Reconstruction of spline spectra-signals from generalized sinc function by finitely many samples

Abstract

Reconstruction of signals by their Fourier (transform) samples is investigated in many mathematical/engineering problems such as the inverse Radon transform and optical diffraction tomography. This paper concerns on the reconstruction of spline-spectra signals in $V(\hbox {sinc}_{a})$ by finitely many Fourier samples, where $\hbox {sinc}_{a}$ is the generalized sinc function. There are two main results on this topic. When the spectra knots are known, the exact reconstruction formula conducted by finitely many Fourier samples is established in the first main theorem. When the spectra knots are unknown, in the second main theorem we establish the approximations to the spline-spectra signals also by finitely many Fourier samples. Numerical simulations are conducted to check the efficiency of the approximation.

Appendix

Proof of Lemma 2.2

By the definition of $\{\xi _{k}\}_{k=1}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}$, we have $-\pi =\xi _{1}<\xi _{2}<\cdots < \xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}=\xi _{1}+\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mu _{0}\le \pi$ and $\lfloor \frac{2\pi }{\mu _{0}}\rfloor \ge s$. Furthermore, if $\lfloor \frac{2\pi }{\mu _{0}}\rfloor =\frac{2\pi }{\mu _{0}}$, then

$$\begin{aligned} \bigcup _{k=1}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor }[\xi _{k},\xi _{k+1})=[-\pi ,\pi ). \end{aligned}$$

(4.1)

And if $\lfloor \frac{2\pi }{\mu _{0}}\rfloor <\frac{2\pi }{\mu _{0}}$, then

$$\begin{aligned} \bigcup _{k=1}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor }[\xi _{k},\xi _{k+1})\bigcup \left[ \xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1},\pi \right) =[-\pi ,\pi ). \end{aligned}$$

(4.2)

We next prove that for any $i\in \{1,2, \ldots , s\}$, there exists at least a point $\xi _{k_{i}}\in [t_{i},t_{i+1})$ where $k_{i}\in \{1, \ldots , \lfloor \frac{2\pi }{\mu _{0}}\rfloor +1\}$. This statement will be proved for the two cases (4.1) and (4.2), respectively.

Case (4.1): For the interval $[t_{s},t_{s+1})$, it follows from (2.3) and (4.1) that $\xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}= t_{s+1}=\pi$ which together with

$$\begin{aligned} t_{s+1}-t_{s}\ge \mu _{0}=\xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}-\xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor }\end{aligned}$$

(4.3)

leads to $\xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor }\ge t_{s}$. Then

$$\begin{aligned} \xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor }\in [t_{s},t_{s+1}). \end{aligned}$$

(4.4)

For $i=1,$ it is clear that

$$\begin{aligned} \xi _{1}=-\pi =t_{1}\in [t_{1},t_{2}). \end{aligned}$$

(4.5)

Allowing for (4.4) and (4.5), we just need to prove the statement for the intervals $[t_{i},t_{i+1}),i=2,\ldots ,s-1$ where $s>2.$ Or else, if there exists $i_{0}\in \{2, \ldots , s-1\}$ such that $\xi _{k}\notin [t_{i_{0}},t_{i_{0}+1})$ for any $k\in \{1, \ldots , \lfloor \frac{2\pi }{\mu _{0}}\rfloor \}$, then by (2.3) and (4.1) we have

$$\begin{aligned} \{\xi _{k}\}_{k=1}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor }\subset [-\pi ,\pi )\setminus [t_{i_{0}},t_{i_{0}+1})= \bigcup _{i=1}^{s}[t_{i},t_{i+1})\setminus [t_{i_{0}},t_{i_{0}+1}).\end{aligned}$$

(4.6)

It follows from (4.6) that there exists $k\in \{1, \ldots , \lfloor \frac{2\pi }{\mu _{0}}\rfloor -1\}$ such that $\xi _{k+1}-\xi _{k}>t_{i_{0}+1}-t_{i_{0}}\ge \mu _{0}$. This contradicts with $\xi _{k+1}-\xi _{k}=\mu _{0}$. Therefore, the statement holds for the case (4.1).

Case (4.2): For the interval $[t_{s},t_{s+1})$, it follows from $\xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}=\xi _{1}+ \lfloor \frac{2\pi }{\mu _{0}}\rfloor \mu _{0}$ and $\xi _{1}=-\pi$ that

$$\begin{aligned} \pi -\xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}=2\pi - \left\lfloor \frac{2\pi }{\mu _{0}} \right\rfloor \mu _{0}= \left( \frac{2\pi }{\mu _{0}}- \left\lfloor \frac{2\pi }{\mu _{0}} \right\rfloor \right) \mu _{0}< \mu _{0},\end{aligned}$$

(4.7)

which together with $t_{s+1}=\pi$ and $t_{s+1}-t_{s}\ge \mu _{0}$ leads to $\xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}>t_{s}$. Then

$$\begin{aligned} \xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}\in [t_{s},t_{s+1}).\end{aligned}$$

(4.8)

For $i=1$, it is easy to prove that (4.5) still holds. By (4.5) and (4.8), we just need to prove the statement for the intervals $[t_{i},t_{i+1}),i=2,\ldots ,s-1$ where $s>2.$ Or else, if there exists $i_{0}\in \{2,\ldots ,s-1\}$ such that $\xi _{k}\notin [t_{i_{0}},t_{i_{0}+1})$ for any $k\in \{1, \ldots , \lfloor \frac{2\pi }{\mu _{0}}\rfloor +1\}$, then it follows from (2.3) and (4.2) that

$$\begin{aligned} \{\xi _{k}\}_{k=1}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}\subset [-\pi ,\pi )\setminus [t_{i_{0}},t_{i_{0}+1})= \bigcup _{i=1}^{s}[t_{i},t_{i+1})\setminus [t_{i_{0}},t_{i_{0}+1}). \end{aligned}$$

(4.9)

Now by (4.9), there exists $k\in \{1, \ldots , \lfloor \frac{2\pi }{\mu _{0}}\rfloor \}$ such that $\xi _{k+1}-\xi _{k}>t_{i_{0}+1}-t_{i_{0}}\ge \mu _{0}$. This contradicts with $\xi _{k+1}-\xi _{k}=\mu _{0}$. Summarizing what has addressed above, for any $i\in \{1,2, \ldots , s\}$ there exists at least a point $\xi _{k_{i}}\in [t_{i},t_{i+1})$, where $k_{i}\in \{1, \ldots , \lfloor \frac{2\pi }{\mu _{0}}\rfloor +1\}$.

Next we consider how to determine $\{d_{i}\}_{i=1}^{s}$. For any $i\in \{1,\ldots , s\}$, by the above statement there exists $\xi _{k_{i}}\in [t_{i},t_{i+1})$ where $k_{i}\in \{1, \ldots , \lfloor \frac{2\pi }{\mu _{0}}\rfloor +1\}$. Then it follows from (2.5) that $\xi _{k_{i}}+ 2m_{k_{i}}\pi \in {\mathcal {C}}^{m_{k_{i}}}_{i}$ with $m_{k_{i}}\in {\mathbb {Z}}$, which together with (2.9) leads to

$$\begin{aligned} {\widehat{f}}(\xi _{k_{i}}+ 2m_{k_{i}}\pi )=d_{i}\widehat{\hbox {sinc}_{a}}(\xi _{k_{i}}+ 2m_{k_{i}}\pi ).\end{aligned}$$

(4.10)

Based on the observed samples $\{{\widehat{f}}(\xi _{k}+2m_{k}\pi )\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}_{k=1}$, define $\{ {\widetilde{d}}_{k}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}_{k=1}$ by

$$\begin{aligned} {\widetilde{d}}_{k}:=\frac{{\widehat{f}}(\xi _{k}+2m_{k}\pi )}{\widehat{\hbox {sinc}_{a}}(\xi _{k}+2m_{k}\pi )}, \end{aligned}$$

(4.11)

where $m_{k}\in {\mathbb {Z}}$. Obviously, by (4.10) and (4.11) we have

$$\begin{aligned} \{d_{i}\}^{s}_{i=1}\subseteq \{ {\widetilde{d}}_{k}\}^{ \big \lfloor \frac{2\pi }{\mu _{0}}\big \rfloor +1}_{k=1}. \end{aligned}$$

(4.12)

It follows from $\xi _{1}\in [t_{1},t_{2})$ that $d_{1}={\widetilde{d}}_{1}$. Note that $d_{i}\ne d_{i+1}$. Then $\{d_{i}\}^{s}_{i=2}$ can be obtained iteratively by

$$\begin{aligned} d_{i}={\widetilde{d}}_{\vartheta (i)}, \end{aligned}$$

(4.13)

where $\vartheta (i)=\min \{\vartheta (i-1)< k\le (\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1) \ \hbox {s.t.} \ {\widetilde{d}}_{k}\ne d_{i-1} \}$ with $\vartheta (1)=1$.

Proof of Proposition 2.3

We first prove that $\{\eta _{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}$ increases as i does. Direct observation on (2.12) (A1-A5) gives us that we just need to prove $\xi _{k}+2\varepsilon (\mathbf{n} -1)<\xi _{k+1}$ for $k=1,2,\ldots ,\lfloor \frac{2\pi }{\mu _{0}}\rfloor$. It follows from $\mathbf{n} =\lceil \frac{\mu _{0}}{2\varepsilon }\rceil$ and $\xi _{k+1}=\xi _{k}+\mu _{0}$ that $\xi _{k+1}-\big (\xi _{k}+2\varepsilon (\mathbf{n} -1)\big )=\mu _{0}-2\varepsilon (\lceil \frac{\mu _{0}}{2\varepsilon }\rceil -1) =(\frac{\mu _{0}}{2\varepsilon }-\lceil \frac{\mu _{0}}{2\varepsilon }\rceil )2\varepsilon +2\varepsilon$. Note that $-1<\frac{\mu _{0}}{2\varepsilon }-\lceil \frac{\mu _{0}}{2\varepsilon }\rceil \le 0$. Then

$$\begin{aligned} 0<\xi _{k+1}-\big (\xi _{k}+2\varepsilon (\mathbf{n} -1)\big )\le 2\varepsilon .\end{aligned}$$

(4.14)

That is, $\xi _{k}+2\varepsilon (\mathbf{n} -1)<\xi _{k+1}$. And $\{\eta _{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}$ increases as i does.

We next prove that $|t_{k}-{\widetilde{t}}_{k}|\le \varepsilon$ for $k=2,\ldots ,s$, where ${\widetilde{t}}_{k}$ is defined by (2.14).

Note that $\{\eta _{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}\supseteq \{\xi _{k}\}_{k=1}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}$. Then it is easy to follow from Lemma 2.2, (2.13) and (4.12) that $\{d_{j}\}^{s}_{j=1}\subseteq \{ {\widehat{d}}_{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}$.

Specifically, by the similar procedure for Lemma 2.2 (4.13), $\{d_{j}\}_{j=1}^{s}$ can be determined iteratively by

$$\begin{aligned}d_{1}={\widehat{d}}_{1}, \ d_{k}={\widehat{d}}_{\tau (k)}, k=2,\ldots ,s, \end{aligned}$$

where

$$\begin{aligned} \tau (k)=\min \{\tau (k-1)< i\le \left( \left\lfloor \frac{2\pi }{\mu _{0}} \right\rfloor \mathbf{n} +1 \right) \ \hbox {s.t.} \ {\widehat{d}}_{i}\ne d_{k-1} \} \ \hbox {with} \ \tau (1)=1. \end{aligned}$$

(4.15)

Direct observation on (2.12) (A1–A5) and (4.14) gives us that for any $k\in \{1, \ldots , \lfloor \frac{2\pi }{\mu _{0}}\rfloor \}$, we have

$$\begin{aligned} \eta _{k\mathbf{n} +1}-\eta _{k\mathbf{n} }=\xi _{k+1}-\big (\xi _{k}+2\varepsilon (\mathbf{n} -1)\big )\le 2\varepsilon .\end{aligned}$$

(4.16)

And for $i=(k-1)\mathbf{n} +1,\ldots ,k\mathbf{n} -1$, we have

$$\begin{aligned} \eta _{i+1}-\eta _{i}=2\varepsilon .\end{aligned}$$

(4.17)

Now for $\varepsilon \in (0,\frac{\mu _{0}}{2}]$, (4.16) and (4.17) lead to

$$\begin{aligned} \eta _{i+1}-\eta _{i}\le 2\varepsilon \le \mu _{0}, \ i=1,2,\ldots , \left\lfloor \frac{2\pi }{\mu _{0}}\right\rfloor \mathbf{n} .\end{aligned}$$

(4.18)

It follows from (4.15) that $d_{1}={\widehat{d}}_{1}=\cdots ={\widehat{d}}_{\tau (2)-1}$, which together with $\eta _{1}=\xi _{1}\in [t_{1},t_{2})$ and (4.18) leads to $\eta _{\tau (2)-1}\in [t_{1},t_{2})$.

On the other hand, note that $\eta _{\tau (2)}-\eta _{\tau (2)-1}\le \mu _{0}$ and ${\widehat{d}}_{\tau (2)}=d_{2}\ne d_{1}$. Then using (4.15) again we have $\eta _{\tau (2)}\in [t_{2},t_{3})$. For $k=2,\ldots ,s$, we can also prove that $\eta _{\tau (k)-1}\in [t_{k-1},t_{k})$ and $\eta _{\tau (k)}\in [t_{k},t_{k+1})$. Consequently,

$$\begin{aligned} t_{k}\in (\eta _{\tau (k)-1},\eta _{\tau (k)}], \ k=2,\ldots ,s. \end{aligned}$$

(4.19)

For ${\widetilde{t}}_{k}$ defined by (2.14), we next prove that $|t_{k}-{\widetilde{t}}_{k}|\le \varepsilon$. We first prove ${\widetilde{t}}_{k}\in (-\pi ,\pi )$. By the definition of $\{\xi _{k}\}_{k=1}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}$, we have $\xi _{1}=-\pi$ and $\xi _{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}=\xi _{1}+\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mu _{0}\le \pi$. Recall that $\{\eta _{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}$ increases as i does. Then by (2.12) we have

$$\begin{aligned} -\pi \le \eta _{i}\le \pi , i=1,2,\ldots , \left\lfloor \frac{2\pi }{\mu _{0}} \right\rfloor \mathbf{n} +1. \end{aligned}$$

(4.20)

It follows from (2.14) and (4.20) that ${\widetilde{t}}_{k}\in (-\pi ,\pi )$. Now by (4.19), it is easy to prove that $|t_{k}-{\widetilde{t}}_{k}|\le \frac{\eta _{\tau (k)}-\eta _{\tau (k)-1}}{2}$. Consequently, the proof can be concluded by (4.18).

Proof of Theorem 2.5

Some lemmas for proving Theorem 2.5

Lemma 4.1

Let $\{\eta _{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}$ be defined as Proposition 2.3 (2.12), where $\mu _{0}=\min \{t_{j+1}-t_{j}: j=1,\ldots ,s\}$ with $\{t_{j}\}_{j=1}^{s+1}$ satisfying (2.3) and $\mathbf{n} =\lceil \frac{\mu _{0}}{2\varepsilon }\rceil$ with $\varepsilon \in (0,\frac{\mu _{0}}{2}]$. If $\{{\widetilde{t}}_{k}\}_{k=2}^{s}$ are constructed via (2.14) such that $|t_{k}-{\widetilde{t}}_{k}|\le \varepsilon$, then

$$\begin{aligned} {\widetilde{t}}_{k+1}-{\widetilde{t}}_{k}\ge 2\varepsilon \ \hbox {for} \ k=2,\ldots ,s-1.\end{aligned}$$

(4.21)

Proof

For any $k\in \{2,\ldots ,s-1\}$, it follows from (2.14) that

$$\begin{aligned} {\widetilde{t}}_{k+1}-{\widetilde{t}}_{k}=\frac{\eta _{\tau (k+1)}+\eta _{\tau (k+1)-1}-\eta _{\tau (k)}-\eta _{\tau (k)-1}}{2}. \end{aligned}$$

(4.22)

Now by Proposition 2.3, $\{\eta _{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}$ increases as i does which together with (4.15) leads to that

$\eta _{\tau (k+1)-1}\ge \eta _{\tau (k)}$. Invoking (4.22), we have

$$\begin{aligned} {\widetilde{t}}_{k+1}-{\widetilde{t}}_{k}\ge \frac{\eta _{\tau (k+1)}-\eta _{\tau (k)-1}}{2}.\end{aligned}$$

(4.23)

We next prove that ${\widetilde{t}}_{k+1}-{\widetilde{t}}_{k}\ge 2\varepsilon$ for the three cases: $\varepsilon =\frac{\mu _{0}}{2}$, $\varepsilon \in (\frac{\mu _{0}}{4},\frac{\mu _{0}}{2})$ and $\varepsilon \in (0,\frac{\mu _{0}}{4}]$.

If $\varepsilon =\frac{\mu _{0}}{2}$ then $\mathbf{n} =\lceil \frac{\mu _{0}}{2\varepsilon }\rceil =1$, which together with (2.12) leads to $\{\eta _{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}=\{\xi _{i}\}_{i=1}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor +1}$.

Note that $\xi _{i+1}-\xi _{i}=\mu _{0}$ for $i=1,\ldots ,\lfloor \frac{2\pi }{\mu _{0}}\rfloor$. Then by (4.15) we have

$$\begin{aligned} \eta _{\tau (k+1)}-\eta _{\tau (k)-1}\ge 2\mu _{0}=4\varepsilon . \end{aligned}$$

(4.24)

It follows from (4.23) and (4.24) that ${\widetilde{t}}_{k+1}-{\widetilde{t}}_{k}\ge 2\varepsilon .$

If $\varepsilon \in (\frac{\mu _{0}}{4},\frac{\mu _{0}}{2})$ then $\mathbf{n} =\lceil \frac{\mu _{0}}{2\varepsilon }\rceil =2$. Allowing for $\mathbf{n} =2$ and (2.12), we have

$$\begin{aligned} \{\eta _{i}\}^{\lfloor \frac{2\pi }{\mu _{0}}\rfloor \mathbf{n} +1}_{i=1}= \left\{ \xi _{1},\xi _{1}+2\varepsilon ,\xi _{2},\xi _{2}+2\varepsilon ,\ldots , \xi _{\big \lfloor \frac{2\pi }{\mu _{0}}\big \rfloor }+2\varepsilon ,\xi _{\big \lfloor \frac{2\pi }{\mu _{0}} \big \rfloor +1} \right\} . \end{aligned}$$

(4.25)

And by (4.19), we have

$$\begin{aligned} \eta _{\tau (k+1)}-\eta _{\tau (k)-1}>t_{k+1}-t_{k}\ge \mu _{0}. \end{aligned}$$

(4.26)

That is, $\eta _{\tau (k+1)}-\eta _{\tau (k)-1}>\mu _{0}$.

Consequently, it follows from (4.25), (4.26) and $\xi _{i+1}-\xi _{i}=\mu _{0}$ for $i=1,\ldots ,\lfloor \frac{2\pi }{\mu _{0}}\rfloor$ that

$$\begin{aligned} \eta _{\tau (k+1)}-\eta _{\tau (k)-1}\ge \mu _{0}+2\varepsilon >4\varepsilon .\end{aligned}$$

(4.27)

Then (4.23) and (4.27) lead to ${\widetilde{t}}_{k+1}-{\widetilde{t}}_{k}>2\varepsilon .$

If $\varepsilon \in (0,\frac{\mu _{0}}{4}]$ then it follows from $|t_{k}-{\widetilde{t}}_{k}|\le \varepsilon$, $|t_{k+1}-{\widetilde{t}}_{k+1}|\le \varepsilon$ and $t_{k+1}-t_{k}\ge \mu _{0}$ that

$$\begin{aligned}{\widetilde{t}}_{k+1}-{\widetilde{t}}_{k}\ge t_{k+1}-t_{k}-2\varepsilon \ge 2\varepsilon . \end{aligned}$$

The proof is concluded. $\square$

Lemma 4.2

Suppose that $f\in V(\hbox {sinc}_{a})$ has the spline spectra such that (1.15) holds with $d_{k}\ne d_{k+1}$ and the spectra knots $t_{i}, i=1,\ldots ,s+1$ satisfying (2.3). Denote ${\widetilde{t}}_{1}:=-\pi$ and ${\widetilde{t}}_{s+1}:=\pi$. Let $\varepsilon$ (being sufficiently small) and $\{{\widetilde{t}}_{k}\}^{s}_{k=2}$ (being the approximation to $\{t_{k}\}^{s}_{k=2}$) be as in Proposition 2.3. Define ${\widetilde{f}}$ via (2.6) with $\{t_{i}\}^{s+1}_{i=1}$ being replaced by $\{{\widetilde{t}}_{i}\}^{s+1}_{i=1}$. Then for any fixed $k\in \{2,\ldots ,s\}$, we have

$$\begin{aligned}&\displaystyle \sum _{m\in {\mathbb {Z}}\setminus \{0\}}\Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}([{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi ))}^{2}\\&\quad \le 2\varepsilon (1+a)^{2}\sum _{m\in {\mathbb {Z}}\setminus \{0\}}a^{4|m|-2}(|d_{k-1}|+|d_{k}|)^{2}. \end{aligned}$$

Proof

We first prove that

$$\begin{aligned}{}[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )\subseteq [t_{k-1},t_{k+1})\end{aligned}$$

(4.28)

and

$$\begin{aligned}{}[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )\subseteq [{\widetilde{t}}_{k-1},{\widetilde{t}}_{k+1})\end{aligned}$$

(4.29)

for $k=2,\ldots ,s.$ Recall in Proposition 2.3 that $\varepsilon \in (0,\frac{\mu _{0}}{2}]$ with

$$\begin{aligned} \mu _{0}=\min \{t_{i+1}-t_{i}: i=1,\ldots ,s\},\end{aligned}$$

(4.30)

and $\{{\widetilde{t}}_{k}\}_{k=2}^{s}$ are constructed via (2.14) such that $|t_{k}-{\widetilde{t}}_{k}|\le \varepsilon \ \hbox {for}\ k=2,\ldots ,s.$ Clearly,

$$\begin{aligned} {\widetilde{t}}_{k}-\varepsilon \le t_{k}\le {\widetilde{t}}_{k}+\varepsilon , \ \hbox {for}\ k=2,\ldots ,s.\end{aligned}$$

(4.31)

It follows from (4.30) that

$$\begin{aligned} t_{i+1}-t_{i}\ge \mu _{0}\ge 2\varepsilon \ \hbox {for}\ i=1,\ldots ,s.\end{aligned}$$

(4.32)

Now by (4.31) and (4.32), it is easy to check that $t_{k-1}\le {\widetilde{t}}_{k}-\varepsilon$ and ${\widetilde{t}}_{k}+\varepsilon \le t_{k+1}$. Then (4.28) holds. We next prove that (4.29) holds. Recall that $\{{\widetilde{t}}_{k}\}_{k=2}^{s}$ are constructed via (2.14) such that $|t_{k}-{\widetilde{t}}_{k}|\le \varepsilon$. Then by Lemma 4.1 (4.21), we have

$$\begin{aligned} {\widetilde{t}}_{i}<{\widetilde{t}}_{i}+\varepsilon \le {\widetilde{t}}_{i+1}-\varepsilon <{\widetilde{t}}_{i+1},\ i=2,\ldots ,s-1, \end{aligned}$$

(4.33)

where $s>2$. For $k=2$, by ${\widetilde{t}}_{1}=-\pi =t_{1}$, $|t_{2}-{\widetilde{t}}_{2}|\le \varepsilon$ and $t_{2}-t_{1}\ge 2\varepsilon$ we have

$$\begin{aligned} {\widetilde{t}}_{2}+\varepsilon -{\widetilde{t}}_{1}\ge t_{2}-t_{1}\ge 2\varepsilon .\end{aligned}$$

(4.34)

Then ${\widetilde{t}}_{1}\le {\widetilde{t}}_{2}-\varepsilon$. Now (4.33) and (4.34) with $i=2$ lead to $[{\widetilde{t}}_{2}-\varepsilon ,{\widetilde{t}}_{2}+\varepsilon )\subseteq [{\widetilde{t}}_{1},{\widetilde{t}}_{3})$. For $k=s$, by ${\widetilde{t}}_{s+1}=\pi =t_{s+1}$, $|t_{s}-{\widetilde{t}}_{s}|\le \varepsilon$ and $t_{s+1}-t_{s}\ge 2\varepsilon$ we have

$$\begin{aligned} {\widetilde{t}}_{s+1}-({\widetilde{t}}_{s}-\varepsilon ) \ge t_{s+1}-t_{s}\ge 2\varepsilon .\end{aligned}$$

(4.35)

Namely, ${\widetilde{t}}_{s}+\varepsilon \le {\widetilde{t}}_{s+1}$. Then (4.33) and (4.35) with $i=s-1$ lead to $[{\widetilde{t}}_{s}-\varepsilon ,{\widetilde{t}}_{s}+\varepsilon )\subseteq [{\widetilde{t}}_{s-1},{\widetilde{t}}_{s+1})$. On the other hand, the direct observation on (4.33) with $i=2,\ldots ,s-1$ gives us that (4.29) holds for $k=3,\ldots ,s-1$. Summarizing what has been addressed above, (4.29) holds.

We next estimate $\sum _{m\in {\mathbb {Z}}\setminus \{0\}}\Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}([{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi ))}^{2}$ for the two cases: ${\widetilde{t}}_{k}\le t_{k}$ and ${\widetilde{t}}_{k}> t_{k}$. By (1.15) and the definition of ${\widetilde{f}}$, for any $\xi \in {\mathbb {R}}$ we have

$$\begin{aligned} \left\{ \begin{array}{llllll} {\widehat{f}}(\xi )=\left( \sum _{l\in {\mathbb {Z}}}\sum _{k=1}^{s}d_{k}\chi _{[t_{k}+2l\pi ,t_{k+1}+2l\pi )}(\xi )\right) \widehat{\hbox {sinc}_{a}}(\xi ),\\ \widehat{{\widetilde{f}}}(\xi )=\left( \sum _{l\in {\mathbb {Z}}}\sum _{k=1}^{s}d_{k}\chi _{[{\widetilde{t}}_{k}+2l\pi ,{\widetilde{t}}_{k+1}+2l\pi )}(\xi )\right) \widehat{\hbox {sinc}_{a}}(\xi ). \end{array}\right. \end{aligned}$$

(4.36)

If ${\widetilde{t}}_{k}\le t_{k}$, then it follows from (4.28) and the first equation in (4.36) that

$$\begin{aligned} \begin{array}{lll} {\widehat{f}}\chi _{[{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi )}(\xi ) &{}=&{}{\widehat{f}}\chi _{[{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+2m\pi )}(\xi ) +{\widehat{f}}\chi _{[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )}(\xi )\\ &{} \ &{}+{\widehat{f}}\chi _{[t_{k}+2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi )}(\xi )\\ &{}=&{}d_{k-1}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+2m\pi )}(\xi )\\ &{}\ &{}+d_{k-1}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )}(\xi )\\ &{} \ &{}+d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[t_{k}+2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi )}(\xi ). \end{array}\end{aligned}$$

(4.37)

And by (4.29) and the second equation in (4.36), we have

$$\begin{aligned} \begin{array}{lll} \widehat{{\widetilde{f}}}\chi _{[{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi )}(\xi ) &{}=&{} d_{k-1}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+2m\pi )}(\xi )\\ &{}\ &{}+d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )}(\xi )\\ &{}\ &{}+d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[t_{k}+2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi )}(\xi ). \end{array}\end{aligned}$$

(4.38)

Correspondingly, (4.37) and (4.38) lead to

$$\begin{aligned} \begin{array}{lll} ({\widehat{f}}-\widehat{{\widetilde{f}}})\chi _{[{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi )}(\xi ) =(d_{k-1}-d_{k})\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )}(\xi ). \end{array}\end{aligned}$$

(4.39)

Direct observation on (4.39) gives us that we need to estimate

$$\begin{aligned}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )}(\xi ).\end{aligned}$$

For any $m\in {\mathbb {Z}}\setminus \{0\}$, define

$$\begin{aligned} T^{1}_{m}:=[(2m-1)\pi ,2m\pi ), \ T^{2}_{m}:=[2m\pi ,(2m+1)\pi ). \end{aligned}$$

(4.40)

From (1.1) and (4.40) we arrive at

$$\begin{aligned} \big (T^{1}_{m}\cup T_{m}^{2}\big )\subset \big ( I_{2|m|-1}\cup I_{2|m|}\big ).\end{aligned}$$

(4.41)

For any fixed $k\in \{2,\ldots ,s\}$, it follows from (2.14) that ${\widetilde{t}}_{k}\in (-\pi ,\pi )$, which together with (2.3) and ${\widetilde{t}}_{k}\le t_{k}$ leads to $-\pi<{\widetilde{t}}_{k}\le t_{k}<\pi$. Then by (4.40) we have

$$\begin{aligned}{}[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )\subset [-\pi +2m\pi ,\pi +2m\pi )=T_{m}^{1}\cup T_{m}^{2}.\end{aligned}$$

(4.42)

Now combining (4.41) and (4.42) we have

$$\begin{aligned}{}[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )\subset \big (I_{2|m|-1}\cup I_{2|m|}\big ).\end{aligned}$$

(4.43)

Note that by (1.2) we have

$$\begin{aligned} \widehat{\hbox {sinc}_{a}}(\xi )=\left\{ \begin{array}{lllll}(1+a)a^{2|m|-1}, &{} \quad \xi \in I_{2|m|-1},\\ (1+a)a^{2|m|}, &{} \quad \xi \in I_{2|m|}, \end{array}\right. \end{aligned}$$

(4.44)

where $a\in (-1,1)$.

Then it follows from (4.43) and (4.44) that

$$\begin{aligned} \widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )}(\xi )\le (1+a)a^{2|m|-1}\chi _{[{\widetilde{t}}_{k}+2m\pi ,t_{k}+2m\pi )}(\xi ).\end{aligned}$$

(4.45)

Consequently, by (4.39) and (4.45) we have

$$\begin{aligned} \begin{array}{lll} &{}\displaystyle \sum _{m\in {\mathbb {Z}}\setminus \{0\}}\Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}([{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi ))}^{2}\\ &{} \quad \displaystyle \le \sum _{m\in {\mathbb {Z}}\setminus \{0\}}|t_{k}-{\widetilde{t}}_{k}|(1+a)^{2}a^{4|m|-2}|d_{k-1}-d_{k}|^{2}\\ &{} \quad \displaystyle \le 2\varepsilon (1+a)^{2}\sum _{m\in {\mathbb {Z}}\setminus \{0\}} a^{4|m|-2}(|d_{k-1}|+|d_{k}|)^{2}. \end{array}\end{aligned}$$

(4.46)

If ${\widetilde{t}}_{k}> t_{k}$, then by the similar procedures for ${\widetilde{t}}_{k}\le t_{k}$, we can prove that (4.46) still holds for any fixed $k\in \{2,\ldots ,s\}$. The proof is concluded. $\square$

Proof of Theorem 2.5

Note that ${\widetilde{t}}_{1}=-\pi$ and ${\widetilde{t}}_{s+1}=\pi$. Then by Lemma 4.2 (4.29), we have $-\pi ={\widetilde{t}}_{1}<{\widetilde{t}}_{2}<\ldots<{\widetilde{t}}_{s}<{\widetilde{t}}_{s+1}=\pi$. Consequently,

$$\begin{aligned}{}[-\pi ,\pi )=\bigcup _{i=1}^{s}[{\widetilde{t}}_{i},{\widetilde{t}}_{i+1}).\end{aligned}$$

(4.47)

Define $Q_{1}:=\cup _{k=2}^{s}[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )$ and

$$\begin{aligned} Q_{2}:=\left\{ \begin{array}{rcllc} \displaystyle [{\widetilde{t}}_{1},{\widetilde{t}}_{2}-\varepsilon )\cup [{\widetilde{t}}_{2}+\varepsilon ,{\widetilde{t}}_{3}), &{}&{}s=2\\ \displaystyle [{\widetilde{t}}_{1},{\widetilde{t}}_{2}-\varepsilon ) \cup [{\widetilde{t}}_{s}+\varepsilon ,{\widetilde{t}}_{s+1})\cup \{\cup _{k=2}^{s-1}[{\widetilde{t}}_{k}+\varepsilon ,{\widetilde{t}}_{k+1}-\varepsilon )\}, &{}&{} s>2, \end{array}\right. \end{aligned}$$

(4.48)

where $\varepsilon \in (0,\frac{\mu _{0}}{2}]$. Incidentally, for (4.33), (4.34) and (4.35), $Q_{2}$ is well defined. By (4.47), it is easy to check that

$$\begin{aligned} \begin{array}{lll} [-\pi ,\pi ) = Q_{1}\cup Q_{2}. \end{array} \end{aligned}$$

(4.49)

Next we prove that

$$\begin{aligned} {\widehat{f}}(\xi )=\widehat{{\widetilde{f}}}(\xi ) \end{aligned}$$

(4.50)

for $\xi \in Q_{2}+2l\pi$ with any $l\in {\mathbb {Z}}$. For $s\ge 2$, recall that $\{{\widetilde{t}}_{k}\}_{k=2}^{s}$ is constructed via (2.14) such that $|t_{k}-{\widetilde{t}}_{k}|\le \varepsilon$. Then by $|t_{2}-{\widetilde{t}}_{2}|\le \varepsilon$ and $|t_{s}-{\widetilde{t}}_{s}|\le \varepsilon$ we have ${\widetilde{t}}_{2}-\varepsilon \le t_{2}$ and $t_{s}\le {\widetilde{t}}_{s}+\varepsilon$, which together with ${\widetilde{t}}_{1}=t_{1}$ and ${\widetilde{t}}_{s+1}=t_{s+1}$ lead to

$$\begin{aligned}{}[{\widetilde{t}}_{1},{\widetilde{t}}_{2}-\varepsilon )\subseteq [t_{1},t_{2}),\ [{\widetilde{t}}_{s}+\varepsilon ,{\widetilde{t}}_{s+1})\subseteq [t_{s},t_{s+1}).\end{aligned}$$

(4.51)

Now combining (4.51) and Lemma 4.2 (4.36) we have

$$\begin{aligned} \left\{ \begin{array}{lll} \widehat{{\widetilde{f}}}\chi _{[{\widetilde{t}}_{1}+2l\pi ,{\widetilde{t}}_{2}-\varepsilon +2l\pi )}(\xi ) &{}={\widehat{f}}\chi _{[{\widetilde{t}}_{1}+2l\pi ,{\widetilde{t}}_{2}-\varepsilon +2l\pi )}(\xi )\\ &{}=d_{1}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{1}+2l\pi ,{\widetilde{t}}_{2}-\varepsilon +2l\pi )}(\xi ),\\ \widehat{{\widetilde{f}}}\chi _{[{\widetilde{t}}_{s}+\varepsilon +2l\pi ,{\widetilde{t}}_{s+1}+2l\pi )}(\xi ) &{}={\widehat{f}}\chi _{[{\widetilde{t}}_{s}+\varepsilon +2l\pi ,{\widetilde{t}}_{s+1}+2l\pi )}(\xi )\\ &{}=d_{s}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{s}+\varepsilon +2l\pi ,{\widetilde{t}}_{s+1}+2l\pi )}(\xi ). \end{array}\right. \end{aligned}$$

(4.52)

That is, (4.50) holds for $s=2$, or for $s>2$ and $\xi \in \Big ([{\widetilde{t}}_{1},{\widetilde{t}}_{2}-\varepsilon ) \cup [{\widetilde{t}}_{s}+\varepsilon ,{\widetilde{t}}_{s+1})\Big )+2l\pi$. By (4.48), to complete the proof of (4.50) we just need to prove it for $s>2$ and

$$\begin{aligned}\xi \in \bigcup _{k=2}^{s-1}[{\widetilde{t}}_{k}+\varepsilon +2l\pi ,{\widetilde{t}}_{k+1}-\varepsilon +2l\pi ). \end{aligned}$$

For $k\in \{2,\ldots ,s-1\}$, it is easy to derive from $|t_{k}-{\widetilde{t}}_{k}|\le \varepsilon$ and $|t_{k+1}-{\widetilde{t}}_{k+1}|\le \varepsilon$ that

$$\begin{aligned}{}[{\widetilde{t}}_{k}+\varepsilon ,{\widetilde{t}}_{k+1}-\varepsilon )\subseteq [t_k, t_{k+1}). \end{aligned}$$

(4.53)

By (4.36), the restriction of ${\widehat{f}}$ on the interval $[t_k, t_{k+1})+2l\pi$ is

$$\begin{aligned}{\widehat{f}}\chi _{[t_{k}+2l\pi ,t_{k+1}+2l\pi )}(\xi ) =d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[t_{k}+2l\pi ,t_{k+1}+2l\pi )}(\xi ),\end{aligned}$$

which together with (4.53) leads to

$$\begin{aligned} {\widehat{f}}\chi _{[{\widetilde{t}}_{k}+\varepsilon +2l\pi ,{\widetilde{t}}_{k+1}-\varepsilon +2l\pi )}(\xi ) =d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}+\varepsilon +2l\pi ,{\widetilde{t}}_{k+1}-\varepsilon +2l\pi )}(\xi ). \end{aligned}$$

(4.54)

Using (4.36) again, the restriction of $\widehat{{\widetilde{f}}}$ on $[{\widetilde{t}}_k, {\widetilde{t}}_{k+1})+2l\pi$ is

$$\begin{aligned}\widehat{{\widetilde{f}}}\chi _{[{\widetilde{t}}_{k}+2l\pi ,{\widetilde{t}}_{k+1}+2l\pi )}(\xi ) =d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}+2l\pi ,{\widetilde{t}}_{k+1}+2l\pi )}(\xi ),\end{aligned}$$

which together with $[{\widetilde{t}}_{k}+\varepsilon ,{\widetilde{t}}_{k+1}-\varepsilon )\subset [{\widetilde{t}}_k, {\widetilde{t}}_{k+1})$ leads to that

$$\begin{aligned} \widehat{{\widetilde{f}}}\chi _{[{\widetilde{t}}_{k}+\varepsilon +2l\pi ,{\widetilde{t}}_{k+1}-\varepsilon +2l\pi )}(\xi ) =d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}+\varepsilon +2l\pi ,{\widetilde{t}}_{k+1}-\varepsilon +2l\pi )}(\xi ). \end{aligned}$$

(4.55)

Then it follows from (4.54) and (4.55) that

$$\begin{aligned} \begin{array}{lll} \widehat{{\widetilde{f}}}\chi _{[{\widetilde{t}}_{k}+\varepsilon +2l\pi ,{\widetilde{t}}_{k+1}-\varepsilon +2l\pi )}(\xi ) ={\widehat{f}}\chi _{[{\widetilde{t}}_{k}+\varepsilon +2l\pi ,{\widetilde{t}}_{k+1}-\varepsilon +2l\pi )}(\xi ). \end{array}\end{aligned}$$

(4.56)

Summarizing what has been addressed above, (4.50) holds.

Next we estimate $\Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}({\mathbb {R}})}^{2}$. By (4.49) and (4.50), it is easy to check that

$$\begin{aligned} \begin{array}{lll} \Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}({\mathbb {R}})}^{2} =P_{1}+P_{2}, \end{array}\end{aligned}$$

(4.57)

where

$$\begin{aligned} \left\{ \begin{array}{lll} P_{1}=\sum _{k=2}^{s}\Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}([{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon ))}^{2},\\ P_{2}=\sum _{m\in {\mathbb {Z}}\setminus \{0\}}\sum _{k=2}^{s}\Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}([{\widetilde{t}}_{k}-\varepsilon +2m\pi ,{\widetilde{t}}_{k}+\varepsilon +2m\pi ))}^{2}. \end{array}\right. \end{aligned}$$

(4.58)

Note that $\{t_{k},{\widetilde{t}}_{k}\}\subset [{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )$ for $k\in \{2,\ldots ,s\}$. Motivated by this, $P_{1}$ is estimated for the two cases: ${\widetilde{t}}_{k}\le t_{k}$ and ${\widetilde{t}}_{k}> t_{k}$. If ${\widetilde{t}}_{k}\le t_{k}$, then it follows from (4.28) and the first equation in (4.36) that

$$\begin{aligned} \begin{array}{lll} {\widehat{f}}\chi _{[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )}(\xi ) &{}=&{}{\widehat{f}}\chi _{[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k})}(\xi )+ {\widehat{f}}\chi _{[{\widetilde{t}}_{k},t_{k})}(\xi ) +{\widehat{f}}\chi _{[t_{k},{\widetilde{t}}_{k}+\varepsilon )}(\xi )\\ &{}=&{}d_{k-1}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k})}(\xi ) +d_{k-1}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k},t_{k})}(\xi )\\ &{}\ &{}+d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[t_{k},{\widetilde{t}}_{k}+\varepsilon )}(\xi ). \end{array}\end{aligned}$$

(4.59)

On the other hand, by (4.29) and the second equation in (4.36) we have

$$\begin{aligned} \begin{array}{lll} \widehat{{\widetilde{f}}}\chi _{[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )}(\xi )&{}=d_{k-1}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k})}(\xi ) +d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k},t_{k})}(\xi )\\ &{} \quad +d_{k}\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[t_{k},{\widetilde{t}}_{k}+\varepsilon )}(\xi ), \end{array}\end{aligned}$$

which together with (4.59) leads to

$$\begin{aligned} ({\widehat{f}}-\widehat{{\widetilde{f}}})\chi _{[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )}(\xi )= (d_{k-1}-d_{k})\widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k},t_{k})}(\xi ). \end{aligned}$$

(4.60)

Direct observation on (4.48) and (4.49) gives us that $[{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )\subset [-\pi ,\pi )$. Recall that $[{\widetilde{t}}_{k},t_{k})\subset [{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon )$. Then $[{\widetilde{t}}_{k},t_{k})\subset [-\pi ,\pi )$. Clearly, by (1.2) with $n=0$ we have

$$\begin{aligned} \widehat{\hbox {sinc}_{a}}(\xi )\chi _{[{\widetilde{t}}_{k},t_{k})}(\xi )=(1+a)\chi _{[{\widetilde{t}}_{k},t_{k})}(\xi ).\end{aligned}$$

(4.61)

Now (4.60) and (4.61) lead to

$$\begin{aligned} \begin{array}{lll} \Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}([{\widetilde{t}}_{k}-\varepsilon ,{\widetilde{t}}_{k}+\varepsilon ))}^{2} &{}=|t_{k}-{\widetilde{t}}_{k}|(1+a)^{2}|d_{k-1}-d_{k}|^{2}\\ &{}\le 2\varepsilon (1+a)^{2}(|d_{k-1}|+|d_{k}|)^{2}. \end{array}\end{aligned}$$

Consequently,

$$\begin{aligned} P_{1}\le 2\varepsilon (1+a)^{2}\sum _{k=2}^{s}(|d_{k-1}|+|d_{k}|)^{2}.\end{aligned}$$

(4.62)

If ${\widetilde{t}}_{k}> t_{k}$, then by the similar procedures for ${\widetilde{t}}_{k}\le t_{k}$, (4.62) still holds. For the second term $P_{2}$, it follows from Lemma 4.2 that

$$\begin{aligned} \begin{array}{lll} P_{2}\le 2\varepsilon (1+a)^{2}\sum _{m\in {\mathbb {Z}}\setminus \{0\}}a^{4|m|-2}\sum _{k=2}^{s}(|d_{k-1}|+|d_{k}|)^{2}. \end{array}\end{aligned}$$

(4.63)

Now from (4.57), (4.62) and (4.63) we arrive at

$$\begin{aligned} \begin{array}{llllll} \Vert f-{\widetilde{f}}\Vert _{L^{2}({\mathbb {R}})}^{2}&{}=&{}\frac{1}{2\pi }\Vert {\widehat{f}}-\widehat{{\widetilde{f}}}\Vert _{L^{2}({\mathbb {R}})}^{2}\\ &{}\le &{}\displaystyle \frac{\varepsilon }{\pi }(1+a)^{2}\left( 1+\sum _{m\in {\mathbb {Z}}\setminus \{0\}}a^{4|m|-2}\right) \sum _{k=2}^{s}(|d_{k-1}|+|d_{k}|)^{2}\\ &{}=&{}\displaystyle \frac{\varepsilon }{\pi }\left\{ \frac{1+a}{1-a}+ \frac{a^{2}(1+a)^{2}}{1+a^{2}}\right\} \sum _{k=2}^{s}(|d_{k-1}|+|d_{k}|)^{2}. \end{array}\end{aligned}$$

(4.64)