I have a question regarding the Continuous Wavelet Transform (CWT) of non finite energy functions, such as $g(t) = a\exp(i\omega_0t)$. We know that the CWT is defined for functions in the Hilbert space $L^2(\mathbb{R})$, but I've found in Chapter 4 of "A Wavelet Tour of Signal Processing" that the CWT of $g(t)$ is $(W_\psi g)(u,s) = a\sqrt{s}\hat{\psi^*}(s\omega_0)\exp(i\omega_0t)$, where $\psi$ is a wavelet. How is this possible ? (since $g \not\in L^2(\mathbb{R})$)

I have a question regarding the Continuous Wavelet Transform (CWT) of non finite energy functions, such as $g(t) = a\exp(i\omega_0t)$. We know that the CWT is defined for functions in the Hilbert space $L^2(\mathbb{R})$, but I've found in Chapter 4 of "A Wavelet Tour of Signal Processing" that the CWT of $g(t)$ is $(W_\psi g)(u,s) = a\sqrt{s}\hat{\psi^*}(s\omega_0)\exp(i\omega_0u)$, where $\psi$ is a wavelet. How is this possible ? (since $g \not\in L^2(\mathbb{R})$)

I have a question regarding the Continuous Wavelet Transform (CWT) of non finite energy functions, such as $g(t) = a\exp(i\omega_0t)$. We know that the CWT is defined for functions in the Hilbert space $L^2(\mathbb{R})$, but I've found in Chapter 4 of "A Wavelet Tour of Signal Processing" that the CWT of $g(t)$ is $(W_\psi f)(u,s) = a\sqrt{s}\hat{\psi^*}(s\omega_0)\exp(i\omega_0t)$, where $\psi$ is a wavelet. How is this possible ? (since $g \not\in L^2(\mathbb{R})$)

I have a question regarding the Continuous Wavelet Transform (CWT) of non finite energy functions, such as $g(t) = a\exp(i\omega_0t)$. We know that the CWT is defined for functions in the Hilbert space $L^2(\mathbb{R})$, but I've found in Chapter 4 of "A Wavelet Tour of Signal Processing" that the CWT of $g(t)$ is $(W_\psi g)(u,s) = a\sqrt{s}\hat{\psi^*}(s\omega_0)\exp(i\omega_0t)$, where $\psi$ is a wavelet. How is this possible ? (since $g \not\in L^2(\mathbb{R})$)