# Thread: Problem of a function

1. ## Problem of a function

Suppose that $f:\mathbb{R}\mapsto\mathbb{R}$ is a continuous, bounded, strictly increasing function.

Questions:

(a) Show that there is a point $a_{1}\epsilon\mathbb{R}$ such that $f(a_{1})>a_{1}$
(b) For each $n\epsilon\mathbb{N}$, define $a_{n+1}=f(a_{n})$. Explain why $a_{2}>a_{1}$. Then explain why $a_{n+1}>a_{n}$ for all $n\epsilon\mathbb{N}$
(c) Explain why the sequence ${a_{n}}$ is bounded above.
(d) Explain why the sequence ${a_{n}}$ converges to some number, L.
(e) Explain why f(L) = L.

My work so far:

(a) f is bounded and strictly increasing, therefore f'(x) > 0, so there is a point $x\epsilon\mathbb{R}$ such that x < f(x). I know this reason is not enough, still working on the proper proof.

(b) Now $a_{2}=a_{1+1}=f(a_{1})>a_{1}$ as defined by part (a).
Let $S = {n \epsilon \mathbb{N}:a_{n+1}>a_{n}}$
$1 \epsilon S$ as $a_{1+1}=a_{2} > a_{1}$
let $k \epsilon S$, then $a_{k+1} > a_{k}$
Now $a_{k+1+1} = f(a_{k+1}) > a_{k+1}$, thus proved $k+1 \epsilon S$
Therefore S is inductive, and proved $a_{n+1}>a_{n}$

(c) Because f is continuous and bounded.

(d) Because f is monotonic, implies ${a_{n}}$ is monotonic, thus ${a_{n}}$ is bounded. All bounded monotonic sequence converges, thus the sequence converges to L.

(e) ${a_{n}}$converges to L, then $f({a_{n}})$ converges to L because f is continuous.

I know I made some mistakes up there, please check, thank you.

KK

2. Originally Posted by tttcomrader
(a) Show that there is a point $a_{1}\epsilon\mathbb{R}$ such that $f(a_{1})>a_{1}$
Let us assume that $f(x)\geq x$ for all $x\in \mathbb{R}$. Then $f(x)$ is not bounded because the function $x$ is not bounded. Which is a contradiction.

Now follow the logic....
FOR ALL $x\in \mathbb{R}$ we have $f(x)\geq x$.

The negation of that (which must be because by contradiction) is,
FOR SOME $x\in \mathbb{R}$ we have $f(x).
Because the negation of a universal quantifier is an existencial quantifier. And the negation of $\geq$ is $<$.

3. Originally Posted by tttcomrader

(a) f is bounded and strictly increasing, therefore f'(x) > 0, so there is a point $x\epsilon\mathbb{R}$ such that x < f(x). I know this reason is not enough, still working on the proper proof.
The problem with that is we do not know whether $f$ is differenciable or not.

(b) Now $a_{2}=a_{1+1}=f(a_{1})>a_{1}$ as defined by part (a).
Let $S = {n \epsilon \mathbb{N}:a_{n+1}>a_{n}}$
$1 \epsilon S$ as $a_{1+1}=a_{2} > a_{1}$
let $k \epsilon S$, then $a_{k+1} > a_{k}$
Now $a_{k+1+1} = f(a_{k+1}) > a_{k+1}$, thus proved $k+1 \epsilon S$
Therefore S is inductive, and proved $a_{n+1}>a_{n}$
I got the same thing. It is inductive because $a_{n}>a_{n-1}$ then $a_{n+1}=f(a_n)>f(a_{n-1})=a_n$ because it is strictly increasing.

4. I'm sorry, I don't really understand what you were doing. Did you proved f(x) < x by contradiction? But I'm trying to prove f(x) > x.

What is negation?

Sorry about my lack of understanding.

KK

5. Originally Posted by tttcomrader
I'm sorry, I don't really understand what you were doing. Did you proved f(x) < x by contradiction? But I'm trying to prove f(x) > x.

What is negation?

Sorry about my lack of understanding.

KK
Sorry I proved a statement that you are not looking for.

6. Originally Posted by ThePerfectHacker
The real number $f(a_1)$ is an upper bound.
Because,
$f(a_1)>a_1>a_2>a_3>...$.
Isn't $a_{n+1}>a_n$? So shouldn't $a_1?

So would you say f is bounded above by $f(a_n)$?

Thanks.

KK

7. Originally Posted by tttcomrader
Isn't $a_{n+1}>a_n$? So shouldn't $a_1?

So would you say f is bounded above by $f(a_n)$?

Thanks.

KK
Sorry, again. I am doing this entire problem with reverse inequalities, igonore those posts.

No you cannot say $f(a_n)$ an an upper bounded because $a_n$ is not a real number.