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**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.