"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?

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).
