# Bounded function

• Jan 13th 2011, 01:00 PM
mus
Bounded function
Can someone assist me in the following:

Let V be the space of continuous functions $f: \mathbb{R} \to \mathbb{R}$, such that:

$\forall \epsilon >0 \exists a\in \mathbb{R}^{+} \forall x \in \mathbb{R}: |x|>a \Rightarrow |f(x)|< \epsilon$

I have shown that $V$ is a vector space.

How do I show that every $f \in V$ is bounded.
• Jan 13th 2011, 01:12 PM
Drexel28
Quote:

Originally Posted by mus
Can someone assist me in the following:

Let V be the space of continuous functions $f: \mathbb{R} \to \mathbb{R}$, such that:

$\forall \epsilon >0 \exists a\in \mathbb{R}^{+} \forall x \in \mathbb{R}: |x|>a \Rightarrow |f(x)|< \epsilon$

I have shown that $V$ is a vector space.

How do I show that every $f \in V$ is bounded.

These are called functions of compact support. They're obviously bounded since there exists some compact set $C\subseteq\mathbb{R}$ such that $\displaystyle \sup_{x\in\mathbb{R}-C}|f(x)|<1$ and so for every $x\in\mathbb{R}$ one has $\displaystyle |f(x)|\leqslant \max\{1,\sup_{x\in C}|f(x)|\}$. So answer me this why is $\displaystyle \sup_{x\in C}|f(x)|<\infty$.
• Jan 13th 2011, 01:38 PM
Jose27
Quote:

Originally Posted by Drexel28
These are called functions of compact support.

Actually, functions with compact support are those which are zero outside of some compact set. This space V (sometimes denoted $C_0(\mathbb{R})$) is the space of functions vanishing at infinity.
• Jan 13th 2011, 02:14 PM
surjective
Could you kindly elaborate on the above. It is too abstract!!
• Jan 13th 2011, 02:47 PM
Plato
Quote:

Originally Posted by mus
Let V be the space of continuous functions $f: \mathbb{R} \to \mathbb{R}$, such that:
$\forall \epsilon >0 \exists a\in \mathbb{R}^{+} \forall x \in \mathbb{R}: |x|>a \Rightarrow |f(x)|< \epsilon$
How do I show that every $f \in V$ is bounded.

In the above let $\epsilon =1$ then there is $a_1>0$ such that $|x|>a_1\Rightarrow ~|f(x)|< 1$.
Now $f$ is bounded on $[-a_1,a_1]$. WHY?
Does that mean that $f$ is bounded on $\mathbb{R}~?$
• Jan 13th 2011, 03:17 PM
Drexel28
Quote:

Originally Posted by Jose27
Actually, functions with compact support are those which are zero outside of some compact set. This space V (sometimes denoted $C_0(\mathbb{R})$) is the space of functions vanishing at infinity.

Of course. Stupid slip of the tongue.
• Jan 13th 2011, 04:49 PM
mus
Don't see how f it is bounded on [-a1;a1]? Please explain!
• Jan 13th 2011, 05:05 PM
Plato
Quote:

Originally Posted by mus
Don't see how f it is bounded on [-a1;a1]? Please explain!

Are saying that you are working on the sorts of problems and do not know that continuous functions on finite closed intervals are bounded? If that is true, then you need to sit down a live tutor and work through you misunderstandings.
• Jan 13th 2011, 05:14 PM
mus
This subject is completely new to me and when your teacher rushes through the book without giving time to study the theorems then what do you expect? By the way I did not post to get talked down to but to be enlightened. So if I lack the basics then either my tutor could give that to me or you could be helpful enough to do so. I could say alot more but I will refrain. Could anyone else be kind enough to help me through this problem without the sarcasm.

Thanks a lot.
• Jan 13th 2011, 05:24 PM
Plato
Quote:

Originally Posted by mus
This subject is completely new to me and when your teacher rushes through the book without giving time to study the theorems then what do you expect? By the way I did not post to get talked down to but to be enlightened. So if I lack the basics then either my tutor could give that to me or you could be helpful enough to do so. I could say alot more but I will refrain. Could anyone else be kind enough to help me through this problem without the sarcasm

There in nothing in my reply to suggest that I meant to make you feel bad. I simply suggested that you have a one-on one, face to face contact with you instructor.
It is simply a fact that: Continuous functions on compact sets are bounded. If you do not understand that, then you need ‘live’ help.