal basis of V_M ⊂ L_2;ξ(− 1/2, 1/2), in the frequency domain signals in V_M are spanned by $\{V_{m}(z)=\hat {\xi }(z-m)\}_{m=-M}^{M}$ (see Lemma 4). We bound V_n(z) by

$$ \left| \hat{b}_{m}(z)\right| = \left| \hat{\xi}(z-m)\right| \leq \left\{ \begin{array}{ccc} D & , & \left| z-m \right| \leq 1 \\ D \left|z-m\right|{~}^{-k} & , & \left|z-m\right|>1 . \end{array} \right. $$

For $\hat {q} = {\sum }_{m=-M}^{M} c_{m} \hat {b}_{m}$, with $\left \|q\right \|{~}_{\xi }=\left \|\{c_{m}\}_{m=1}^{N}\right \|{~}_{2}$, we have $\left |c_{m}\right |\leq \left \|q\right \|{~}_{\xi }$ for all m. Thus

$$ \left|\hat{q}(z)\right| \leq \sum\limits_{m=-M}^{M} \left|c_{m}\right| \left|\hat{b}_{m}(z)\right| \leq \sum\limits_{m=-M}^{M} \left\|q\right\|{~}_{\xi} \left\{ \begin{array}{ccc} D & , & \left|z-m\right|\leq 1 \\ D \left|z-m\right|{~}^{-k} & , & \left|z-m\right|>1 \end{array} \right. $$

(88)

For $\left |z\right |\leq M+2$, (88) leads to the bound

$$ \left|\hat{q}(z)\right| \leq 2D\left\|q\right\|{~}_{\xi}+ \left\|q\right\|{~}_{\xi}\sum\limits_{-M\leq m\leq M, m\neq \left\lceil z \right\rceil ,\left\lfloor z\right\rfloor} D \left|z-m\right|{~}^{-k}, $$

(89)

and (89) is maximized by taking z = 0. We extend the sum to m between $-\infty$ and $\infty$, and without loss of generality increase the value of the sum by choosing z = 0. Thus, since k > 2,

$$ \begin{array}{@{}rcl@{}} \left|\hat{q}(\omega)\right| & \!\leq\! &2D\left\|q\right\|{~}_{\xi}+ 2\left\|q\right\|{~}_{\xi} D \sum\limits_{m=1}^{\infty} m^{-k} \!\leq\! 2D\left\|q\right\|{~}_{\xi}+ 2\left\|q\right\|{~}_{\xi} D \Big(1 + {\int}_{1}^{\infty}m^{-k} dm \Big)\\ & \leq& 4\left\|q\right\|{~}_{\xi} D + 2\left\|q\right\|{~}_{\xi} D(k-1)^{-1} \leq 6 D \left\|q\right\|{~}_{\xi}. \end{array} $$

Now, for $\left |z\right |>M+2$, without loss of generality consider z > 0. By (88) we have

$$ \begin{array}{@{}rcl@{}} \left|\hat{q}(z)\right| & \leq& \left\|q\right\|{~}_{\xi} \sum\limits_{m=-M}^{M} D (z-m)^{-k} \leq \left\|q\right\|{~}_{\xi} {\int}_{-\infty}^{M+1} D (z-m)^{-k} dm\\ & \leq& \left\|q\right\|{~}_{\xi} D (z-M-1)^{-k+1}(k-1)^{-1}. \end{array} $$

Overall,

$$ \frac{\left|\hat{q(z)}\right|}{\left\|q\right\|{~}_{\xi}} \leq \hat{E}(z) := \left\{ \begin{array}{ccc} 6 D & , & \left|z\right|\leq M+2 \\ 6 D \big(\left|z\right|-M-1\big)^{-k+1}& , & \left|z\right|> M+2 \end{array} \right. . $$

(90)

We consider the domain $G_{M}^{\prime } = \{(x,\omega )\ |\ W^{-1}M^{-1} < \omega <AM \}$. Recall that restricting to $G_{M}^{\prime }$ is equivalent to restricting to G_M. Let ψ_M denote by abuse of notation the projection in phase space that restricts functions to the domain $G_{M}^{\prime }$. Next, we bound the error $\left \|(I-\psi _{M})V_{f}[q]\right \|{~}_{2}$ for signals in V_M. Note that for every $\omega \in \mathbb {R}$

$$ {\int}_{\mathbb{R}}\left|V_{f}[q](\omega,x)\right|{~}^{2} dx = {\int}_{\mathbb{R}} \left|\hat{q}(z)\right|{~}^{2}\omega^{-1}\left|\hat{f}(\omega^{-1}z)\right|{~}^{2} dz. $$

(91)

Indeed, by Lemma 4

$$ \begin{array}{ll} V_{f}[q](x,\omega)& = {\int}_{\mathbb{R}} \hat{q}(z)\omega^{-1/2}\overline{\hat{f}(\omega^{-1}z) e^{-2\pi i x z}} dz \\ & =\mathcal{F}^{-1}\Big(\hat{q} \omega^{-1/2}\overline{\hat{f}\big(\omega^{-1}(\cdot)\big)}\Big) (x), \end{array} $$

so by Plancherel’s identity

$$ {\int}_{\mathbb{R}}\left|V_{f}(q)(\omega,x)\right|{~}^{2} dx = {\int}_{\mathbb{R}} \left|\hat{q}(z) \omega^{-1/2}\overline{\hat{f}\big(\omega^{-1}z\big)}\right|{~}^{2} dz. $$

(92)

Let us now bound the right-hand side of (92) for ω such that $(x,\omega )\notin G_{M}^{\prime }$. We then integrate the result for $\omega \in (WM,\infty )$, and for ω ∈ (0, W^− 1M^− 1), to bound the error in phase space truncation by ψ_M. Negative ω are treated similarly.

We start with ω > WM. Here, we decompose the integral along z into two integrals with boundaries in 0, M + 2 + (0.5(ω − M − 2))^1/2. For each segment of the integral along z, by additivity of the integral, we integrate ω along $(WM,\infty )$ and show that the resulting value is small. The first segment gives

$${\int}_{0}^{M+2 + \big(0.5(\omega-M-2)\big)^{1/2}} \left|\hat{q}(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1} z)\right|{~}^{2}dz$$

$$\leq {\int}_{0}^{M+2 + \big(0.5(\omega-M-2)\big)^{1/2}} \left|\hat{q}(z)\right|{~}^{2} dz \max_{0\leq z <M+2 + \big(0.5(\omega-M-2)\big)^{1/2}}\left|\omega^{-1/2}\hat{f}(\omega^{-1} z)\right|{~}^{2}.$$

By Lemma 28, the dilated mother wavelet satisfies

$$ \omega^{-1/2}\hat{f}(\omega^{-1} z) \leq 2\pi S \left\|f\right\|{~}_{1} \left|z\right| \omega^{-1.5} , $$

(93)

so

$$ \begin{array}{@{}rcl@{}} && {\int}_{0}^{M+2 + \big(0.5(\omega-M-2)\big)^{1/2}} \left|q(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1} z)\right|{~}^{2}dz \end{array} $$

(94)

$$ \begin{array}{@{}rcl@{}} && \leq \left\|q\right\|{~}_{2}^{2} 4\pi^{2} S^{2} \left\|f\right\|{~}_{1}^{2} \Big(M+2 + \big(0.5(\omega-M-2)\big)^{1/2}\Big)^{2} \omega^{-3} = 4\pi^{2} S^{2} \left\|q\right\|{~}_{2}^{2} \left\|f\right\|{~}_{1}^{2} G(\omega) \end{array} $$

for G(ω) = (M + 2 + (0.5(ω − M − 2))^1/2)²ω^− 3. We study G(ω) for different values of ω. When ω > M², the value (M + 2 + (0.5(ω − M − 2))^1/2)² is bounded by 4ω for large enough M, so

$$ G(\omega) \leq 4\omega^{-2}. $$

(95)

When ω < M²,(M + 2 + (0.5(ω − M − 2))^1/2)² is bounded by 4M² for large enough M, so

$$ G(\omega) \leq 4M^{2}\omega^{-3}. $$

(96)

For ω > M², integrating via the bound (95) gives

$$ {\int}_{M^{2}}^{\infty}{\int}_{0}^{M+2 + \big(0.5(\omega-M-2)\big)^{1/2}} \left|q(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1} z)\right|{~}^{2}dz d\omega =o_{M}(1)\left\|q\right\|{~}_{2}^{2}, $$

(97)

where o_M(1) is a function that converges to zero as $M\rightarrow \infty$. For WM < ω < M², integrating via the bound (96) gives

$$ {\int}_{WM}^{M^{2}}{\int}_{0}^{M+2 + \big(0.5(\omega-M-2)\big)^{1/2}} \left|q(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1} z)\right|{~}^{2}dz d\omega =O(W^{-2})\left\|q\right\|{~}_{2}^{2} $$

(98)

Overall, (98) and (97) give

$$ {\int}_{WN}^{\infty}{\int}_{0}^{M+2 + \big(0.5(\omega-M-2)\big)^{1/2}} \left|\hat{q}(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1} z)\right|{~}^{2}dz d\omega =\big(O(W^{-2}) + o_{M}(1)\big)\left\|q\right\|{~}_{2}^{2}. $$

(99)

Next, we study $z\in \Big (M+2 + \big (0.5(\omega -M-2)\big )^{1/2},\infty \Big )$. Here, we take the maximum of the signal squared, and take the 2 norm of the window. By (90) we obtain for large enough M

$$ \begin{array}{@{}rcl@{}} && {\int}_{M+2 + \big(0.5(\omega-M-2)\big)^{1/2}}^{\infty} \left|q(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1} z)\right|{~}^{2}dz \\ && \leq\left\|f\right\|{~}_{2}^{2} 36D^{2} \Big(M+2 + \big(0.5(\omega-M-2)\big)^{1/2}- M-1\Big)^{-2k+2} \left\|q\right\|{~}_{\xi}^{2} \\ && < \left\|f\right\|{~}_{2}^{2} 36D^{2} 2(\omega-M-2)^{-k+1} \left\|q\right\|{~}_{\xi}^{2} \end{array} $$

(100)

Integrating the bound (100) along $w\in (WM,\infty )$ gives

$$ {\int}_{WM}^{\infty}{\int}_{M+2 + \big(0.5(\omega-M-2)\big)^{1/2}}^{\infty} \left|q(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1} z)\right|{~}^{2}dz d\omega = o_{M}(1)\left\|q\right\|{~}_{\xi}^{2}=o_{M}(1)\left\|q\right\|{~}_{2}^{2}, $$

(101)

since k > 2 and $\left \|q\right \|{~}_{\xi }\leq E\left \|q\right \|{~}_{2}$.

Last, we integrate ω ∈ (0, W^− 1M^− 1). By (90)

$$\left\|\hat{q}\right\|{~}_{\infty} \leq 6D\left\|q\right\|{~}_{\xi}.$$

Thus

$$ {\int}_{-\infty}^{\infty}\left|\hat{q}(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1}z)\right|{~}^{2}dz \leq \left\|\hat{q}\right\|{~}_{\infty}^{2}\left\|\hat{f}\right\|{~}^{2}_{2} \leq 36D^{2} \left\|f\right\|{~}_{2}^{2} \left\|q\right\|{~}_{\xi}^{2} = O(1) \left\|q\right\|{~}_{2}^{2}. $$

(102)

Thus, the integration of the bound (102) for ω ∈ (0, W^− 1M^− 1) gives

$$ {\int}_{0}^{W^{-1}M^{-1}}{\int}_{-\infty}^{\infty}\left|\hat{q}(z)\right|{~}^{2}\left|\omega^{-1/2}\hat{f}(\omega^{-1}z)\right|{~}^{2}dz = O(M^{-1})\left\|q\right\|{~}_{2}^{2}. $$

(103)

Summarizing the estimates (99), (101), and (103), together with the analogue bounds for ω < 0, we obtain

$$\left\|(I-\psi_{M})V_{f}[q]\right\|{~}_{2}= \big(O(W^{-1}) + o_{M}(1)\big)\left\|q\right\|{~}_{2}.$$

Last, since by the frame assumption $\left \|q\right \|{~}_{2}^{2} \leq A^{-1}\left \|V_{f}(q)\right \|{~}_{2}^{2}$, we have

$$\frac{\left\|(I-\psi_{M})V_{f}[q]\right\|{~}_{2}}{\left\|V_{f}[q]\right\|{~}_{2}}= \mathcal{J}(W,M)= O(W^{-1}) + o_{M}(1).$$

This means that given 𝜖 > 0, we may choose W large enough to guarantee $\mathcal {J}(W,M) < \epsilon$ up from some large enough M₀, and also guarantee for every $M\in \mathbb {N}$

$$\left\|\psi_{M}\right\|{~}_{1}\leq C^{\epsilon} M$$

with C^𝜖 = 3W by (42). □

1.3 2.3 Proofs of Section 5.2

Recall that every q ∈ V_{M, R} is a trigonometric polynomials supported on L²(−R/2, R/2)

$$q(x) = \sum\limits_{m=-M}^{M} c_{n} R^{-1/2}\exp\big(\frac{2\pi i}{R} n x\big)$$

with $\left \|\mathbf {c}\right \|{~}_{2}=\left \|q\right \|{~}_{2}$. Since the Fourier transform of the indicator function of [− 1/2, 1/2] is the sinc function $\text {sinc}(z) := \frac {\sin \limits (\pi x)}{\pi x}$, the normalized indicator function of [−R/2, R/2] is given in the frequency domain by

$$R^{1/2}\text{sinc}(Rz) \leq \frac{R^{-1/2}}{\pi\left|z\right|}. $$

Hence,

$$ \hat{q}(z)=R^{1/2} \sum\limits_{n=-M}^{M} c_{n} \text{sinc}\big(R(z-R^{-1}n)\big) $$

(104)

recall that (45) reads: for every z > Y or z < −Y

$$ \hat{f}(z) \leq C^{\prime}\left|z\right|{~}^{-\kappa}. $$

(105)

Recall that STFT envelope G_{M, R} of (47) is defined by

$$ G_{M,R}=[-R/2-S/2,R/2+S/2]\times [-WM/R,WM/R]. $$

(106)

In the proof of Proposition 21, we use the following simple fact that can be shown by a direct calculation.

Lemma 31

Let V_f be the STFT based on the window $f\in L^{2}(\mathbb {R})$, and let $q\in L^{2}(\mathbb {R})$ be a signal. Then

$$ {\int}_{\mathbb{R}}\left|V_{f}[q](\omega,x)\right|{~}^{2} dx = {\int}_{\mathbb{R}} \left|\hat{q}(z)\right|{~}^{2}\left|\hat{f}(z-\omega)\right|{~}^{2} dz. $$

(107)

The following lemma will be used in the proofs of Propositions 21 and 22.

Lemma 32

Let $f\in L^{2}(\mathbb {R})$ be supported in [−S, S] and satisfy (50). Let q ∈ V_{M, R}, with R = O(M). Then for every W > 4,

$$ {\int}_{[-WM/R,WM/R]^{c}}{\int}_{\mathbb{R}} \left|\hat{q}(z)\right|{~}^{2}\left|\hat{f}(z-\omega)\right|{~}^{2} dzd\omega = o_{W}(1)\left\|q\right\|{~}_{2}^{2}, $$

(108)

where [−WM/R, WM/R]^c is the set $\{\omega \in \mathbb {R}\ |\ w\notin [-WM/R,WM/R]\}$, and o_W(1) is a function that decays to zero as $W\rightarrow \infty$.

Proof

We consider ω > 0 and z > 0, and note that the other cases are similar. For each value of ω > WM/R, we decompose the integral (107) along z into the two integrals in

$$z\in(0,(W^{1/2}M/R+\omega)/2)\quad \text{and} \quad z\in((W^{1/2}M/R+\omega)/2,\infty).$$

For z ∈ (0, (W^1/2M/R + ω)/2), since ω ≥ MW/R and z ≤ (W^1/2M/R + ω)/2,

$$z-\omega \leq(W^{1/2}M/R-\omega)/s \leq -\frac{1}{2}(W-W^{1/2})M/R <0,$$

so $\left |z-\omega \right |{~}^{-2\kappa }$ obtains its maximum at z = (W^1/2M/R + ω)/2. Thus, by (105),

$${\int}_{0}^{(W^{1/2}M/R+\omega)/2} \left|\hat{q}(z)\right|{~}^{2}\left|\hat{f}(z-\omega)\right|{~}^{2} dz \leq \left\|q\right\|{~}_{2}^{2} \max_{0 \leq z \leq (W^{1/2}M/R+\omega)/2}C^{\prime 2}\left|z-\omega\right|{~}^{-2\kappa}$$

$$ =\left\|q\right\|{~}_{2}^{2} C^{\prime 2}\left|(W^{1/2}M/R+\omega)/2-\omega\right|{~}^{-2\kappa}= \left\|q\right\|{~}_{2}^{2} C^{\prime 2}\left|(W^{1/2}M/R-\omega)/2\right|{~}^{-2\kappa} $$

(109)

Integrating the bound (109) for $\omega \in (WM/R,\infty )$ gives

$$ \begin{array}{@{}rcl@{}} {\int}_{WM/R}^{\infty}{\int}_{0}^{(W^{1/2}M/R+\omega)/2} \left|\hat{q}(z)\right|{~}^{2}\left|\hat{f}(z-\omega)\right|{~}^{2} dz d\omega &=& (W-W^{1/2})^{1-2k}M^{1-2\kappa}R^{2\kappa-1}\left\|q\right\|{~}_{2}^{2} O(1) \\ & =& o_{W}(1)(M/R)^{1-2\kappa} \left\|q\right\|{~}_{2}^{2}. \end{array} $$

(110)

Note that (M/R)^{1 − 2κ} = O(1) since R = O(M) and κ > 1/2.

For $z\in \big ((W^{1/2}M/R+\omega )/2,\infty \big ), \hat {q}$ decays like M^1/2(z − M)^− 1. Indeed, since z > M

$$ \begin{array}{@{}rcl@{}} R^{1/2} \sum\limits_{n=-M}^{M} c_{n} \text{sinc}\big(R(z-R^{-1}n)\big) & \leq& R^{-1/2}\left\|\{c_{n}\}\right\|{~}_{2} \sqrt{ \sum\limits_{n=-M}^{M} \frac{1}{(z-R^{-1}n)^{2}} } \\ && \leq R^{-1/2}\left\|q\right\|{~}_{2} \sqrt{ \sum\limits_{n=-M}^{M} \frac{1}{(z-M/R)^{2}} } \\ && \leq 2R^{-1/2}\left\|q\right\|{~}_{2}\sqrt{M} (z-M/R)^{-1}. \end{array} $$

(111)

Now, by (111),

$$ \begin{array}{@{}rcl@{}} && {\int}_{(W^{1/2}M/R+\omega)/2}^{\infty} \left|\hat{q}(z)\right|{~}^{2}\left|\hat{f}(z-\omega)\right|{~}^{2} dz \\ && \leq 4R^{-1}\left\|f\right\|{~}_{2}^{2}\left\|q\right\|{~}_{2}^{2} M \max_{(W^{1/2}M/R+\omega)/2 \leq z < \infty} (z-M/R)^{-2}\\ && =16R^{-1}\left\|f\right\|{~}_{2}^{2}\left\|q\right\|{~}_{2}^{2} M \big(\omega+(W^{1/2}-2)M/R\big)^{-2}. \end{array} $$

(112)

Integrating the bound (112) for $\omega \in (WM/R,\infty )$ gives

$$ {\int}_{WM/R}^{\infty}{\int}_{(W^{1/2}M/R+\omega)/2}^{\infty} \left|\hat{q}(z)\right|{~}^{2}\left|\hat{f}(z-\omega)\right|{~}^{2} dz d\omega = (W+W^{1/2}-2)^{-1}\left\|q\right\|{~}_{2}^{2} O(1) . $$

(113)

The bounds (110) and (113) give together (108). □

Proof of Proposition 21

Let q ∈ V_{M, R} Lemmas 31 and 32, are combined to give $\left \|{(I-\psi ^{W}_{M,R})V_{f}[q]}\right \|{~}_{2}= o_{W}(1)\left \|q\right \|{~}_{2}$, so by the frame inequality

$$\frac{\left\|(I-\psi^{W}_{M,R})V_{f}[q]\right\|{~}_{2}}{\left\|V_{f}[q]\right\|{~}_{2}}= o_{W}(1).$$

This means that given 𝜖 > 0, we may choose W large enough to guarantee $\frac {\left \|{(I-\psi ^{W}_{M,R})V_{f}[q]}\right \|{~}_{2}}{\left \|{V_{f}[q]}\right \|{~}_{2}} < \epsilon$, and also guarantee that for every $R,M\in \mathbb {N}, \left \|\psi _{M,R}\right \|{~}_{1}\leq C_{\epsilon } M$, with C_𝜖 = 2W(1 + 2S), by (106), for R = O(M). □

The proof of Proposition 22 is similar to that of Proposition 21. We start with an analogous lemma to Lemma 31, based on (30) and Lemma 4.

Lemma 33

Let V_f be the LTFT (Definition 11), and let $q\in L^{2}(\mathbb {R})$ be a signal. Then, for any (ω, τ) such that $\left |\omega \right |>b^{M,R}_{\tau }$.

$$ {\int}_{\mathbb{R}}\left|V_{f}[q](x,\omega,\tau)\right|{~}^{2} dx = {\int}_{\mathbb{R}} \left|\hat{q}(z)\right|{~}^{2}\left| [\mathcal{D}(b^{M,R}_{\tau}/\tau)\hat{f}](z-\omega)\right|{~}^{2} dz. $$

(114)

Proof of Proposition 22

Let 𝜖 > 0. Note that since V_f[s_{M, R}] is supported in the x direction in $[R/2- \tau _{2}/a^{M,R}_{\tau }, R/2+ \tau _{2}/a^{M,R}_{\tau }]$, it is enough to show that restricting the frequency direction of G to − WM/R < ω < WM/R results in an error less than 𝜖. Note that all of the truncated atoms, with $\left |\omega \right |\geq WM/ R$, are high frequency STFT atoms.

By Lemma 33, we must study the error

$$ {\int}_{\tau_{1}}^{\tau_{2}} {\int}_{[-WM/R,WM/R]^{c}}{\int}_{\mathbb{R}} \left|\hat{s}_{M,R}(z)\right|{~}^{2}\left| [\mathcal{D}(b^{M,R}_{\tau}/\tau)\hat{f}](z-\omega)\right|{~}^{2} dz d\omega d\tau. $$

(115)

Note that all atoms $[\mathcal {D}(b^{M,R}_{\tau }/\tau )\hat {f}]$ in (115) satisfy (105) with some constant $Y^{\prime },C^{\prime \prime }$ instead of $Y,C^{\prime }$. Hence, by Lemma 32, the error (115) is of order $o_{W}(1)\left \|s_{M,R}\right \|{~}^{2}_{2}$. The rest of the proof is the same as the proof of Proposition 21. □