# Thread: Question with a Gigantic Hint that makes no sense

1. ## Question with a Gigantic Hint that makes no sense

Hello there, I have this question that comes with a hint; however the hint doesnt seem to make any sense to me whatsoever so I would appreciate some insight into this.

Let f$\displaystyle : \mathbb{R} \rightarrow \mathbb{R}$ be a function. We define $\displaystyle f(x) \rightarrow$0 as $\displaystyle x \rightarrow \infty$to mean that

$\displaystyle \forall \epsilon > 0 \exists K > 0$such that $\displaystyle x > K \implies |f(x)| < \epsilon$

Similarly define $\displaystyle f(x) \rightarrow 0$as $\displaystyle x \rightarrow - \infty$to mean that

$\displaystyle \forall \epsilon > 0 \exists K > 0$ such that $\displaystyle x < - K \implies |f(x)| < \epsilon$

Now suppose that $\displaystyle f: \mathbb{R} \rightarrow [0, \infty )$ is continuous and has $\displaystyle f(x) \rightarrow 0$ as $\displaystyle x \rightarrow \infty$and $\displaystyle f(x) \rightarrow 0$ as $\displaystyle x \rightarrow -\infty$. Show that f is bounded and attains its maximal value on $\displaystyle \mathbb{R}$ , i.e. there is some $\displaystyle v \in \mathbb{R}$ with $\displaystyle f(x) \le f(v) \forall x \in \mathbb{R}$.

[HINT: If f(x) = 0 for all x, there is nothing to prove (why not? [well because f(x)= a constant so it is bounded by that and that is its maximal value for all x] so we can assume that there is some $\displaystyle x \in \mathbb{R}$with f(x) > 0. Then take $\displaystyle \epsilon = \frac{f(x)}{2}$ and find $\displaystyle K_1, K_2 > 0$such that $\displaystyle -K_1 < x < K_2$and $\displaystyle |f(y)| < \epsilon$ for $\displaystyle y < - K_1$ or $\displaystyle y > K_2$. Now use the extreme value theorem ]

Solution: or as much as I can do...

Now $\displaystyle \exists x \in \mathbb{R}$ such that $\displaystyle f(x) > 0$.

Let $\displaystyle \epsilon = \frac{f(x)}{2}$. Choose $\displaystyle K_1 > 0$ such that $\displaystyle y<-K_1 \implies |f(y)| < \epsilon = \frac{f(x)}{2}$.
Choose $\displaystyle K_2 > 0$ such that $\displaystyle y> K_2 \implies |f(y)| < \epsilon = \frac{f(x)}{2}$ .

But $\displaystyle x< -K_1 \implies |f(x)| = f(x) < \frac{f(x)}{2}$ which is a contradiction.

And $\displaystyle x> K_2 \implies |f(x)| = f(x) < \frac{f(x)}{2}$ which is a contradiction.

Hence $\displaystyle -K_1 \le x \le K_2$.

But this doesn't seem to show me anything?? Any help with this would be deeply appreciated.

2. Simply observe that f attains its maximum (that we know which is $\displaystyle \geq f(x))$ in the compact interval $\displaystyle [K_1,K,2]$

3. Originally Posted by slevvio
Hello there, I have this question that comes with a hint; however the hint doesnt seem to make any sense to me whatsoever so I would appreciate some insight into this.

Let f$\displaystyle : \mathbb{R} \rightarrow \mathbb{R}$ be a function. We define $\displaystyle f(x) \rightarrow$0 as $\displaystyle x \rightarrow \infty$to mean that

$\displaystyle \forall \epsilon > 0 \exists K > 0$such that $\displaystyle x > K \implies |f(x)| < \epsilon$

Similarly define $\displaystyle f(x) \rightarrow 0$as $\displaystyle x \rightarrow - \infty$to mean that

$\displaystyle \forall \epsilon > 0 \exists K > 0$ such that $\displaystyle x < - K \implies |f(x)| < \epsilon$

Now suppose that $\displaystyle f: \mathbb{R} \rightarrow [0, \infty )$ is continuous and has $\displaystyle f(x) \rightarrow 0$ as $\displaystyle x \rightarrow \infty$and $\displaystyle f(x) \rightarrow 0$ as $\displaystyle x \rightarrow -\infty$. Show that f is bounded and attains its maximal value on $\displaystyle \mathbb{R}$ , i.e. there is some $\displaystyle v \in \mathbb{R}$ with $\displaystyle f(x) \le f(v) \forall x \in \mathbb{R}$.

[HINT: If f(x) = 0 for all x, there is nothing to prove (why not? [well because f(x)= a constant so it is bounded by that and that is its maximal value for all x] so we can assume that there is some $\displaystyle x \in \mathbb{R}$with f(x) > 0. Then take $\displaystyle \epsilon = \frac{f(x)}{2}$ and find $\displaystyle K_1, K_2 > 0$such that $\displaystyle -K_1 < x < K_2$and $\displaystyle |f(y)| < \epsilon$ for $\displaystyle y < - K_1$ or $\displaystyle y > K_2$. Now use the extreme value theorem ]

Solution: or as much as I can do...

Now $\displaystyle \exists x \in \mathbb{R}$ such that $\displaystyle f(x) > 0$.

Let $\displaystyle \epsilon = \frac{f(x)}{2}$. Choose $\displaystyle K_1 > 0$ such that $\displaystyle y<-K_1 \implies |f(y)| < \epsilon = \frac{f(x)}{2}$.
Choose $\displaystyle K_2 > 0$ such that $\displaystyle y> K_2 \implies |f(y)| < \epsilon = \frac{f(x)}{2}$ .

But $\displaystyle x< -K_1 \implies |f(x)| = f(x) < \frac{f(x)}{2}$ which is a contradiction.

And $\displaystyle x> K_2 \implies |f(x)| = f(x) < \frac{f(x)}{2}$ which is a contradiction.

Hence $\displaystyle -K_1 \le x \le K_2$.

But this doesn't seem to show me anything?? Any help with this would be deeply appreciated.

Well I think the hint is a huge one, and in fact it is practically the solution: in general and from what they give you, you have that that $\displaystyle |f(y)| \leq e=\frac{f(x)}{2} \,,\,\forall y<-K_1\,\,and\,\,\,\forall y>K_2$ , so the function is bounded in $\displaystyle (-\infty,-K_1)\cup (K_2,\infty)$ , and we're left only to worry about boundness of the function in $\displaystyle [-K_1,K_2]$...but this is a closed bounded interval, and f is continuous, so...

Tonio

4. What does compact mean ?

5. Originally Posted by slevvio
What does compact mean ?

In $\displaystyle \mathbb{R}$ this is the same as closed and bounded.

Tonio

6. Originally Posted by slevvio
What does compact mean ?
If you don't know this concept, at least to solve the problem yo must know that in a closed and bounded interval $\displaystyle [K_1,K_2]$ each continuous function attains its maximum. It's the unique thing you have to add to your argument to conclude.

7. Thanks for the help I knew I was on the cusp of solving it hehe

8. ## Simple Question about the Extreme Value Theorem

ok, managed to prove it now thanks