In this section, we introduce concrete examples according to whether the order of vanishing moments is finite or infinite.
• Case of finite vanishing moments) Let us consider the function and the wavelet
$$f(x) = \operatorname {sech}(h x), \quad \quad\psi(x) =\frac{d }{dx} \operatorname {sech}(h x). $$
In particular, when \(h = \sqrt{\frac{\pi}{2}}\), it holds that \(\hat{f}( \xi) = f(\xi)\), and we also see that \(\hat{\psi}( \xi)=i\xi f(\xi) \) and \(\psi\in S^{1, \infty}_{1,h' }(\mathbf{R}) \) with \(0< h' <h\). By the change of variables \(t =e^{2hax}\), we have the wavelet transform
$$\begin{aligned} W_{\psi}f (a,b) = &\sqrt{\frac{a }{ C_{\psi}}} \int_{\mathbf{R}} \psi(x)f(ax+b) \,dx \\ = &-4h\sqrt{\frac{a }{ C_{\psi}}} \int_{\mathbf{R}} \frac{e^{hx} -e^{-hx}}{(e^{hx} +e^{-hx})^{2}}\cdot\frac {1}{e^{h(ax+b) } +e^{-h(ax+b)} } \,dx \\ = &- \frac{2 }{e^{hb} \sqrt{C_{\psi}a}} \int_{0}^{\infty}\frac{t^{1/a}-1}{(t^{1/a}+1)^{2}} \cdot \frac{dt}{t^{\frac{a-1}{2a} } (t +e^{-2hb})} , \end{aligned}$$
where
$$C_{\psi}= \int_{\mathbf{R}} \frac{ \vert \hat{\psi}(\xi) \vert ^{2}}{ \vert \xi \vert } \,d\xi\quad ( =2\log2 ) . $$
Using the Hölder inequality \(A +B \geq\frac {p}{(p-1)^{1-1/p}}A^{1/p}B^{1-1/p}\) with
$$1< \frac{2}{1+ \frac{h' }{h } }< p_{\pm}:=\frac{2}{1\pm\frac{h' }{h\max \{1,a\} } } < \frac{2}{1- \frac{h' }{h } } < \infty \quad \text{for } \operatorname {sgn}b =\pm1 \text{ respectively} , $$
we obtain the estimate (from above)
$$\begin{aligned}& \bigl\vert W_{\psi}f (a,b) \bigr\vert \\& \quad \leq \frac{2 }{e^{hb} \sqrt {C_{\psi}a}} \int_{0}^{\infty}\frac{1}{t^{1/a}+1 } \cdot \frac{dt}{t^{\frac{a-1}{2a} } \cdot\frac{p_{\pm}}{(p_{\pm}-1)^{1-1/ p_{\pm}}} t^{1/p_{\pm}} e^{-2hb(1-1/p_{\pm})}} \\& \quad = \frac{2 (p_{\pm}-1)^{1-1/p_{\pm}}e^{ hb(1-2/p_{\pm})} }{p_{\pm}\sqrt {C_{\psi}a}} \int_{0}^{\infty}\frac{1}{t^{1/a}+1 } \cdot \frac{dt}{t^{\frac{a-1}{2a} +\frac{1}{p_{\pm}} } } \\& \quad \leq \frac{C_{h' }e^{ hb(1-2/ p_{\pm})} }{ \sqrt{ a}} \int_{0}^{\infty}\frac{1}{t^{1/a}+1 } \cdot \frac{dt}{t^{ 1- \frac{1}{2a} \pm\frac{h' }{2h \max\{1,a\} } }} \\& \quad \leq \frac{C_{h' }e^{ hb(1-2/ p_{\pm})} }{ \sqrt{ a}} \biggl\{ \int_{0}^{1}\frac{1}{0+1 } \cdot \frac{dt}{t^{ 1- \frac{1}{2a} \pm\frac{h }{2h\max\{1,a\} } }} + \int_{1}^{\infty}\frac{1}{t^{1/a}+0 } \cdot \frac{dt}{t^{ 1- \frac{1}{2a} \pm\frac{h' }{2h \max\{1,a\} } }} \biggr\} \\& \quad = \frac{C_{h' }e^{ hb(1-2/ p_{\pm})} }{ \sqrt{ a}} \biggl\{ \frac {1}{\frac{1}{2a} \pm\frac{h' }{2h \max\{1,a\} } } +\frac{1}{\frac {1}{2a} \mp\frac{h' }{2h \max\{1,a\} }} \biggr\} \\& \quad = \frac{C_{h' }e^{ hb(1-2/ p_{\pm})} }{ \sqrt{ a}} \frac{4a }{1- ( \frac{h' a }{h \max\{1,a\} })^{2}} \\& \quad \leq C_{ h' }a^{1/2}e^{ -h' \vert b \vert / \max\{1,a\} } , \end{aligned}$$
(5)
here we used the fact that
$$1- \frac{1}{2a} \pm\frac{h' }{2h \max\{1,a\} } < 1 , \quad \quad 1+\frac {1}{2a} \pm \frac{h' }{2h \max\{1,a\} } >1. $$
Then (i)′ and (ii)′ in Theorem 2.4 become
$$\begin{aligned}& (\mathrm{i})' \quad \bigl\Vert e^{ h' \vert b/(a+1) \vert } W_{\psi}f \bigr\Vert _{L^{\infty}(\mathbf{R}_{+} \times \mathbf{R}) } \leq C \bigl\Vert e^{h \vert x \vert }f \bigr\Vert _{L^{\infty}(\mathbf{R}) }, \\& (\mathrm{ii})'\quad \bigl\Vert a^{-1/2} e^{ h' } W_{\psi}f \bigr\Vert _{L^{\infty}(\mathbf{R}_{+} \times \mathbf{R}) } \leq C \bigl\Vert e^{h \vert \xi \vert }\hat{f} \bigr\Vert _{L^{\infty}(\mathbf{R}) } . \end{aligned}$$
From estimate (5) it is possible that this example is the near critical case of \((\mathrm{i})'\) and \((\mathrm{ii})'\) since \(\vert b/(a+1) \vert \sim \vert b \vert / \max\{1,a\} \).
Remark 4.1
If we consider the typical example of the Mexican hat wavelet
$$\psi(x) =\frac{2}{\pi^{1/4}\sqrt{3}} \bigl(1 -x^{2} \bigr) e^{-x^{2}/2}, \quad \quad\hat{\psi}(\xi) =\frac{2\sqrt{2\pi}}{\pi^{1/4}\sqrt{3}}\xi^{2} e^{-\xi^{2}/2}, $$
we see that \(\psi\in S^{1/2, \infty}_{1/2,h'}(\mathbf{R}) \) with \(0< h' <h=1/2\). In particular, when \(f(x) =e^{-x^{2}/2}\in S^{1/2, \infty}_{1/2,h}(\mathbf {R}) \), the wavelet transform is computed as
$$W_{\psi}f (a,b) =\frac{2\sqrt{2}\pi^{1/4}a^{5/2}(a^{2}-1-b^{2})}{\sqrt{3 C_{\psi}}(a^{2}+1)^{5/2}}e^{-b^{2}/(2a^{2}+2)}. $$
Then (i)′ in Theorem 2.4 becomes
$$(\mathrm{i})' \quad \bigl\Vert e^{ h' 2^{-1} \vert b/(a+1) \vert ^{2} } W_{\psi}f \bigr\Vert _{L^{\infty}(\mathbf{R}_{+} \times \mathbf{R}) } \leq C \bigl\Vert e^{h \vert x \vert ^{2}}f \bigr\Vert _{L^{\infty}(\mathbf{R}) }. $$
The exponent \(-b^{2}/(2a^{2}+2)\) is not a critical case of \((\mathrm{i})'\) with \(0< h' <h=\frac{1}{2}\) since \(h'2^{-1} \vert b/(a+1) \vert ^{2} \sim b^{2}/(4a^{2}+4)\). Therefore, we gave the new wavelet \(\psi(x) =\frac{d }{dx} \operatorname {sech}(h x) \in S^{1, \infty}_{1,h' }(\mathbf{R}) \) with \(0< h' <h= \sqrt{\frac{\pi}{2}}\).
• Case of infinite vanishing moments) Firstly we prove the following.
Proposition 4.2
The inverse Fourier transform of
\(e^{-\xi^{2}-t^{2}\xi^{-2}}\)
is given by
$$ {\mathcal {F}}^{-1} \bigl[e^{-\xi^{2}-t^{2}\xi^{-2}} \bigr](x) = \frac{1}{ \sqrt{2}} \sum_{n=0}^{\infty}\frac{(-2t)^{n}}{n!} {}_{1}F_{1} \biggl(\frac{1-n}{2}, \frac{1}{2}, -\frac{x^{2}}{4} \biggr), $$
(6)
where
\({}_{1}F_{1} (a,b,z )\)
is the confluent hypergeometric function of the first kind.
Remark 4.3
The change of variables also yields
$$\begin{aligned} {\mathcal {F}}^{-1} \bigl[e^{-\xi^{2}/2-t^{2}\xi^{-2}} \bigr](x) =& \sqrt{ \frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}/2-t^{2}\xi^{-2}}\cos x\xi \,d\xi \\ =& \frac{2}{\sqrt{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-(t/\sqrt{2})^{2}\xi^{-2}}\cos(\sqrt{2}x)\xi \,d \xi \\ = & \sqrt{2} {\mathcal {F}}^{-1} \bigl[e^{-\xi^{2}-(t/\sqrt{2})^{2}\xi^{-2}} \bigr](\sqrt {2}x) \\ = & \sum_{n=0}^{\infty}\frac{(-\sqrt{2} t)^{n}}{n!} {}_{1}F_{1} \biggl(\frac{1-n}{2},\frac{1}{2}, - \frac{x^{2}}{2} \biggr). \end{aligned}$$
Proof of Proposition 4.2
Let us put
$$I(t,x):={\mathcal {F}}^{-1} \bigl[e^{-\xi^{2}-t^{2}\xi^{-2}} \bigr](x) = \sqrt{ \frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-t^{2}\xi^{-2}}\cos x\xi \,d\xi. $$
Differentiating \(I(t,x)\) in x, we have
$$\begin{gathered} \partial_{x} I(t,x)=- \sqrt{\frac{2}{\pi}} \int^{\infty}_{0} e^{-\xi^{2}-t^{2}\xi^{-2}}\xi\sin x\xi \,d \xi, \\ \partial_{x}^{2} I(t,x)=- \sqrt{\frac{2}{\pi}} \int^{\infty}_{0} e^{-\xi^{2}-t^{2}\xi^{-2}}\xi^{2} \cos x\xi \,d\xi. \end{gathered} $$
On the other hand, differentiating I in t, we also have
$$\partial_{t} I(t,x)=- \sqrt{\frac{2}{\pi}} 2t \int^{\infty}_{0} e^{-\xi^{2}-t^{2}\xi^{-2}}\xi^{-2} \cos x\xi \,d\xi. $$
Moreover, the integration by parts yields
$$\begin{aligned} \partial_{t} I(t,x) =&- \sqrt{\frac{2}{\pi}} t^{-1} \int^{\infty}_{0} \bigl\{ e^{ -t^{2}\xi^{-2}} \bigr\} 'e^{-\xi^{2} }\xi\cos x\xi \,d\xi \\ =& \sqrt{\frac{2}{\pi}} t^{-1} \int^{\infty}_{0} e^{ -\xi^{2}-t^{2}\xi^{-2}} \bigl\{ \bigl(1-2 \xi^{2} \bigr) \cos x\xi- x\xi\sin x\xi \bigr\} \,d\xi. \end{aligned}$$
Thus, we see that \(I(t,x)\) satisfies the partial differential equation
$$ \partial_{t} I (t,x) = 2 t^{-1} \biggl\{ \frac{1}{2} I (t,x) +\partial_{x}^{2} I (t,x) + \frac{x}{2}\partial_{x} I (t,x) \biggr\} . $$
(7)
We may suppose that \(x \geq0\) since \(I(t,x) = \sqrt{\frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-t^{2}\xi^{-2}}\cos x\xi \,d\xi\) is an even function in x. Now we consider the point \(x=2\sqrt{-y} \) (\(y \leq0\)) and get for \(J(t,y):=I(t, 2\sqrt{-y})\)
$$\partial_{y}J(t,y) =\frac{-1}{ \sqrt{-y}}(\partial_{x}I) (t, 2\sqrt{-y}), \quad\quad \partial^{2}_{y}J(t,y) =- \frac{1}{ y} \bigl(\partial_{x}^{2}I \bigr) (t,2 \sqrt{-y})-\frac{1}{2y }\partial_{y}J(t,y) . $$
Therefore, by the change of variables \(x=2\sqrt{-y} \), it holds that
$$\partial_{t} J(t,y) = 2 t^{-1} \biggl\{ \frac{1}{2} J(t,y) -y\partial_{y}^{2} J(t,y) + \biggl(y- \frac{1}{2} \biggr) \partial_{y} J(t,y) \biggr\} . $$
To solve this partial differential equation, we shall use the method of separation of variables. By putting \(J (t,y)= \sum_{n =0}^{\infty}L_{n}(t)K_{n}(y)\), we obtain
$$\frac{t\partial_{t} L_{n}(t)}{2L_{n}(t)}=\frac{-y \partial^{2}_{y}K_{n}(y) - (\frac {1}{2} -y ) \partial_{y} K_{n}(y)+\frac{1}{2}K_{n}(y)}{K_{n}(y)}=:\lambda_{n}. $$
We immediately see that \(L_{n}(t) =t^{2\lambda_{n} }L_{n}(1)\). It is known that
$$ \begin{aligned}[b] I(t,0)&= \sqrt{\frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-t^{2}\xi^{-2}} \,d\xi \\ &= \frac{1}{ \sqrt{2} } e^{-2 t}\quad \biggl( \equiv\frac{1}{ \sqrt{2} } \biggl\{ 1 +\frac{(-2)}{1!}t^{1}+\frac{(-2)^{2}}{2!}t^{2}+\cdots \biggr\} \biggr). \end{aligned} $$
(8)
We note that
$$I(t,0)=J (t,0) = \sum_{n =0}^{\infty}L_{n}(t)K_{n}(0)= \sum_{n =0}^{\infty}L_{n}(t) , $$
here we may take \(K_{n}(0)=1\) for all \(n \in\mathbf{N}\) by choosing the suitable \(L_{n}(t)\). Hence we see that \(\lambda_{n} =\frac{n}{2}\) and
$$L_{n}(t) =t^{n }L_{n}(1)= \frac{1}{ \sqrt{2} } \frac{(-2)^{n}}{n!} t^{n}. $$
Meanwhile, the eigenvalue problem
$$-y \partial^{2}_{y}K_{n}(y) - \biggl( \frac{1}{2} -y \biggr) \partial_{y} K_{n}(y)+ \frac{1}{2}K_{n}(y)=\frac{n}{2} K_{n} (y) $$
with \(K_{n}(0)=1\) has
$$K_{n}(y) = {}_{1}F_{1} \biggl(\frac{1-n}{2}, \frac{1}{2}, y \biggr). $$
Thus it follows that
$$J(t,y)= \sum_{n=0}^{\infty}L_{n}(t)K_{n}(y)= \sum_{n=0}^{\infty}\frac{1}{ \sqrt{2} } \frac{(-2)^{n}}{n!} t^{n} {}_{1}F_{1} \biggl(\frac{1-n}{2},\frac{1}{2}, y \biggr) , $$
which gives
$$ I(t,x)= \frac{1}{ \sqrt{2} } \sum_{n=0}^{\infty}t^{n} \frac{(-2)^{n}}{n!} {}_{1}F_{1} \biggl( \frac{1-n}{2},\frac{1}{2}, -\frac{x^{2}}{4} \biggr). $$
(9)
We knew that \(I(t,x)\) is an even function in advance and supposed that \(x\geq0\). The last representation also implies that \(I(t,x)\) is an even function in x. So, (9) holds for all \(x \in\mathbf{R}\).
We have derived (9) by solving the partial differential equation. To avoid confusion, let us denote the solution represented as in (9) by \(\tilde{I}(t,x) \). It remains to show the uniqueness of \(\tilde{I}(t,x) = \frac{1}{ \sqrt{2} } \sum_{n=0}^{\infty}t^{n} \frac{(-2)^{n}}{n!} {}_{1}F_{1} (\frac{1-n}{2},\frac{1}{2}, -\frac{x^{2}}{4} )\) and \(I(t,x)= \sqrt{\frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-t^{2}\xi^{-2}}\cos x\xi \,d\xi\) except the case of \(t=0\). Instead of \(I(t,x)\), we consider for \((s,x) \in(0,\infty) \times \mathbf{R}\)
$${\mathcal {I}} (s,x) \bigl(=I(\sqrt {s},x) \bigr)= \sqrt{\frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-s\xi^{-2}}\cos x\xi \,d\xi $$
for the differentiation with respect to s. Then, by Stirling’s formula, we obtain
$$\begin{aligned} \bigl\vert \partial_{s}^{m}\partial_{x}^{j} {\mathcal {I}} (s,x) \bigr\vert \leq& \sqrt{\frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}/2}\cdot e^{-s\xi^{-2}}\xi^{-2m}\cdot e^{-\xi^{2}/2} \bigl( \xi^{2} \bigr)^{j/2}\,d\xi \\ \leq& C \sup_{\eta\geq0} e^{-s\eta}\eta^{m} \cdot\sup_{\mu\geq0}e^{-\mu/2}\mu^{j/2} \\ \leq& C e^{-m } \biggl( \frac{m}{s} \biggr)^{m} \cdot e^{-j/2 }j^{j/2} \\ \leq& C r_{s}^{m+j}m!(j!)^{1/2} \quad \bigl(\leq C r_{s}^{m+j}m!j! \bigr). \end{aligned}$$
This implies that \({\mathcal {I}} (s,x)\) is analytic for \((s,x) \in [s_{0},\infty) \times\mathbf{R}\) with arbitrarily fixed \({s_{0}>0}\). Therefore, we see that \(I (t,x)= \sqrt{\frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-t^{2}\xi^{-2}}\cos x\xi \,d\xi\) is analytic for \((t,x) \in {(0,\infty) \times\mathbf{R}}\). □
Remark 4.4
Probably \(I(t,x)\) would be analytic also at \(t=0\). But \({\mathcal {I}} (s,x)\) (\(=I(\sqrt {s},x)\)) loses the analyticity at \(s=0\). Indeed, we find that \(I(\sqrt {s},0) = \frac{1}{ \sqrt{2} } e^{-2 \sqrt {s}}= \frac{1}{ \sqrt{2} } \{ 1 +\frac {(-2)}{1!}\sqrt {s} +\frac{(-2)^{2}}{2!}s+\cdots \}\).
The Taylor expansion around a point \(t=T>0\) gives
$$I(t,x) \biggl(= \sqrt{\frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-t^{2}\xi^{-2}}\cos x\xi \,d\xi \biggr)= \sum_{n\geq0, k\geq0}a_{n,k}(t-T)^{n}x^{2k}, $$
since \(I(t,x) \) is an even function in x. By (9) we also get another Taylor expansion
$$\tilde{I}(t,x) = \frac{1}{ \sqrt{2} } \sum_{n=0}^{\infty}\bigl\{ (t-T)+T \bigr\} ^{n} \frac{(-2)^{n}}{n!} {}_{1}F_{1} \biggl(\frac{1-n}{2},\frac{1}{2}, -\frac{x^{2}}{4} \biggr) = \sum _{n\geq0, k\geq0}\tilde{a}_{n,k}(t-T)^{n}x^{2k}. $$
Then \(U(t,x):= I(t,x)-\tilde{I}(t,x)= \sum_{n\geq0, k\geq0} u_{n,k}(t-T)^{n}x^{2k}\) satisfies
$$\partial_{t} U(t,x) = 2 t^{-1} \biggl\{ \frac{1}{2} U (t,x) +\partial_{x}^{2} U (t,x) +\frac{x}{2} \partial_{x} U(t,x) \biggr\} , $$
and by (8)
Therefore, we get \(u_{n,0}=0\) for all \(n \geq0\) and
$$ \sum_{n\geq1, k\geq0}nu_{n,k}t(t-T)^{n-1} x^{2k} = \sum_{n\geq0, k\geq0}(2k+1) \bigl\{ u_{n,k}+4(k+1) u_{n,k+1} \bigr\} (t-T)^{n}x^{2k}, $$
(10)
here we used that
$$\partial_{x}^{2} I= \sum_{n\geq0, k\geq1}2k(2k-1)u_{n,k}(t-T)^{n}x^{2k-2}= \sum_{n\geq0, k\geq0}2(k+1) (2k+1)u_{n,k+1}(t-T)^{n}x^{2k} . $$
Moreover, the left-hand side of (10) is changed into
$$\sum_{n\geq1, k\geq0}nu_{n,k}t(t-T)^{n-1} x^{2k} = \sum_{n\geq0, k\geq0} \bigl\{ nu_{n,k}+ (n+1) u_{n+1,k}T \bigr\} (t-T)^{n} x^{2k} . $$
Thus, it holds that
$$nu_{n,k}+ (n+1) u_{n+1,k}T = (2k+1) \bigl\{ u_{n,k}+4(k+1)u_{n,k+1} \bigr\} . $$
Hence, when \(u_{n,0}=0\) for all \(n \geq0\), we find that \(u_{n,1}=0\) for all \(n \geq0\), and recursively \(u_{n,k}=0\) for all \(n \geq0 \) and \(k \geq0\). So, we have
$$U(t,x)= \sum_{n\geq0, k\geq0}u_{n,k}(t-T)^{n}x^{2k} \equiv0. $$
This concludes that \(I(t,x)\) (\(= \sqrt{\frac{2}{\pi}} \int^{\infty}_{0}e^{-\xi^{2}-t^{2}\xi^{-2}}\cos x\xi \,d\xi \)) must coincide with \(\tilde{I}(t,x)\) (\(= \frac{1}{ \sqrt{2} } \sum_{n=0}^{\infty}t^{n} \frac {(-2)^{n}}{n!} {}_{1}F_{1} (\frac{1-n}{2},\frac{1}{2}, -\frac{x^{2}}{4} ) \)) for \((t,x) \in(0,\infty) \times\mathbf{R}\). □
As an application of Proposition 4.2, we can compute the Fourier transform and the wavelet transform of concrete functions in the Gelfand-Shilov spaces. So, now let us take \(\hat{\psi}(\xi) =\hat{f} (\xi) = e^{-\xi^{2}- \xi^{-2}}\). We see that \(\psi, f \in S^{1/2,1/2}_{3/2,h} (\mathbf{R})\) for some \(h>0\) since \(e^{-\xi^{2}} \) gives \(\mu=1/2\) and the Gevrey function \(e^{- \xi ^{-2}} \) gives \(\delta=1/2\) and \(\nu=3/2\) by the Paley-Wiener theorem. Then by (2) it follows that
$$\begin{aligned} W_{\psi}f (a,b) = &2\sqrt{ \frac{a}{ C_{\psi}}} \int^{\infty}_{0}e^{-(1+a^{2})\xi^{2}- (1+ a^{-2}) \xi^{-2}}\cos b\xi \,d\xi \\ =& 2\sqrt{ \frac{a}{ C_{\psi}(1+a^{2})} } \int^{\infty}_{0}e^{-\omega^{2}- (1+a^{-2})(1+a^{2}) \omega^{-2}}\cos \frac{b}{\sqrt{1+a^{2} }}\omega \,d\omega \\ =& \sqrt{ \frac{a}{ C_{\psi}(1+a^{2})} } \int^{\infty}_{-\infty} e^{-\omega^{2}- (a +1/a)^{2} \omega^{-2}}e^{i \frac {b}{\sqrt{1+a^{2} }}\omega} \,d\omega \\ = & \sqrt{ \frac{2\pi a}{ C_{\psi}(1+a^{2})} } {\mathcal {F}}^{-1} \bigl[e^{-\omega^{2}- (a +1/a)^{2} \omega^{-2}} \bigr] \biggl( \frac {b}{\sqrt{1+a^{2} }} \biggr) . \end{aligned}$$
By the Paley-Wiener theorem, we find that for some \(\rho>0\)
$$\bigl\vert W_{\psi}f (a,b) \bigr\vert \leq Ce^{ -\rho \vert b/ \sqrt{ 1+a^{2}} \vert ^{2/3}} \sim Ce^{ -\rho \vert b/( 1+a ) \vert ^{2/3}} . $$
This implies that the order (i) in Theorem 2.4 is almost optimal with respect to a and b. Using Proposition 4.2 with \(t=1\) and \(t = a +1/a \), we have the following.
Theorem 4.5
Let
\(\hat{\psi}(\xi) =\hat{f} (\xi) = e^{-\xi^{2}- \xi^{-2}}\)
for
\(\xi \neq0\)
and =0 for
\(\xi= 0\). Then
$$\psi(x) =f(x) = \frac{1}{ \sqrt{2} } \sum_{n=0}^{\infty}\frac{(-2 )^{n}}{n!} {}_{1}F_{1} \biggl(\frac{1-n}{2}, \frac{1}{2}, -\frac{x^{2}}{4} \biggr) \in S^{1/2,1/2}_{3/2,h}( \mathbf{R}) $$
for some
\(h>0\), and the wavelet transform is given by
$$W_{\psi}f (a,b) = \sqrt{ \frac{\pi a}{ C_{\psi}(1+a^{2})} }\sum _{n=0}^{\infty}\frac{\{-2 (a +1/a)\}^{n}}{n!} {}_{1}F_{1} \biggl(\frac{1-n}{2},\frac{1}{2}, -\frac{b^{2}}{4(1+a^{2})} \biggr), $$
where
\({}_{1}F_{1} (a,b,z )\)
is the confluent hypergeometric function of the first kind.
Remark 4.6
Especially when \(b=0\), we also find
$$ \bigl\vert W_{\psi}f (a,0) \bigr\vert = \sqrt{ \frac{\pi a}{ C_{\psi}(1+a^{2})} }e^{-2 (a +\frac{1}{a})} . $$
(11)
Then (iii)′ in Theorem 2.4 becomes
$$\bigl\Vert a^{-1/2} e^{h' 2\max\{ a , a^{-1}\}}W_{\psi}f \Vert _{L^{\infty}(\mathbf{R}_{+} \times \mathbf{R})} \leq C \Vert e^{h\max \{ \vert \xi \vert ^{2}, \vert \xi \vert ^{-2}\} }\hat{f} \bigr\Vert _{L^{\infty}(\mathbf{R}) } . $$
(11) implies that \(\max\{ a , a^{-1}\} \) in \((\mathrm{iii})'\) cannot be improved anymore since \(h' \sim1\) and
$$h' 2 \max \bigl\{ a , a^{-1} \bigr\} \sim2 \biggl(a + \frac{1}{a} \biggr) . $$
Remark 4.7
As introduced in Remark 1.1, the Bessel wavelet \(\psi(x) \) satisfies \(\hat{\psi}(\xi) = e^{-\xi- \xi^{-1}}\) for \(\xi> 0\) and \(\hat{\psi}(\xi)=0\) for \(\xi\leq 0\) belongs to \({\mathcal {S}}^{1,+}_{2}(\mathbf{R})\). Hence, we also see that
$$\psi(x)=\frac{1}{\pi\sqrt{1-ix}}K_{1}(2\sqrt{1-ix})+\frac{1}{\pi\sqrt{1+ix}}K_{1}(2 \sqrt{1+ix}) $$
satisfies \(\hat{\psi}(\xi) = e^{- \vert \xi \vert - \vert \xi \vert ^{-1}}\) for \(\xi\neq0\) and \(\hat{\psi}(\xi)=0\) for \(\xi=0\) belongs to \({\mathcal {S}}^{1}_{2}(\mathbf{R})\) and \(S^{1 ,1 }_{ 2,h}(\mathbf{R}) \) for some \(h>0\).