# Functions in Normed Linear Space

1. Dec 8, 2011

### bugatti79

If we have $\{f,g \in \mathbb{C} [a,b]: p\ge || f(x)-g(x)|| \le q \forall x \in [a,b]\}$

If we know the functions ie, f(x)=x^3+3 and g(x)=x-1. Will the sup norm norm involve max{|f(x)|,|g(x)| for plotting?
And if so, how would one get the 'max' for each function?

Thanks

2. Dec 8, 2011

### LCKurtz

Is that first inequality supposed to be ≤?
If you are asking for the sup norm of f-g, no, you wouldn't get it in terms of $\|f\| \hbox{ and }\|g\|$. You would get it by calculating max|f(x)-g(x)| by usual calculus methods. And I don't see what your question has to do with your inequality above.

3. Dec 8, 2011

### bugatti79

Yes, you are correct regards the inequality. What calculus methods are used to evaluate this? Is the inequality telling us the limits between p and q for this norm, right?

4. Dec 8, 2011

### LCKurtz

For one thing, you need to be more careful with your notation. You have functions f and g ε C[a,b]. You define a norm on this space:$$\|f\| = max_{[a,b]}|f(x)|$$ The notation $\|f(x)\|$ doesn't mean anything. $\|f\|$ is the norm of the function and |f(x)| is the value at of the function f at x. Presumably your original inequality should have read$$p\le|f(x) - g(x)|\le q$$ for all x in [a,b], and presumably p ≥ 0.

It still isn't clear to me what you are actually trying to calculate. Certainly that inequality implies that $\|f-g\|\le q$ but you can't claim equality. And you can not calculate $\|f-g\|$ exactly without knowing specific formulas for f and g. And if you have specific formulas for f and g, you don't need the p-q inequality for anything.

I don't know if I have answered your question because I don't know what the real question is.

5. Dec 9, 2011

### bugatti79

Ok, thanks for your reply. The task is to draw the region in which these functions exist. I am choosing arbitrary numbers and functions so I can understand how to draw this region with these functions etc.

So if we let $f(x)=x^3+3$ and $g(x)=x-1$ and p=3, q=6 say, how would I evaluate

$\|f-g \|$?

Is this norm written as $\|f-g \|=max_{[a,b]}|f(x)-g(x)|$?

Thanks

6. Dec 9, 2011

### I like Serena

I'm trying to reverse engineer what your question is.
What you wrote in the opening post makes no mathematical sense.
Do you perhaps have the literal problem statement as given to you?

Right now, I'm guessing that you're asked to draw the region where p≤|f(x)-g(x)|≤q for the given functions f, g and the given values p, q.
You would also need values for a and b though...

Actually, you should also specify how the norm ||*|| is defined.
The use of double vertical lines suggests it is not simply the absolute value...
Do you have any information on that?

Last edited: Dec 9, 2011
7. Dec 9, 2011

### bugatti79

Im basically trying to consruct an exercise from what I have in my notes that will help me understand this type of task.

yes, this is the task but with double lines and not single as you hav above...If we define a=0 and b=1 lets say.

The norm is defined as I understand it as follows

$$\{f,g \in \mathbb{C} [a,b]: p\ge || f(x)-g(x)|| \le q \forall x \in [a,b]\}$$

where C[a,b] is the function space with sup norm...

Makes sense now I hope...?

8. Dec 9, 2011

### Fredrik

Staff Emeritus
It doesn't. The notation $\{x\in S:P(x)\}$ is used to define a set. It denotes the set of all x in S such that condition P(x) is satisfied. So your statement is read as "the set of all f,g in $\mathbb C[a,b]$ such that <condition>". The condition almost makes sense, but it's not clear what $\|f(x)-g(x)\|$ means. The part that makes the least sense is "the set of all f,g". There should only be one symbol there, not two. Also, what do you mean by $\mathbb C[a,b]$? Continuous complex-valued functions defined on [a,b]? I think the standard notation is to use a regular C, not the mathbb C. Does your book/professor use $\mathbb C$?

9. Dec 9, 2011

### bugatti79

Hi Fredrik,

Thanks for the clear explanation. You are correct, it should be just regular C. Ok, If I rearrange the question like this then

$$\{g \in C[a,b]: p \le || f(x)-g(x)|| \le q \forall x \in [a,b]\}$$ with

f(x) = x^3+3 and a=0 and b=1. Would this type of question make sense in which I am asked to plot the function within this space?

10. Dec 9, 2011

### Fredrik

Staff Emeritus
You can use a function $f\in C[0,1]$ to define a subset of $C[0,1]$ by $$\{g\in C[0,1]:p\leq|f(x)-g(x)|\leq q\ \forall x\in[0,1]\}$$ or by $$\{g\in C[0,1]:p\leq\|f-g\|\leq q\}.$$ These appear to be two different sets.

What function do you want to plot? Both sets contain many functions.

11. Dec 9, 2011

### bugatti79

Now we are getting somewhere. I would like to undestand both in terms of plotting. If we concentrate on the second one and let p=2 and q =6 say.

What approach do I take to evaluate and plot ||f-g||. wouldnt p and q be on the real line?

Thanks

12. Dec 9, 2011

### Fredrik

Staff Emeritus
Since we're using the supremum norm, defined by $$\|h\|=\sup_{t\in[0,1]}|h(t)|$$ for all $h\in C[0,1]$, it's not too hard to visualize what the condition $\|f-g\|\leq q$ is saying. First plot f. Then draw two copies of the graph of f a distance q above and below the graph of f. If g is a member of the set we have defined, then the graph of g must be contained in the region between the two new graphs.

The condition $p\leq\|f-g\|$ is not quite that easy to visualize, but if you think about it, you should be able to figure out a way to do it.

13. Dec 9, 2011

### micromass

Staff Emeritus
Plot what?? $\|f-g\|$ is just a number. How would you plot it??

And to evaluate it, you need explicit information about f and g.

14. Dec 9, 2011

### Fredrik

Staff Emeritus
Good point. I didn't even notice.

(I didn't even look closely at that sentence, and just assumed that he just wanted to know something about the graphs of the functions in the set).

15. Dec 9, 2011

### bugatti79

Is f(x)=x^3+3 not explicit information? PerhapsI meant 'draw the region' in f and g locate their graphs not 'plot the functions' which is different I guess?

How did you determine that ||f-g|| was just a number?

16. Dec 9, 2011

### Fredrik

Staff Emeritus
A norm is by definition a function that takes vectors to numbers (and satisfies a few conditions that you should already be familiar with). So the norm of any vector is just a number. In this case, we can also use the explicit definition $\|h\|=\sup_{t\in[0,1]}|h(t)|$, which is clearly just a number, no matter what function h is.

17. Dec 9, 2011

### bugatti79

So to draw the regions in which the functions contain their graphs would just be a number between p=2 and q=6 on the real line. Sounds too simple. What about the function g in C[0,1] and f(x)=x^2-1?

18. Dec 9, 2011

### Fredrik

Staff Emeritus
As I said, the graph of a member of that set would be contained in the region between two copies of the graph of f, one of them 6 (i.e. q) units above the graph of f, and one of them 6 units below the graph of f. That's just what we get from the condition $\|f-g\|\leq q$. What we get from the condition $p\leq \|f-g\|$ is more complicated. I haven't thought it through.

19. Dec 9, 2011

### Ray Vickson

Your notation $||h(x)||$ makes no sense in an one-dimensional problem (i.e. where h(x) = f(x)-g(x) is a NUMBER). In that case you need to use just |h(x)|.

Your seeming desire to use the notation $||h(x)||$ would make sense in a higher-dimensional context, for example, in mappings from $R$ to $R^3.$ In such a case, if
$$h(x) = \left[ \begin{array}{c} h_1(x) \\ h_2(x) \\h_3(x) \end{array} \right],$$
we could have several definitions of ||.||, such as (i) $$||h(x)|| = \max(|h_1(x)|, |h_2(x)|, |h_2(x)|),$$ or (ii)
$$||h(x) ||= \sum_{i=1}^3 h_i(x)^2,$$
or using some other vector norm ||.|| in 3 dimensions. However, to repeat: THIS DOES NOT OCCUR IN ONE DIMENSION.

RGV

20. Dec 9, 2011

### I like Serena

Hehe... just one point that is at least at distance p and is continuously connected.
I've got no clue how to visualize that in a graph!
However I visualize it, it looks the same as when p=0.

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