-1
$\begingroup$

I am retreating back on this statement, after some explorations and calculation Bow to Willie and others who were skeptical on this. Main difficulty can be seen in this reference. But I must mention that my quest for jump discontinuities has not seen a dead end but a new light after this failure. I am very much interested in this class of functions and this kind of jumps, and have found a new way to deal with them in 2 dimensions. My new pursuit: A direction dependent jump can be converted to jump of a 1-d function via Radon transform, reflecting jumps in different directions at some point into separate 1-d jumps via Radon transform. Via Fourier slice theorem, such 1-d jumps can be dealt with different slices of FT of the image using a 1-d technique ). Cant wait to lay my hands on image signal processing.

The initial question was in $N$ dimensions which follows after the heading "Initial question". But in view of Willie's comments, I realized it has flaws, but I am still optimistic, and hope they are not fatal.

So I'd like to formulate and pose the question for the simple case of $N=2$, with which I am at comfort, considering my limitations with higher math, and then hope it can be generalized to $N>2$ by mathematicians. I seek their help in this regard.

Case of $N=2$

Main motivation is monochromatic image signal or any 2-d signal. Mathematically We represent it as a function $f:\mathbb{R}^2 \to \mathbb{R}$. I know image signal has compact support, but for mathematical convenience, we can always place zeros where ever it is not defined and extend its domain to $\mathbb{R}^2$. (Atleast I do this blindly just for mathematical beauty).

Definition-1 A function $f:\mathbb{R}^2 \to \mathbb{R}$ is said to be of bounded variation, if for every rectifiable curve in $\mathbb{R}^2$, the 1-d function obtained by restricting the domain of $f$ to the curve, is a function of bounded variation in 1-d sense.

(Motivation for this definition : For the following definition to make sense, and for the limits defined in it to exist always.)

Definition-2 : Directional limits. Given a point $\hat{x}_0 = (x,y) \in \mathbb{R}^2$, for every $\theta \in \mathbb{R}$ we define a directional limit as $$u_{\theta}(\hat{x}_0) = \lim_{r\to 0+} f(x+r\cos(\theta),y+r\cos(\frac{\pi}{2}-\theta))$$ where $r \in \mathbb{R}$.

Definition-3 : Limit function. We define a limit function $J_{\hat{x}_0} : \mathbb{R} \to \mathbb{R}$ for every point $\hat{x}_0 \in \mathbb{R}^2$ given as $$J_{\hat{x}_0}(\theta) = u_{\theta}(\hat{x}_0)$$

Definition-4 Class of functions $\mathcal{V}$ consists of all functions of the form $f:\mathbb{R}^2\to\mathbb{R}$ which satisfy the following two conditions.

  1. $f$ should be a function of bounded variation as per Definition-1.
  2. For every $\hat{x}_0 \in \mathbb{R}^2$, the associated limit function $J_{\hat{x}_0}(\theta)$ (as in def-3) should be a function of bounded variation when its domain is restricted to $[0,2\pi)$.

Problem

Let $f \in \mathcal{V}$ be a square integrable function and let its Fourier transform be $\hat{f}$. Given any $\theta \in \mathbb{R}$ and $r\in\mathbb{R}$ we define a directional partial sum $S^{\theta}_r : \mathbb{R}^2 \to \mathbb{R}$ as $$S^{\theta}_r(\hat{x}) = \int\limits_{-rcos(\pi/2-\theta)}^{r\cos(\pi/2-\theta)} \int\limits_{-rcos\theta}^{r\cos\theta} \hat{f}(k_x,k_y) e^{i(xk_x+yk_y)} \mathrm{d}{k_x} \mathrm{d}{k_y} $$ where $\hat{x} = (x,y) \in \mathbb{R}^2$.

Is the following statement true?

Given any $\hat{x}_0 \in \mathbb{R}^2$ and any $\theta \in [0,\pi)$ $$\lim_{r\to\infty} S^{\theta}_r(\hat{x}_0) = \frac{1}{2}(u_{\theta}(\hat{x}_0) + u_{(\pi-\theta})(\hat{x}_0) )$$


Higher dimensional case revisited

Definition 1 Let $\Sigma\subset \mathbb{R}^N$ be a smooth submanifold. A function $f:\Sigma\to \mathbb{R}$ is said to have bounded variation if for every rectifiable curve $\gamma\subset \Sigma$, the composition $f\circ\gamma$ has bounded variation in the usual one-dimensional sense.

Now, let $\Sigma \subset\mathbb{R}^N$ be a smooth submanifold and let $x\in \Sigma$. We write $\exp_x: T_x\Sigma \to \Sigma$ to denote the exponential map.

Definition 2 Let $\Sigma$ be a $d+1$ dimensional submanifold of $\mathbb{R}^N$. Let $f:\Sigma\to \mathbb{R}$. Fix $x\in \Sigma$ and $\omega\in T_x\Sigma$ a unit vector. We write the directional limit $$ f_\omega(x) := \lim_{t \to 0+} f\circ \exp_x(t\omega) $$ whenever the limit on the right hand side exists. If for every unit vector $\omega\in \mathbb{S}^d \subset T_x\Sigma$ the directional limit $f_\omega(x)$ exists, we write $J_x: \omega \mapsto f_\omega(x)$. Note that $J_x:\mathbb{S}^d \to \mathbb{R}$.

Definition 3 Let $\Omega$ be a smooth submanifold of $\mathbb{R}^N$. We say that a function $f$ belongs to the class $\mathcal{V}(\Omega)$ iff $f$ has bounded variation and that at every point $x\in \Omega$ its blow-up $J_x$ has bounded variation, both in the sense of Definition 1.

Definition 4 Given $\omega\in \mathbb{S}^{N-1} \subset \mathbb{R}^N$, we write $\omega_i$ to be the $i$th coordinate value of $\omega$ relative to the standard rectangular coordinate system. We can define the rectangle $R_r^\omega$ for $r > 0$ to be $$ R_r^\omega = (-r |\omega_1|, r|\omega_1|) \times (-r |\omega_2|,r|\omega_2|) \times \cdots \times (-r|\omega_N|, r|\omega_N|)~,$$ in other words the rectangle with sides parallel to the standard hyperplanes and with diagonal $r\omega$. We also define $\mathrm{sgn}(\omega) = \prod_{i = 1}^N \mathrm{sgn}(\omega_i)$.

Now let $f\in \mathcal{V}(\mathbb{R}^N)\cap L^1(\mathbb{R}^N)$. Denote by $\hat{f}$ its Fourier transform. Fix $\omega\in \mathbb{S}^{N-1}$. Write $$ S_r^\omega f(x) = \int_{R_r^\omega} \hat{f}(\xi) \exp(i x\cdot \xi) \mathrm{d}\xi~. $$

Question

Is it true that

$$ \lim_{r \to\infty} S^\omega_r f(x) = \frac12 [J_x(\omega) + J_x(-\omega)] $$

for every $x\in \mathbb{R}^N$ and $\omega\in \mathbb{S}^{N-1}$?


Initial question

This is a question about pointwise convergence of a Fourier transform of functions of the form $f: \mathbb{R}^N \to \mathbb{R}$, which is potentially a $N$-dimensional generalization to pointwise convergence of $1$ dimensional Fourier transform. This question arose when I am trying to generalize this statement to $N$-dimensions.

Definition 1 : Functions of bounded variation in $\mathbb{R}^N$. Given any rectifiable curve in $\mathbb{R}^N$, if the function $f:\mathbb{R}^N \to \mathbb{R}$ evaluated on this curve is a function of bounded variation in the 1-d sense, then we say $f$ is a function of bounded variation.

Definition : Directional Limits.

for a function $f:\mathbb{R}^N \to \mathbb{R}$, given a unit vector $\bf{\hat{a}}$, we define the directional limit of $f$ at a point $\bf{x_0} \in \mathbb{R}^N$ along $\bf{\hat{a}}$ as $u_{\bf{\hat{a}}}(\bf{x_0}) = \lim_{\alpha \to 0^+}f(\bf{x_0 + \alpha \hat{a}})$.

Limit function at a point $x_0$ denoted as $J_{\bf{x_0}}:S^{N-1} \to\mathbb{R}$ is defined as $$J_{\bf{x_0}}(\bf{\hat{a}}) = u_{\bf{\hat{a}}}(\bf{x_0})$$. We denote this jump function in $\theta$-coordinates as $J^{\theta}_{\bf{x_0}}:[0,2\pi)^{N-1} \to \mathbb{R}$

Definition of a class of functions (This definition is recursive)

Given $\Omega$ an open subset of $\mathbb{R}^N$, we define a set of functions $\mathcal{V}(\Omega)$ with the following properties.

Iff $f \in \mathcal{V}(\Omega)$ then

  1. $f:\Omega \to\mathbb{R}$ is square integrable and function and of bounded variation as per Definition 1.

  2. With an additional constraint that the limit function in $\theta$-coordinates, at any point $P \in \Omega$ , denoted as $J^\theta_P: [0,2\pi)^{N-1}\to \mathbb{R}$ also belongs to the class of functions $\mathcal{V}([0,2\pi)^{N-1})$.

Fourier partial sum

Consider a function $f \in \mathcal{V}(\mathbb{R}^N))$, and let its Fourier transform be $\hat{f}$. Given any unit vector $\bf{\hat{a}} \in \mathbb{R}^N$, and a positive real number $R$, we define Fourier partial sum as $$S^{\bf{\hat{a}}}_R : \mathbb{R}^N \to \mathbb{R}$$ defined as $$S^{\bf{\hat{a}}}_R(\bf{x}) = \int_{-R\cos(\theta_1)}^{R\cos(\theta_1)} \int_{-R\cos(\theta_2)}^{R\cos(\theta_2)} ...\int_{-R\cos(\theta_{N-1})}^{R\cos(\theta_{N-1})} \int_{-R\cos(\phi)}^{R\cos(\phi)} \hat{f}(k_1,k_2,k_3,...k_N) e^{i(k_1x_1+k_2x_2+...+k_Nx_N)} \mathrm{d}{k_1}\mathrm{d}{k_2}...\mathrm{d}{k_N}$$ where $[\theta_1,\theta_2,...\theta_{N-1}]$ is $\bf{\hat{a}}$ expressed in $\theta$-coordinates, and $\phi = \frac{\Phi_{N-1}}{2^N} - \sum\limits_{j = 1}^{N-1} \theta_j$ where $\Phi_{N-1}$ is the total solid angle subtended by the full surface of a unit $(N-1)$-sphere given as $$\Phi_{N-1} = \frac{2\pi^{\frac{N-1}{2}}}{\Gamma(\frac{N-1}{2})}$$

$\bf{k} = [k_1,k_2,...k_N] \in \mathbb{R}^N$ and $\bf{x} = [x_1,x_2,...x_n] \in \mathbb{R}^N$

Statement

Question is that whether the following statement is true?

Given any point $\bf{x} \in \mathbb{R}^N$, and any unit vector $\bf{\hat{a}} \in \mathbb{R}^N$, $$\lim_{R\to \infty} S^{\bf{\hat{a}}}_R(\bf{x}) = \frac{u_{\bf{\hat{a}}}(\bf{x}) + u_{\bf{-\hat{a}}}(\bf{x})}{2}$$

$\endgroup$
13
  • $\begingroup$ @TerryTao and other Harmonic analysis experts, what do you think about this problem. Also let me know the defects in it, especially the formula for $\phi$, I am not sure I got it what I intended it to. Thanks $\endgroup$
    – Rajesh D
    Nov 5, 2014 at 6:26
  • $\begingroup$ A few random comments: (1) the domain of definition of $J_P^\theta$ is not an open subset of $\mathbb{R}^{N-1}$. (2) What is $\theta$-coordinate? You are identifying $\mathbb{S}^{N-1}$ with essentially $\mathbb{T}^{N-1}$ and that is problematic in my opinion. (3) What's up with $\phi$? By your definition when $N = 3$ you have $\phi = \frac{4\pi}{8} - \sum_{j = 1}^2 \theta_j$ so when $\theta_1 + \theta_2 > \pi/2$, which happens for a large chunk of $[0,2\pi)^2$ your $\phi$ is negative. // Did you come up with all these definitions yourself? If so please include motivations on why such defn. $\endgroup$ Nov 5, 2014 at 9:27
  • $\begingroup$ Maybe I should clarify that my point (1) above is meant to say that it makes no sense to define $\mathcal{V}(\Omega)$ recursively because $[0,2\pi)^{N-1}$ is not open. $\endgroup$ Nov 5, 2014 at 9:27
  • 2
    $\begingroup$ A too-short answer, but it seems that any comment would likely get lost in the others... Also, not responding directly to the literal question, but to the context: the notion of "wave-front set" would seem to me to be one of the concepts the questioner might find useful in refining the formulation of the issue (e.g., refining to the point that the assertions are not easily shown faulty in various ways, e.g., coordinate-(in)dependence as @WillieWong comments). $\endgroup$ Nov 7, 2014 at 14:22
  • 2
    $\begingroup$ Please explain the downvotes! $\endgroup$
    – Rajesh D
    Nov 15, 2014 at 7:02

1 Answer 1

4
$\begingroup$

The answer to your question is: No.

Consider the 2D case. Let $\theta = n \pi / 2$ for $n \in \{0,1,2,3\}$. From your definition $S_r^\theta f = 0$, since on the RHS of its definition you are integrating over a null set. But clearly the corresponding $u_\theta(x) + u_{-\theta}(x)$ doesn't always vanish: consider in particular any non-trivial Schwartz function.

$\endgroup$
17
  • $\begingroup$ Thanks for pointing it out. It still does not kill the spirit of the problem. In this problem, for N=2, for each element (unit vector) in $S^1$ we are giving out (contemplating) a statement to hold. If I recall correctly, to cover $S^1$ we need atleast $4$ co-ordinate charts. $\endgroup$
    – Rajesh D
    Nov 6, 2014 at 12:24
  • $\begingroup$ (CONTINUED..)...So we need to make the statement $4$ times each for each co-ordinate chart, and when we change co-ordinate chart, we lso need to take care to rotate all the co-ordinate systems associated with the problem appropriately (including domains of $f$ and $\hat{f}$ and then make the statement. When we do that, I hope each of the directions $n\pi/2, n=0,1,2,3$ will be covered in one of the co-ordinate charts without the problem of integral becoming zero as you have mentioned in the answer. $\endgroup$
    – Rajesh D
    Nov 6, 2014 at 12:24
  • $\begingroup$ (CONTINUED..)... I don't have the ease to make such statements in mathematical language and I hope you appreciate the spirit and if meaningful, write those statements in a new version of the problem. It would be great if all comes out beautifully. $\endgroup$
    – Rajesh D
    Nov 6, 2014 at 12:25
  • 2
    $\begingroup$ If your result is coordinate dependent, then it'd be useless, especially since the right hand side of the putative equation does not depend on coordinate choice. And no, you don't need 4 charts to cover $\mathbb{S}^1$. You are free to investigate further, but I don't think your line of reasoning is likely to go anywhere. Essentially it seems to me that you want to be able to say that when the inverse Fourier integral is not absolutely convergent, that you can somehow pick out directional dependence by taking the limit in different ways. That idea in itself is not impossible.... $\endgroup$ Nov 6, 2014 at 15:24
  • 1
    $\begingroup$ ... but I don't see any reason why the particular method you take to enforce this directional dependence is the preferred one. It is especially suspicious since the method itself is not invariant under rotations of $\mathbb{R}^N$. $\endgroup$ Nov 6, 2014 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.