Construct a continuous real valued function which takes zero on integers and such that image of function is not closed.

Question

I am trying to construct a continuous real valued function $f:\mathbb{R}\to \mathbb{R}$ which takes zero on all integer points(that is $f(k)=0$ for all $k\in \mathbb{Z}$) and Image(f) is not closed in $\mathbb{R}$

I had $f(x)=\sin(\pi x) $ in mind. But image of $f(x)$ is closed.

I have a feeling that we can use some clever idea to modify this function such that it satisfy our given condition.

Answer 1

The simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $1$ in the $\pm$infinite limit: $$ f(x) = \frac{1+|x|}{2+|x|}\cdot\sin(\pi\cdot x) $$ Plotted together with the asymptotes:

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Incidentally, since you just said “not closed” but not whether it should be bounded, we could also just choose

$$ f\!\!\!\!/(\!\!\!\!/x\!\!\!\!/)\!\!\!\!/ =\!\!\!\!/ x\!\!\!\!/\cdot\!\!\!\!/\sin\!\!\!\!/\!\!\!\!\!\!\!\!/(\pi\!\!\!\!/\cdot\!\!\!\!/ x\!\!\!\!/) $$

^{(The image of this is all of $\mathbb{R}$ which is actually closed, as the commenters remarked.)}

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A more interesting example that just occured to me:

$$ f(x) = \sin x \cdot\sin(\pi\cdot x) $$ Why does this work? Well, this function never reaches $1$ or $-1$, because for that to happen you would simultaneously need $x$ and $\pi\cdot x$ to be an odd-integer multiple of $\tfrac\pi2$. But that can never coincide because $\pi$ is irrational! It does however get arbitrarily close to $\pm1$, in fact it gets close to $-1$ quite quickly due to $\tfrac\pi2 \approx 1.5 = \tfrac32$. But it never actually reaches either boundary.

Answer 2

Using piecewise linear functions (instead of $\sin (\pi x)$) makes this simpler. For each $n \neq 0$ draw the triangle with vertices $(n,0),(n+1,0)$ and $(n+\frac 1 2, 1-\frac 1 {|n|})$. You will immediately see how to construct an example. [You will get a function (piecewise linear function) whose range contains $1-\frac 1 {|n|}$ for each $n$ but does not contain $1$].

Answer 3

If $f$ verifies the desired propery, its restriction $f|_{[n,n+1]}$ gives a continuous function on $[n,n+1]$ that is zero on the edges of the interval, for any $n \in \mathbb{Z}$. Reciprocally, if we have $f_n : [n,n+1] \to \mathbb{R}$ continuous with $f_n(n) = f_n(n+1) = 0$ for each integer $n$, by the gluing lemma this gives a continuous function $f: \mathbb{R} \to \mathbb{R}$ with $f(n) = f_n(n) = 0$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $[n,n+1]$. Now, the function

$$ f_n (t) = \mu_n\sin(\pi(t-n)) $$

takes values on $\mu_n[-1,1] = [-\mu_n,\mu_n]$ and $f_n(n) = f_n(n+1) = 0$. Thus, the family $(f_n)_n$ induces a continuous function $f$ that vanishes at $\mathbb{Z}$ and

$$ f(\mathbb{R}) = \bigcup_{n \in \mathbb{Z}} f_n([n,n+1]) = \bigcup_{n \in \mathbb{Z}}[-\mu_n,\mu_n] $$

so the problem reduces to choosing a sequence $(\mu_n)_n$ so that the former union is open. One possible choice is $\mu_n = 1-\frac{1}{|n|}$ so that

$$ f(\mathbb{R}) = \bigcup_{n\in \mathbb{Z}}[-\mu_n,\mu_n] = \bigcup_{n\in \mathbb{N}}[-1+\frac{1}{n},1-\frac{1}{n}] = (-1,1). $$

Answer 4

The image of $\sin(\pi x)$ is closed because the peaks all reach 1 and -1. A way to make it not closed is for the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $\tanh(x)$. Thus, we can use the function $$ f(x) = \sin(\pi x)\tanh(x), $$ which is obviously continuous, zero at each integer, and can be easily shown to have image $(-1,1)$

Answer 5

Consider

$$f(x) = \sin^2 (\pi x)\frac{x^2}{1+x^2}.$$

Then $f$ is continuous, $f=0$ on the integers, but $f(\mathbb R) = [0,1).$