# Tag Info

7

We can express the form $g(\tau)$ as $$g(\tau)=\sum_{d\mid N}\mu(d)\sum_{\substack{n\in\mathbb{N}\\d\mid n}} a(n)\,q^n.$$ In the notation of Atkin-Lehner (Hecke operators on $\Gamma_0(m)$, Math. Ann. 185 (1970), 134-160), the inner sum is $(f\mid U_d)\mid B_d$, which lies in $M(k,\Gamma_0(dN))$ by Lemmata 2, 6, 14 in the paper. This implies that $$g\in ... 4 No. It is an old (unpublished) result of Kolmogorov that one can reorder the trigonometric system so that there is an L^2 function whose Fourier series diverges on a set of positive measure. This is often refereed to as Kolmogorov's rearrangement theorem. For a proof, see Theorem 2 in Olevskii's book Fourier Series with Respect to General Orthogonal ... 4 For almost all (t_1, t_2, \ldots, t_N, t'_1, t'_2, \ldots, t'_N), just one Fourier term suffices. As I will explain below, I think this good enough for your goal of showing that all knots are Fourier-(1,1,2). Claim Suppose that t_1, t_2, ..., t_N, t'_1, t'_2, ..., t'_N and \pi are linearly independent over \mathbb{Q}. Then there is an ... 4 Since S^n \cong \mathrm{SO}(n+1)/\mathrm{SO}(n), the zonal spherical harmonics you are asking for arise as the spherical functions for the compact Gelfand pair (\mathrm{SO}(n+1),\mathrm{SO}(n)), which are known to be the Gegenbauer (also called ultra-spherical) polynomials. A good reference, including background on Gelfand pairs, is (Section 7.2 of) the ... 4 (Too long for a comment, sorry). If f is in L^1 then the integral defining the Fourier transform is defined at every point and defines a continuous function vanishing at infinity. Now if the transform also happens to be in L^1, the inverse Fourier transform is also a continuous function vanishing at infinity, which coincides a.e. with f. Does this ... 3 Yes. The triangle wave s(x):=\min_{k\in\mathbb{Z}}\big|x-k\big| has an absolutely convergent Fourier series$$s(x)=\frac{1}{4}-\frac{2}{\pi^2}\sum_{k=0}^\infty\frac{1}{(2k+1)^2}\cos\big(2\pi (2k+1)x\big)\, ,\qquad x\in\mathbb{R}.$$Therefore$$T(x):=\sum_{n=0}^\infty2^{-n}s(2^nx)= ...

3

What's your motivation? The Fourier transform is a *-homomorphism for the natural adjoint on $\ell^1(\mathbb Z)$ and so if $f$ is the transform of $a\in\ell^1(\mathbb Z)$ then $|f|^2f$ is the transform of $a^*aa$. Similarly let $g$ and $b$ be related. As $\|a\|,\|b\| \leq M$ we perform a simply triangle-inequality argument, which is valid in any Banach ...

3

$$g(\tau)=f(\tau)\otimes \left(\tfrac{N^2}{\cdot}\right)=\sum_{n\in \mathbb{N}}\left(\tfrac{N^2}{n}\right)a(n)\,q^n \quad \text{where}\left(\tfrac{N^2}{\cdot}\right) \text{ is the Kronecker symbol}.$$ This means that $g(\tau)$ is just the twist of $f(\tau)$ by a principle character. Indeed we have $$g(\tau)\in M(k,\Gamma_0(N^3)).$$

3

If all $t_i$, $t'_i$ are rational numbers with common denominator $d$, then it is enough to examine frequencies $m, n \le d$. After David's answer I decided that it could be worthwhile to look for counterexamples for small $d$ and used an optimization routine to minimize the sum over $\max(0,f(t'_i)-f(t_i))$ for each combination of $m, n$. Counterexamples ...

2

Let $f,g$ be tempered distributions on $\mathbb R^n$ such that $g\in \mathscr O_M$, the so-called multipliers space: $g$ is a smooth function such that $$\forall \alpha, \exists N_\alpha\ge 0,\quad \sup_x\vert(\partial^\alpha g)(x)\vert(1+\vert x\vert)^{-N_\alpha}<+\infty.$$ Then the product $fg$ makes sense as a tempered distribution and we may define ...

1

One very cheap bound for the error is $$c h^2 \sum_n \sup_{|x-n| \le 1} |f''(x)|\;,$$ for a suitable $c$. (I believe that $c=1/3$ works, but haven't checked very carefully.) This still works if your function satisfies that bound only piecewise, as long as there are finitely many pieces and the function is continuous. (You lose an additional term of order ...

1

One class of functions you could consider is those $f$ for which $|f(x)|+|{\hat f}(x)| \le C(1+|x|)^{-1-\delta}$ for some constant $C$ and some $\delta>0$. Here ${\hat f}(x) = \int_{-\infty}^{\infty} f(t) e^{-2\pi i tx} dt$ is the Fourier transform. For this class of functions, one can use the Poisson summation formula, which gives  h\sum_{n\in ...

1

This is Theorem V (page 16) from: A. Beurling, On the spectral synthesis of bounded functions. Acta Math. 81 (1948). In fact, Beurling proves the stronger statement: Theorem Let $f(x) = \sum_{n=-\infty}^{\infty} a_n e(nx)$ (with $a_0=0$) have an absolutely convergent Fourier series such that $|a_{\pm n}| \leq a_n^{*}$ where $a_n^{*}$ is a ...

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