Functions in Normed Linear Space

[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

 

[LCKurtz]

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

Is that first inequality supposed to be ≤?

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

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.

 

[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?

 

[LCKurtz]

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

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.

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?

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.

 

[bugatti79]

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.

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

 

[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?

 

[bugatti79]

Do you perhaps have the literal problem statement as given to you?

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

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...

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

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?

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...?

 

[Fredrik]

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...?

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 $\mathbb C$?

 

[bugatti79]

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 $\mathbb C$?

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?

 

[Fredrik]

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.

 

[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||. wouldn't p and q be on the real line?

Thanks

 

[Fredrik]

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.

 

[micromass]

What approach do I take to evaluate and plot ||f-g||. wouldn't p and q be on the real line?

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.

 

[Fredrik]

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

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).

 

[bugatti79]

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.

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?

 

[Fredrik]

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

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.

 

[bugatti79]

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.

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?

 

[Fredrik]

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.

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.

 

[Ray Vickson]

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

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

 

[I like Serena]

What we get from the condition $p\leq \|f-g\|$ is more complicated. I haven't thought it through.

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.

 

[bugatti79]

Ok guys, thanks for the replies. I am just wondering if i rearranged the question does it make any difference, its not g(x) but just g.

consider C[0,1] with sup norm. Let f:[0,1] with the function f(x)=x^3+3

let $D =\{g \in C[0,1]:3 \le||f-g|| \le 6\}$...? plot the region in which D is present

Cheers

 

[Fredrik]

Plot the graphs of the functions
\begin{align}
& x\mapsto x^3+9\\
& x\mapsto x^3+6\\
& x\mapsto x^3\\
& x\mapsto x^3-3
\end{align} and label them 1,2,3,4 respectively. Every g in D is contained in the region between graphs 1 and 4, and goes outside of the region between graphs 2 and 3 at least once.

 

[bugatti79]

Plot the graphs of the functions
\begin{align}
& x\mapsto x^3+9\\
& x\mapsto x^3+6\\
& x\mapsto x^3\\
& x\mapsto x^3-3
\end{align} and label them 1,2,3,4 respectively. Every g in D is contained in the region between graphs 1 and 4, and goes outside of the region between graphs 2 and 3 at least once.

Hmm...so there is a difference when we specify g and not g(x)?
How did you arrive at those different y intercepts, ie -3, 6 and 9?

Thanks

 

[Fredrik]

I don't even know what it would mean to specify g(x). Those numbers follow immediately from the conditions $$3\leq\sup_{t\in[0,1]}|f(t)-g(t)|\leq 6$$ that you chose. The vertical distance between the graphs of f and g is never greater than 6, and must at some point be greater than 3.

 

[bugatti79]

I don't even know what it would mean to specify g(x). Those numbers follow immediately from the conditions $$3\leq\sup_{t\in[0,1]}|f(t)-g(t)|\leq 6$$ that you chose. The vertical distance between the graphs of f and g is never greater than 6, and must at some point be greater than 3.

If I understand you correctly, could we have as in this link

http://www.wolframalpha.com/input/?i=plot+x%5E3%2C+x%5E3%2B6%2Cx%5E3%2B9%2C+x%5E3%2B12+between+0+and+1

If this is wrong then I don't understand how you get those intercepts. I mean, why did you leave out x^3 +3 which is the function itself.

 

[Fredrik]

I don't know where you got that +12 from. Have you thought at all about what the inequalities in post #24 mean? You can certainly plot f as well, but why would you?

 

[bugatti79]

I don't even know what it would mean to specify g(x). Those numbers follow immediately from the conditions $$3\leq\sup_{t\in[0,1]}|f(t)-g(t)|\leq 6$$ that you chose. The vertical distance between the graphs of f and g is never greater than 6, and must at some point be greater than 3.

Completely clueless now. Where did you get the -3 and +9 from? Shouldnt there be just 2 graphs, one at +3 and the other at +6. The difference being 3, ie greater than equal to 3 and less than or equal than 6...

I find it difficult to apply the 'the supremum or least upper bound of a set S of real numbers is denoted by sup(S) and is defined to be the smallest real number that is greater than or equal to every number in S' in this example.

 

[Fredrik]

Where did you get the -3 and +9 from?

I got "+3" from your definition of f. Because of the condition $\|f-g\|\leq 6$, I added and subtracted 6 from that (as I said that we need to do in post #12), to get 9 and -3 respectively.

Shouldnt there be just 2 graphs, one at +3 and the other at +6. The difference being 3, ie greater than equal to 3 and less than or equal than 6...

No.

I find it difficult to apply the 'the supremum or least upper bound of a set S of real numbers is denoted by sup(S) and is defined to be the smallest real number that is greater than or equal to every number in S' in this example.

In this case, sup S=max S. Maybe that makes things easier.

 

[bugatti79]

I got "+3" from your definition of f. Because of the condition $\|f-g\|\leq 6$, I added and subtracted 6 from that (as I said that we need to do in post #12), to get 9 and -3 respectively.

Why do you subtract as well as add for this condition? The condition just says <=6, hence you just add 6...

So, you are implying for the condition $3 \ge ||f-g||$ we must add and subtract 3 on f(x)?

And then wherever $3 \ge ||f-g||\le 6$ intersect, is where g lies?

 

[Mark44]

Why do you subtract as well as add for this condition? The condition just says <=6, hence you just add 6...

So, you are implying for the condition $3 \ge ||f-g||$ we must add and subtract 3 on f(x)?

And then wherever $3 \ge ||f-g||\le 6$ intersect, is where g lies?

You have part of the inequality messed up (again).

It should be $3 \le ||f-g||\le 6$.

This is saying that the norm of the difference of the functions is between 3 and 6.

 

[Fredrik]

So, you are implying for the condition $3 \ge ||f-g||$ we must add and subtract 3 on f(x)?

The conditions we've been discussing are $3\leq \|f-g\|$ and $\|f-g\|\leq 6$. The first condition implies that the distance between the graphs along a vertical line must at some point be ≥3. The second implies that the distance between the graphs along a vertical line must never be ≥6.

Why do you subtract as well as add for this condition? The condition just says <=6, hence you just add 6...

What the condition says about a given function g is that 6 is an upper bound for the set $\{|f(t)-g(t)|:t\in[0,1]\}$, and that no upper bound for that set is less than 6. Since 6 is an upper bound, we have $|f(t)-g(t)|\leq 6$, which implies $-6\leq f(t)-g(t)\leq 6$ for all t in [0,1]. So for all t in [0,1], f(t)-6 ≤ g(t) ≤ f(t)+6.

And then wherever $3 \ge ||f-g||\le 6$ intersect, is where g lies?

You don't seem to be thinking at all about what the notation means. $\|f-g\|$ is a number, as mentioned a couple of times above. So even if you turn the first inequality sign the right way, you would still be asking about where a pair of inequalities "intersects". That doesn't make any sense.

Edit: We posted almost at the same time. I had obviously not seen the post below when I wrote this one. I think what I said in this one answers what you asked in the post below. I'll add one more thing here, a minor correction to this:

Plot the graphs of the functions
\begin{align}
& x\mapsto x^3+9\\
& x\mapsto x^3+6\\
& x\mapsto x^3\\
& x\mapsto x^3-3
\end{align} and label them 1,2,3,4 respectively. Every g in D is contained in the region between graphs 1 and 4, and goes outside of the region between graphs 2 and 3 at least once.

g doesn't have to go outside of the region between graphs 2 and 3 (the graphs of f+3 and f-3). It just has to touch one of those graphs at least once.

 

[bugatti79]

You have part of the inequality messed up (again).

It should be $3 \le ||f-g||\le 6$.

This is saying that the norm of the difference of the functions is between 3 and 6.

Yes, I keep messing up that. I keep reading the expression from the right but writing from the left lol ! :-)

Yes, I am trying to plot which leads me onto the following

Plot the graphs of the functions
\begin{align}
& x\mapsto x^3+9\\
& x\mapsto x^3+6\\
& x\mapsto x^3\\
& x\mapsto x^3-3
\end{align} and label them 1,2,3,4 respectively. Every g in D is contained in the region between graphs 1 and 4, and goes outside of the region between graphs 2 and 3 at least once.

I can see now where you get these 4 expression, the question I have now is why do we add and subtract 6 and 3 instead of just addiing 6 and subtract 3 on f(x)?

Sorry, I didnt see that post 31.

 

[bugatti79]

You don't seem to be thinking at all about what the notation means. $\|f-g\|$ is a number, as mentioned a couple of times above. So even if you turn the first inequality sign the right way, you would still be asking about where a pair of inequalities "intersects". That doesn't make any sense.

No, I am just trying to 'draw' the region in which the functions have their graphs which you have demonstrated in post 22, thanks.

So its just the last little bit that I am not getting particularly where you suddenly throw in the '-6 bit' in this expression below...?

$-6\leq f(t)-g(t)\leq 6$...

I appreciated your attention, thanks

What the condition says about a given function g is that 6 is an upper bound for the set $\{|f(t)-g(t)|:t\in[0,1]\}$, and that no upper bound for that set is less than 6. Since 6 is an upper bound, we have $|f(t)-g(t)|\leq 6$, which implies $-6\leq f(t)-g(t)\leq 6$ for all t in [0,1]. So for all t in [0,1], f(t)-6 ≤ g(t) ≤ f(t)+6.

 

[Fredrik]

|f(t)-g(t)|≤6 means exactly that -6 ≤ f(t)-g(t) ≤ 6.

 

[bugatti79]

|f(t)-g(t)|≤6 means exactly that -6 ≤ f(t)-g(t) ≤ 6.

I guess that is by definition and taken as given? Its not intuitive particularly of the norm of the difffernce is just a number...