Is there a way to somehow turn a function on an infinite range into a function on a finite range or vice versa?

Question

First I am going to define something.

Definition: For any function $f$ we define the continuous sequence* from $a$ to $b$ to be the function $g(x) = f(ax + b(1-x))$ where $g$ is defined on the range $[0,1)$.

Speaking purely in terms of limits and appealing to intuitivity, we know that $\lim_{x \to 1} g(x) = \lim_{x \to b} f(x)$. I would argue this is true regardless of whether the limit exists but that is more of a geometric equivalence rather than equivalence by value. Either way the existance is irrelevant. This is the motivation behind why I defined this construct.

Now consider $\lim_{x \to \infty} h(x)$. Once again, the existence is irrelevant. My question is whether we can produce a continuous sequence for the interval $[a,\infty)$? Alternatively can we "stretch" a continuous sequence across the domain $[0,\infty)$ so that it becomes a function that has the continuous sequence as a continuous sequence? If it is not possible could one provide sufficient counter-examples?

---

• I name it that because sequences are the standard way of defining discrete limits. Therefore I just thought the name related in that sense. Don't worry sbout it too much. It is a self-defined term.

Answer 1

If I understand your question, the answer is yes.

For any function $f$, define the "continuous sequence from $a$ to infinity" to be

$$g(x) = f\left(a+x/(1-x)^2\right),$$ where $g$ is defined on the range $[0,1)$. Then (conceptually at least),

$$\lim_{x\rightarrow 0} g(x) = \lim_{x\rightarrow a} f(x)$$ $$\lim_{x\rightarrow 1} g(x) = \lim_{x\rightarrow\infty} f(x)$$

---

In formal terms, I might describe your question like this:

The function $h:[0,1)\rightarrow [a,b)$ defined by $h(x) = xb + (1-x)a$ continuously maps the unit interval onto $[a,b)$. As a result, for any function $f$ defined on $[a,b]$, $\lim_{u\rightarrow 1}(f\circ h)(u) = \lim_{x\rightarrow b} f(x)$.

Can we perform a similar trick with $+\infty$ in place of $b$? That is, can we find a continuous function $k:[0,1)\rightarrow [a,\infty)$ such that for all relevant functions $f$, we have $\lim_{u\rightarrow 1}(f\circ k)(u) = \lim_{x\rightarrow \infty}f(x)$?