Consider the Weierstrass Function (somewhat generalized for arbitrary wavelengths $\,\lambda > 0$ ): $$ W(x) = \sum_{n=1}^\infty \frac{\sin\left(n^2\,2\pi/\lambda\,x\right)}{n^2} $$ $W(x)$ is an example of a function that is everywhere continuous but nowhere differentiable (Or, for the nitpickers among us: differentiable only on a set of points of measure zero). We seek to remove the pathological character of the function $W(x)$ in very much the same way as has been tried with removing singularities in Could this be called Renormalization? , namely by convolution with a "broadened" Dirac-Delta function, in our case a Gaussian: $$ \int_{-\infty}^{^\infty} W(\xi) \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}(x-\xi)^2/\sigma^2}\,d\xi =\sum_{n=1}^\infty \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{^\infty} \frac{e^{-\frac{1}{2}(x-\xi)^2/\sigma^2}\left[e^{i\,n^2\,2\pi/\lambda\,\xi}-e^{-i\,n^2\,2\pi/\lambda\,\xi}\right]}{2\,i\,n^2} $$ Exponents are rewritten as follows: $$ -\frac{1}{2}(x-\xi)^2/\sigma^2 \pm i\,n^2\frac{2\pi}{\lambda}\xi = - \frac{1}{2}\left(\frac{\xi}{\sigma}\right)^2 + \frac{x}{\sigma}\frac{\xi}{\sigma} \pm i\,n^2\frac{2\pi}{\lambda}\sigma\frac{\xi}{\sigma} - \frac{1}{2}\left(\frac{x}{\sigma}\right)^2 = -\frac{1}{2}\left[\frac{\xi}{\sigma}-\left(\frac{x}{\sigma} \pm i\,n^2\frac{2\pi}{\lambda}\sigma\right)\right]^2 -\frac{1}{2}\left(n^2\frac{2\pi}{\lambda}\sigma\right)^2 \pm i\,n^2\frac{2\pi}{\lambda}x $$ Giving for the convolution integral: $$ \sum_{n=1}^\infty \int_{-\infty}^{^\infty} e^{-\frac{1}{2}\left[\xi/\sigma-(x/\sigma \pm i\,n^2\,2\pi/\lambda\,\sigma)\right]^2}\,d(\xi/\sigma)\times\sigma\times\frac{1}{\sigma\sqrt{2\pi}}\;(=1)\\ \times e^{-\frac{1}{2}\left(n^2\,2\pi/\lambda\,\sigma\right)^2}\frac{e^{i\,n^2\,2\pi/\lambda\,x} - e^{-i\,n^2\,2\pi/\lambda\,x}}{2i\,n^2}\qquad \Longrightarrow \\ \overline{W}(x) = \sum_{n=1}^\infty e^{-\frac{1}{2}\left(n^2\,2\pi/\lambda\,\sigma\right)^2}\frac{\sin\left(n^2\,2\pi/\lambda\,x\right)}{n^2} $$ The idea is that a monster function which is "smoothed" in this way might be differentiable as well as continuous. A possible argument is that the derivative of the convolution integral can also be calculated in the following way: $$ \frac{d}{dx} \int_{-\infty}^{^\infty} W(\xi) \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}(x-\xi)^2/\sigma^2}\,d\xi = \int_{-\infty}^{^\infty} W(\xi) \frac{1}{\sigma\sqrt{2\pi}} \left(-\frac{x-\xi}{\sigma^2}\right) e^{-\frac{1}{2}(x-\xi)^2/\sigma^2}\,d\xi $$ But I'm not sure if this counts as a proof. Hence the question: is the Weierstrass monster function "tamed" by this "renormalization" procedure and has it become differentiable everywhere indeed?
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Is there any claim about the magnitude of $|W(x)-\overline W(x)|$? – Sergio Parreiras May 21 '14 at 14:20
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7The convolution always copies the differentiability properties of the "better" function, i.e. if $f$ is smooth and $g$ is say, only continuous, then $f*g$ is smooth. Put $f$ to be your Weierstrass function and $g$ the Gaussian. The proof goes like you started but there is no need to write down these monstrous formulas, just take $f$, $g$ arbitrary. Whats missing in your argument is why you can take the differential under the integral. – J.R. May 21 '14 at 14:20
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@YourAdHere: (+1) for your comment. But alas, if I could fill in the "missing" parts and thus answer my own question, then I wouldn't have asked for an another answer. – Han de Bruijn May 21 '14 at 19:18
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Your Ad Here's suggestion works fine. Alternatively, $\overline{W}\in C^{\infty}$ also follows from the fact that if you differentiate the new series term by term (any number of times), you obtain another uniformly and absolutely convergent series. – Aug 05 '14 at 18:53
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1@Han de Bruijn: This is one of the most fundamental properties of the convolution operation. You should find it discussed in any analysis textbook that covers convolution or Fourier analysis. – Nate Eldredge Aug 06 '14 at 06:01
1 Answers
Since we have: $$\overline{W}(x) = \sum_{n=1}^{+\infty}e^{-An^2}\frac{\sin(Bn^2 x)}{n^2}$$ it follows that: $$\overline{W}(x)-\overline{W}(y)=\sum_{n=1}^{M}e^{-An^2}\frac{\sin(Bn^2 x)-\sin(Bn^2 y)}{n^2}+O(e^{-AM^2}(x-y)),$$ so by assuming $|x-y|\leq\frac{1}{M^2}$ and exploiting the Lipschitz condition for the sine function: $$\left|\frac{\overline{W}(x)-\overline{W}(y)}{x-y}\right|\leq \sum_{n=1}^{M}B e^{-An^2}+O(e^{-AM^2})=O(1),$$ hence the smoothed Weierstrass function is a Lipschitz function, and $$\overline{W}'(x)=\sum_{n=1}^{+\infty}Be^{-An^2}\cos(Bn^2 x)$$ since the RHS is an absolutely convergent series. Iterating the same argument over and over we have that $\overline{W}$ is a $C^{\infty}$ function, since the sine and cosine function are always bounded and the coefficients $B^k n^{2k}e^{-An^2}$ decay pretty fast as soon as $n$ is big enough.
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