# Thread: A proof involving continuity?

1. ## A proof involving continuity?

Okay, so heres the problem:

Suppose $\displaystyle f$ is the function satisfying $\displaystyle |f(x)| \leq |x|$ for all $\displaystyle x$. Show that $\displaystyle f$ is continuous at $\displaystyle 0$.

Okay, this is my work so far:

We need to show $\displaystyle \lim_{x \to 0} f(x) = f(0)$, I started by figuring the value of $\displaystyle f(0)$. To find that value, I proceeded with:

Since:

$\displaystyle x = 0$

and:

$\displaystyle |f(x)| \leq |x|$

then:

$\displaystyle |f(0)| \leq |0| \;\; \Rightarrow \;\; |f(0)| \leq 0$

but:

$\displaystyle |a| \geq 0 \;\;\;\; for \;\; all \;\; a$

So we must have:

$\displaystyle |f(0)| \geq 0$

But we also have the restriction that:

$\displaystyle |f(0)| \leq 0$

Therefore it must be true that:

$\displaystyle |f(0)| = 0 \;\; \Rightarrow \;\; f(0) = 0$

So, now i just need to show that:

$\displaystyle 0 < |x-0| < \delta \;\; \Rightarrow \;\; |f(x) - f(0)| < \epsilon$

but since $\displaystyle f(0) = 0$; we need only show:

$\displaystyle 0 < |x| < \delta \;\; \Rightarrow \;\; |f(x)| < \epsilon$

This is where i'm stuck. I betting the answer is right under my nose, and is really obivous, and I'm just not seeing it. I appreciate any help that anybody can give.

2. Just let $\displaystyle \varepsilon = \delta$

3. Originally Posted by Plato
Just let $\displaystyle \varepsilon = \delta$

$\displaystyle 0 < |x-0| < \delta$

and letting:

$\displaystyle \delta = \epsilon$

Implies that $\displaystyle |f(x) - f(0)| = |f(x)| < \epsilon$

??

I'm guessing the answer is probably blatantly obvious or something, but I'm new to these proofs so I'm still learning. Thanks

4. Prove something stronger:

If $\displaystyle f(x)$ defined on $\displaystyle (a,b)$, and suppose there exist $\displaystyle L\in\mathbb{R}$ so to every two points $\displaystyle x_1$ and $\displaystyle x_2$ in $\displaystyle (a,b)$,

$\displaystyle |f(x_1)-f(x_2)|\leq L|x_1-x_2|$

then $\displaystyle f(x)$ continuous on $\displaystyle (a,b)$

5. $\displaystyle \left| {f(x) - f(0)} \right| = \left| {f(x) - 0} \right| = \left| {f(x)} \right| \leqslant \left| x \right| < \delta = \varepsilon$

6. Originally Posted by mfetch22

$\displaystyle 0 < |x-0| < \delta$

and letting:

$\displaystyle \delta = \epsilon$

Implies that $\displaystyle |f(x) - f(0)| = |f(x)| < \epsilon$

??

I'm guessing the answer is probably blatantly obvious or something, but I'm new to these proofs so I'm still learning. Thanks
$\displaystyle |f(x)| \le |x| < \delta$

EDIT: Darn you Plato. Beaten again.