Continuity implies boundedness in an interval proof

[tylerc1991]

Homework Statement

If a function f is continuous at a point x, then f is bounded on some interval centered at x. That is, \exists M \geq 0 s.t. \forall y, if |x - y| < \delta, then |f(y)| \leq M

Homework Equations

The Attempt at a Solution

Let \varepsilon > 0. Since f is continuous at x, \exists \delta > 0 s.t. \forall y, if |x - y| < \delta, then |f(x) - f(y)| < \varepsilon. Now,

|f(x) - f(y)| < \varepsilon \iff

- \varepsilon < f(x) - f(y) < \varepsilon \iff

f(x) - \varepsilon < f(y) < f(x) + \varepsilon.

Stated differently,

f(y) \in (f(x) - \varepsilon, f(x) + \varepsilon).

We are trying to find an M \geq 0 s.t. |f(y)| < M, or f(y) \in (-M, M). This is where I get a little stuck. I realize that we may choose any \mu > 0 and say that M = f(x) + \varepsilon + \mu. But I run into trouble when doing this. I want something like M = f(x) \pm (\varepsilon + \mu). This would cover the interval that I am trying to get M into, but I don't exactly know how to say it (except by the way just mentioned of course). Is there a way to write this as a SINGLE value (as opposed to the \pm showing up)?

EDIT: What about M = f(x) + |\varepsilon + \mu|?

Thank you for your help!

 

[micromass]

Can't you just do

M=\max\{f(x)+\varepsilon,-f(x)+\varepsilon\}

I see no use for the \mu...

 

[Petek]

If you have to prove that a function is continuous, you have to show that certain conditions are satisfied for every value of \epsilon > 0. However, if you're given that a function is continuous and you have to prove something about it, you can select a specific, convenient value for \epsilon > 0. So let \epsilon = 1. Since f is continuous at x, \exists \delta > 0 such that if |x - y| < \delta, then |f(x) - f(y)| < 1. Can you take it from there?

 

[tylerc1991]

Can't you just do
M=\max\{f(x)+\varepsilon,-f(x)+\varepsilon\}
[/itex]...

Since M \geq 0, where is the negative f(x) coming from and how would it fit into choosing an M?

... if you're given that a function is continuous and you have to prove something about it, you can select a specific, convenient value for \epsilon > 0. So let \epsilon = 1. Since f is continuous at x, \exists \delta > 0 such that if |x - y| < \delta, then |f(x) - f(y)| < 1. Can you take it from there?

Right, but I could just plug in \varepsilon = 1 into what I had written so far and I would still be in the same boat (unless I am just completely misunderstanding). I am getting stuck on picking an M that will work. I can see how to do it if everything is positive (M = f(x) + \varepsilon), but I don't see how to handle it if it starts to dip into the negatives.

 

[Petek]

An additional hint:

|f(y)| = |f(y) - f(x) + f(x)| \leq |f(y) - f(x)| + |f(x)|

 

[micromass]

Since M \geq 0, where is the negative f(x) coming from and how would it fit into choosing an M?

The negative f(x) is to be able to deal with negative function values.

If |f(y)|\leq -f(x)+\varepsilon, then f(x)-\varepsilon\leq f(y). So the f(y) is bounded below.