Proving a function has a compact support

Question

I'm trying to understand Willie Wong's answer which shows that for each $t\in \mathbb{R}$, $g(\tau , t - \tau)$ has a compact support, i.e., the set $A_t = \{ \tau \in \mathbb{R}: g(\tau , t - \tau) = 1\}$ is bounded. Let $S = \{ (-2)^k : k \in \mathbb{N} \}$ and choose $0 < \epsilon \ll 1$. Define $$\chi(t) = \begin{cases} 1 & \exists s\in S, \quad |t-s| < \epsilon \\ 0 & \text{otherwise} \end{cases}$$ and $g(x,y) = \chi(x) \chi(y)$. It's clear that $g(\tau , t - \tau) = 1$ is equivalent to $$\exists k \in \mathbb{N}: |\tau - (-2)^k| \lt \epsilon \ \ \text{and} \ \ \exists l \in \mathbb{N}:|t-\tau - (-2)^l|\lt \epsilon.$$ So we have the implication, $$g(\tau , t - \tau) = 1 \implies \exists k,l \in \mathbb{N}: |t - ((-2)^k+(-2)^l)| \lt 2\epsilon$$ and its contraposition $$\forall k,l \in \mathbb{N}: |t - ((-2)^k+(-2)^l)| \ge 2\epsilon \implies g(\tau , t - \tau) = 0.$$ So we should suppose that $$\exists k_0,l_0 \in \mathbb{N}: |t - ((-2)^{k_0}+(-2)^{l_0})| \lt 2\epsilon \tag{1}$$ and check when $g(\tau , t - \tau) = 1$ occurs or equivalently, $$\exists k \in \mathbb{N}: |\tau - (-2)^k| \lt \epsilon \ \ \text{and} \ \ \exists l \in \mathbb{N}:|t-\tau - (-2)^l|\lt \epsilon. \tag{2}$$ I don't know how to proceed further. It's obvious that we should use $(1)$ and $(2)$ to prove that the set of $\tau$'s is bounded but I don't see how to do it. It seems that Willie Wong has used the following fact for that purpose:

$(-2)^k + (-2)^l = (-2)^m + (-2)^n$ if and only if either $\{k,\ell\} = \{m,n\}$ or $k = \ell = \min(m,n) - 1 = \max(m,n) - 2$ or vice versa.

Answer 1

Now, the formal write up actually doesn't 100% correspond to the intuition in the picture. But it is close enough. (See my other answer for the intuition.)

Case 1: $t$ is such that $g(\tau, t-\tau)\equiv 0$

Nothing to do here

Case 2: $t$ is such that for some $\tau$, $g(\tau,t-\tau) = 1$

By your equation (1), you have $| t - ((-2)^{k_0} + (-2)^{\ell_0})| < 2 \epsilon$. You are looking for $\tau$ such that equation (2) holds, which using equation (1) would require

$$ |\tau - (-2)^{k}| < \epsilon, \quad | (-2)^{k_0} + (-2)^{\ell_0} - \tau - (-2)^\ell| < 3 \epsilon $$

which implies (by triangle inequality)

$$ | (-2)^{k_0} + (-2)^{\ell_0} - (-2)^\ell - (-2)^k | < 4 \epsilon $$

Now suppose $\epsilon < \frac14$, then as we are looking on the LHS at differences between integers, for the difference to be less than 1 the term has to be zero. Hence we must have

$$ (-2)^{k_0} + (-2)^{\ell_0} - (-2)^\ell - (-2)^k = 0 \tag{A}$$

By the "fact" about powers of $(-2)$ I proved for you, for a fixed $(k_0,\ell_0)$ this equation has finitely many (in fact at most 3) solutions.

Hence a necessary condition for both equations (1) and (2) to hold is that $\tau$ is within $\epsilon < \frac14$ of $(-2)^k$, where $(k,\ell)$ is one of the no more than 3 solutions to equation $A$.

This shows that the set of $\tau$ satisfying the requirement is bounded.

Answer 2

I've already given you quite a bit of details in the other answer. I think what you are having trouble with is the visualization. So let me add a picture and talk a bit about the proof using it.

The picture shown above is a plot of the function $g(x,y)$. The teal squares are where $g(x,y) = 1$; I've set $\epsilon = 1/4$ for this picture (which is small enough for the result to hold, but for writing up a proof in words it is more convenient to use something even smaller like $1/10$ so you don't have to worry about applying Pythagorean theorem and such). At the middle of the picture you see a square with cross hairs, that is the square corresponding to $k = \ell = 0$, so $x \approx y \approx 1$.

Given a fixed $t$, the function $\tau \mapsto g(\tau, t-\tau)$ can be thought of as a cross section of the function $g$ along a line with slope $-1$ (the line $x + y = t$). The compact support property is the same as saying that the line intersects finitely many squares. I've marked 4 such $t$ values in the picture.

• In blue is the case $t = 0$. This is the "generic case" where the line doesn't intersect any of the teal squares at all. In this case $g(\tau,t-\tau)$ is identically zero.
• In purple is the case $t = -10$. Notice that this line intersects two squares. (It will also not intersect any further squares outside of the plotted region.)
• In red are the cases $t = 2$ and $t = 8$. Notice that they each intersect 3 squares.

So now let me explain aspects of the proof.

Reduction to integers

Each square can be labelled by $(k,\ell)$, where $(x,y) = ((-2)^k, (-2)^\ell)$ is the center of the square.

We can partition the $\mathbb{N}^2 = \{(k,\ell)\}$ into equivalence classes, with each class corresponding to those $(k,\ell)$ with the same $(-2)^k + (-2)^\ell$ sum.

From the picture it should be intuitive that as long as $\epsilon$ is small enough (in fact, as long as $\epsilon < \frac34 \sqrt{2}$ is good enough), it is impossible for a single $x +y = t$ line to pass through squares of different classes. So whether $g(\tau,t-\tau)$ has compact support reduces to whether each equivalence class has only finitely many members.

The two different types of intersections

The difference between the red and purple lines can be summarized thus: in the purple case $t / 2$ is not a power of $(-2)$. So it can only be written as $(-2)^k + (-2)^\ell$ for $k \neq \ell$. And the elementary number theory fact that I proved for you tells you that for each such $t$, only two solutions are available (basically you are allowed to swap $k$ and $\ell$).

In the red case you have that $t / 2$ is a power of $(-2)$, or $t = 2(-2)^m$. In this case, there are three possibilities for the $(k,\ell)$ squares, either $k = \ell = m$, or $k = m + 2$ and $\ell = m +1$, or $k = m + 1$ and $\ell = m + 2$.

(As illustration, the purple curve shown in the picture passes through $(k,\ell) \in \{(1,3), (3,1)\}$; the top most red curve passes through $(k,\ell) \in \{(2,2), (3,4), (4,3)\}$.