# Are the derivatives of $\delta(x-1)$ cyclic?

Question 1: Is it true the derivatives of $\delta(x-1)$ are cyclic as illustrated below?

$\quad\delta(x-1)$

$\quad\delta'(x-1)$

$\quad\delta''(x-1)=-\delta(x-1)$

$\quad\delta^{(3)}(x-1)=-\delta'(x-1)$

$\quad\delta^{(4)}(x-1)=\delta(x-1)$

• I think not. First, the translation by $-1$ is not significant. Second, the second derivative of $\delta$ is not the negative of $\delta$, if for no other reason than there are many test functions $f$ with $f''(0)\not=-f(0)$. (Or, $f''(1)\not=-f(1)$, if one thinks that maybe the translation does something...) What is the larger context? E.g., this is not overtly about number theory. – paul garrett Jul 6 '17 at 21:39
• No. Note that $$\langle \delta_1^{(n)},\phi\rangle =(-1)^n\phi^{(n)}(1)$$ – Mark Viola Jul 6 '17 at 21:54
• The larger context is evaluation of Fourier and Mellin convolutions such as $\delta(x)\,*\,g(x)=\int_0^\infty\delta(x)\,g(y-x)\,dx=g(y)$, $\delta(x-1)\,*_{\mathcal{M}_1}\,g(x)=\int_0^\infty\frac{\delta(x-1)}{x}\,g\left(\frac{y}{x}\right)\,dx=g(y)$, and $\delta(x-1)\,*_{\mathcal{M}_2}\,g(x)=\int_0^\infty\delta(x-1)\,g(y\,x)\,dx=g(y)$. Note the Fourier series representation of the Dirac comb consists of $cos$ terms which have cyclic derivatives similar to above. If the derivatives of the distributional representation of the Fourier comb are not cyclic, then doesn't this create a discrepancy? – Steven Clark Jul 6 '17 at 22:02
• Ah, yes, the Dirac comb is $\delta_{\mathbb Z}=\sum_{n\in\mathbb Z}1\cdot e^{2\pi inx}$ (or trig function equivalent), but/and when we take derivatives, there is a multiplication by $2\pi i n$ (and renormalization to try get rid of the $2\pi$ accomplishes nothing). Thus, $\delta_{\mathbb Z}''=(2\pi i)^2\cdot \sum_n n^2\cdot e^{2\pi inx}$. (And, yes, the expansion for $\delta_{\mathbb Z}$ does converge in the Sobolev spaces $H^{-1/2-\varepsilon}(\mathrm{circle})$, and that for the second derivative converges in $H^{-5/2-\varepsilon}(\mathrm{circle})$, for all $\varepsilon>0$.) – paul garrett Jul 6 '17 at 22:14
• Your right of course. $\cos(x)$ has periodic derivatives, but not $\cos(2\,k\,\pi\,x)$. I don't seem to be thinking very clearly today. – Steven Clark Jul 6 '17 at 23:23