Is the following theorem true?

Question

So, I am working on some analysis homework and created a lemma to help me prove something. The problem is I don't know if it's true or false, and I don't want to waste a whole bunch of time attempting to prove a false lemma. Is the following lemma true? Proofs are not required, but greatly appreciated.

Theorem: Let $f$ be a one-to-one and continuous function on the interval $[a,b]$. Then, $\lim_{x \rightarrow t} \phi(x) = \lim_{f(x) \rightarrow f(t)} \phi(x)$, where $\lim_{x \rightarrow t} \phi(x)$ exists.

Note: For those of you confused, see Daron's post below. He helps clarify the lemma and provides a possible proof.

The question can be written a little more simply as "is $\lim_{f(x)\rightarrow f(t)}x=t$" true. — Michael Burr, Dec 03 '15 at 14:43
I suppose this would make more sense if $f:[a,b]\to[a,b]$, but even still it would be odd. — Rocket Man, Dec 03 '15 at 14:44
$\lim_{f(x)\to f(t)}$ doesn't actually mean anything. So you'd first have to define that. Also, you don't say what the range of $f$ is, or what $\phi$ is. — Thomas Andrews, Dec 03 '15 at 14:50

score 3 · Accepted Answer · answered Dec 03 '15 at 14:54

3

Your lemma is true when suitable interpreted.

Of course you need $\phi$ to be a continuous function on $[a,b]$.

Then you define $\displaystyle \lim _{f(x) \to f(t)} \phi(x) = L$ to mean that for any $\epsilon >0$ there is $\delta > 0$ such that $|f(x)-f(t)|< \delta \implies |\phi(x)-L|<\epsilon$.

The fact that this is true relies on how $[a,b]$ is closed and bounded. Or in other words compact. One consequence of this is the inverse function to $f$ is also continuous. In fact you could replace $[a,b]$ with any closed and bounded subset and probably get away with the lemma.

answered Dec 03 '15 at 14:54

Daron

10,300

Thanks, this is exactly what I was looking for! I will make sure to mention you above so that hopefully others will see your post and interpret it correctly. – Birdman2246 Dec 03 '15 at 15:04
Follow up question: If $\phi$ is continuous at $x$ instead of being continuous, does this theorem still hold? – Birdman2246 Dec 03 '15 at 15:07
It should still hold if $\phi$ is continuous at $x$ because, with your set-up, you're choosing a different shaped neighborhood around $x$ for your limit. – Michael Burr Dec 03 '15 at 15:18

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

From the comments above, I will provide my guess as to what the OP meant by $\lim_{f(x)\rightarrow f(t)}\phi(x)=\phi(t)$ means: $$ \forall \varepsilon>0,\exists\delta>0\text{ s.t. }\forall x\text{ with }|f(x)-f(t)|<\delta,|\phi(x)-\phi(t)|<\varepsilon. $$

The original statement is true (provided $\phi$ is reasonably continuous). The statement hinges on the fact that for $f:[a,b]\rightarrow\mathbb{R}$ continuous, injective, and monotonic, then $$ \lim_{x:f(x)\rightarrow f(t)}x=t. $$ The limit is taken over $x$, but $f(x)$ is approaching $f(t)$.

In other words, the question is $$ \lim_{s\rightarrow f(t)}f^{-1}(s)=t. $$

From this problem, although it is more general than what you need, it follows that $f^{-1}$ is continuous. Therefore, the limit follows.

Is the following theorem true?

2 Answers2