# A y such that f(x) <= f(y) all x

1. Feb 26, 2013

### R_beta.v3

1. The problem statement, all variables and given/known data
if $f$ is a continuous function, $f(x) > 0$ for all x, and
$$\lim_{x\rightarrow +\infty} {f(x)} = 0 = \lim_{x\rightarrow -\infty} {f(x)}$$
Prove: There is a y such that $f(x) \leq f(y)$ for all x

I'm assuming that the domain of f is ℝ.

2. Relevant equations
3. The attempt at a solution

Since
$$\lim_{x\rightarrow +\infty} {f(x)} = 0 = \lim_{x\rightarrow -\infty} {f(x)}$$
-There is an $N_1 < 0$ such that, for all x, if x < $N_1$ then $|f(x)| = f(x) < f(-1)$
-There is an $N_2 > 0$ such that, for all x, if x > $N_2$ then $|f(x)| = f(x) < f(+1)$

I'm not 100% sure but, this is what I have in mind:
Let $a$ be any number such that $a < N_1$, $- 1$ and $b$ any number such that $b > N_2$, $1$.
Since f is continuous on [a, b], there is a y in [a, b], such that, $f(x) \leq f(y)$ for all x in [a, b]
Now, if $x < a$ then $f(x) < f(-1) \leq f(y)$ and if $x > b$ then $f(x) < f(1) \leq f(y)$

So $f(x) \leq f(y)$ for all x

2. Feb 26, 2013

### haruspex

Why single out -1? What's special about that?

3. Feb 26, 2013

### R_beta.v3

Well, nothing. I used -1 and 1 because they are simple. The "proof" is basically the same with other numbers.

4. Feb 26, 2013

### jbunniii

I think your proof is fine, but it can be decluttered a bit. I don't see why you need both $f(-1)$ and $f(1)$. Wouldn't it still work if you replaced them both with $f(0)$? Also, why not just use the interval $[N_1, N_2]$ instead of introducing $a$ and $b$?

5. Feb 26, 2013

### R_beta.v3

Of course!!! I can't believe that I didn't do that.

Thanks.