"There are no smooth, symmetric, compactly supported wavelets." In All of Nonparametric Statistics by Larry Wasserman, page 205 states:

For example, in 1992 Ingrid Daubechies constructed a smooth, compactly
  supported “nearly” symmetric1 wavelet called a symmlet.

There is a footnote:

1 There are no smooth, symmetric, compactly supported wavelets.

Why not? Is there some deep truth to be understood here?  Can someone provide some intuition regarding this statement?
 A: This claim is only true for orthonormal wavelet systems, and the difficulty can be avoided by going to either a biorthogonal wavelet system or a wavelet frame. I'm only considering the case of real filter coefficients here.
The conditions imposed require us to find a trigonometric polynomial (it only has finitely many nonzero coefficients because of the compact support) which satisfies $R(0)=1$ (lowpass), $R(\omega)=R(-\omega)$ (symmetry), $|R(\omega)|^{2}+|R(\omega+\pi)|^{2}=1$, and $R(\omega)\overline{R(\omega+\pi)}+R(\omega+\pi)\overline{R(\omega)}=0$ (conditions for an orthonormal wavelet system).
By symmetry, $R(\omega)=c_{0}+\sum_{k=1}^{n}c_{k}\cos(k\omega)$, so $R(\omega+\pi)=c_{0}+\sum_{k=1}^{n}c_{k}(\cos(k\omega)\mathbf{1}_{2\mathbb{Z}}(k)-\cos(k\omega)\mathbf{1}_{2\mathbb{Z}+1}(k))$. So if we let $R_{e}(\omega)=c_{0}+\sum_{k\text{ even}}c_{k}\cos(k\omega)$ and $R_{0}(\omega)=\sum_{k\text{ odd}}c_{k}\cos(k\omega)$, we have $$R(\omega)=R_{e}(\omega)+R_{o}(\omega),\quad R(\omega+\pi)=R_{e}(\omega)-R_{o}(\omega).$$
Therefore $|R(\omega)|^{2}+|R(\omega+\pi)|^{2}=2(|R_{e}(\omega)|^{2}+|R_{o}(\omega)|^{2}),$ and $R(\omega)\overline{R(\omega+\pi)}+R(\omega+\pi)\overline{R(\omega)}=2(|R_{e}(\omega)|^{2}-|R_{o}(\omega)|^{2}).$ The second condition for an orthonormal wavelet system forces $|R_{e}(\omega)|^{2}=|R_{o}(\omega)|^{2},$ and because these are trigonometric polynomials (which are just restrictions of polynomials to the unit circle in $\mathbb{C}$), this forces $R_{e}(\omega)=e^{i\theta\omega}R_{o}(\omega).$ By symmetry, this phase factor must be $\pm 1,$ and since $1=R(0)=R_{o}(0)+R_{e}(0),$ we see that $R_{e}(\omega)=R_{o}(\omega),$ but since one is a trigonometric polynomial with only even frequencies and the other is a trigonometric polynomial with only odd frequencies, this is impossible.
If we use a less restrictive meaning for symmetry (there is some $s$ such that $R(\omega)=|R(\omega)|e^{is\omega}$), then we can accept the Haar scaling function (which is not smooth!). In the case of Haar, we have $e^{-i\omega/2}R(\omega)=\cos(\omega/2),$ so $|R(\omega)|^{2}+|R(\omega+\pi)|^{2}=\cos^{2}(\omega/2)+\sin^{2}(\omega/2)=1,$ and \begin{align*}R(\omega)\overline{R(\omega+\pi)}+R(\omega+\pi)\overline{R(\omega)}&=e^{-i\pi/2}\cos(\omega/2)\sin(\omega/2)+e^{i\pi/2}\cos(\omega/2)\sin(\omega/2)\\&=(-i+i)\cos(\omega/2)\sin(\omega/2)=0,\end{align*} so all is well. The proof in the case of this weaker notion of symmetry should go very similarly to the case I showed, but you'll have $e^{-is\omega}R(\omega)=\sum_{k=0}^{n}c_{k}\cos((k+1/2)\omega)$ ($s$ must be in $(1/2)\mathbb{Z}$ since $R$ is a trigonometric polynomial).
