1
$\begingroup$

enter image description hereLet $g(y)$ be a continuous function defined for sufficiently large real numbers and $\displaystyle \lim_{y \to \infty} g(y) = \infty$.

Does $\displaystyle \lim_{y \to \infty} f(g(y)) = a$ implies $\displaystyle \lim_{x \to\infty} f(x) = a$ ?

I asked this question because I wanted to solve the above exercise from Lang's Undergraduate Analysis.($f(x) = x^x$. The solution is by Rami Shakarchi.)

I think Rami Shakarchi uses the result of my question.

$\endgroup$
  • $\begingroup$ You don't need continuity of $g$, but rather the existence of inverse $h$ of $g$ such that $\lim_{x\to\infty} h(x) =\infty$. $\endgroup$ – Paramanand Singh Sep 6 '17 at 1:46
  • $\begingroup$ @ParamanandSingh If $g$ is continuous, then you do not need it to be injective. The proof follows from the intermediate value theorem without any need for $g^{-1}$ to be well-defined. $\endgroup$ – Michael Lee Sep 6 '17 at 2:06
  • $\begingroup$ @ParamanandSingh That is not true at all. Consider $g(x) = x+2\sin(x)$. $\endgroup$ – Michael Lee Sep 6 '17 at 2:07
  • $\begingroup$ @MichaelLee: got it and thanks for clearing the confusion. +1 given for excellent answer. Will delete my comment about my useless hunch. $\endgroup$ – Paramanand Singh Sep 6 '17 at 2:09
  • $\begingroup$ Please do not include text as an inline image in your question. It is impossible for it to be read by screen readers or internet search engines. $\endgroup$ – Michael Lee Sep 6 '17 at 4:32
1
$\begingroup$

Consider some $M$ such that $M = g(y_0)$ for some $y_0\in \mathbb{R}$. Then, consider any $\{x_n\}_{n=1}^{\infty}\subset \mathbb{R}$ (without loss of generality, let $x_n$ be increasing and let $x_n > M$ for all $n$).$^1$ As $\lim_{y\to \infty} g(y) = \infty$, we have some $y_0'\in \mathbb{R}$ such that $y_0' > y_0$ and $g(y_0') > x_1$. Then, there is a $y_1\in (y_0, y_0')$ such that $g(y_1) = x_1$ by the intermediate value theorem. Do this inductively: for all $n$, choose a $y_n'$ such that $y_n' > y_n$ and $g(y_n') > x_{n+1}$, and let $y_{n+1}\in (y_n, y_n')$ such that $g(y_{n+1}) = x_{n+1}$. Then, $$\lim_{n\to \infty} f(x_n) = \lim_{n\to \infty} f(g(y_n)) = a$$ Since this holds for any $\{x_n\}$, we have that $\lim_{x\to \infty} f(x) = a$.


$^1$We can enforce the first condition here because every sequence that diverges to $\infty$ has a subsequence that increases monotonically. If $\{x_n\}$ is a general sequence such that $x_n\to \infty$, then we can choose any subsequence of $\{x_n\}$, and that subsequence will have a further subsequence that is monotonically increasing and which when mapped under $f$ will have the limit $a$ by the above logic. If every subsequence of a sequence has a further subsequence that converges to the same limit $a$, then the sequence converges to $a$, so $f(x_n)\to a$. We can enforce the second condition because any subsequence that diverges to $\infty$ only has a finite number of terms that are less than or equal to $M$, so we can throw these away because they will not affect the behavior of $f(x_n)$ as $n\to \infty$.

$\endgroup$
  • $\begingroup$ Thank you very much, Mr. Lee. $\endgroup$ – tchappy ha Sep 6 '17 at 3:33
0
$\begingroup$

Yes.

Suppose not. Then we can take some $\varepsilon$ such that no matter how big $K$ is, $f(t)$ is more than $\varepsilon$ away from $a$ for some $t>K$.

Now by assumption, there exists $M$ big enough that $f(g(y))$ is within $\varepsilon$ of $a$ when $y$ is bigger than $M$. Now let $K=f(M)$. By our first statement, there exists $t>K$ such that $f(t)$ is more than $\varepsilon$ away from $a$. But since $g(x)$ approaches infinity, there must be a point $y>M$ where $g(y)>t$. By the Intermediate Value Theorem, there is some $x\in(M,y)$ such that $g(x)=t$.

And now we have a contradiction: Since $x>M$, $f(g(x))$ is within $\varepsilon$ of $a$. However, $g(x)=t$, so by the definition of $t$, we know that $f(g(x))$ is more than $\varepsilon$ away from $a$.

Incidentally, the continuity is really important. Otherwise, we could let $g$ be the floor function, and let $f$ be some function that gets close to $a$ for integers, but moves far away for values between integers.

$\endgroup$
  • $\begingroup$ Thank you very much, D-Slo. $\endgroup$ – tchappy ha Sep 6 '17 at 3:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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