From appendix A of the Mallat paper linked to in the comment by J.M., the smoothing spline figure is the Fourier transform of the following function:
$\hat{\psi}(\omega) = i \omega (\mathrm{sinc}(\omega/4) )^4$
where
$\mathrm{sinc}(\omega) = \frac{\sin(\omega)}{\omega}$ and $\omega = 2 \pi f$
The FT of a sinc function is simply a rectangle of width 1
$\mathcal{F}^{-1}\{sinc(\omega)\} = Rect(\omega)$
Multiplication becomes convolution:
$\mathcal{F}^{-1}\{sinc(\omega)^4\} = Rect(t) \star Rect(t) \star Rect(t) \star Rect(t)$
Working out these four convolutions gives a curve defined piecewise from $t \epsilon [-2,2]$:
$\mathcal{F}^{-1}\{sinc(\omega)^4\} =$
$t^3/6 + t^2 + 2t + 4/3,\text{ } -2 \le t \le -1$
$-t^3/2 - t^2 + 2/3,\text{ } -1
$t^3/2 - t^2 + 2/3,\text{ } 0
$-t^3/6 + t^2 - 2t + 4/3,\text{ } 1
The FT of a scaled sinc is a scaled Rect having width 1/4 and height 4:
$\mathcal{F}^{-1}\{sinc(\omega/4)\} = 4 Rect(4t)$
The extra factor of $i\omega$ is equivalent to a derivative in the time domain according to:
$\frac{d}{dt}(f(t) ) = i \omega \mathcal{F}^{-1}\{\hat{f}(\omega)\}$
Scaling the convolved sincs gives a curve defined piecewise from $t \epsilon [-1/2,1/2]$, which we then take a derivative of to get the final result:
$\mathcal{F}^{-1}\{i \omega sinc(\omega)^4\} =$
$4*((4t)^2/2 + 2*(4t) + 2), \text{ } -1/2 \le t \le -1/4$
$4*(-3/2*(4t)^2 - 2(4t) ) ,\text{ } -1/4
$4*(3/2*(4t)^2 - 2*(4t)) ,\text{ } 0
$4*(-(4t)^2/2 + 2*(4t) - 2),\text{ } 1/4
The plot looks like this:

The problem with this spline is that it is half the width of the plot shown in figure 2 of Mallat's paper. I believe the difference between the two is just a scaling factor though.