Functions in Normed Linear Space

[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...

 

[Fredrik]

That's not a norm. |f(t)-g(t)| is the absolute value of the real number f(t)-g(t).

 

[bugatti79]

That's not a norm. |f(t)-g(t)| is the absolute value of the real number f(t)-g(t).

If that is the absolute value, it could be less than or equal to 6 fair enough, ie the RHS but how could it be '=' to '-'6, ie on the LHS since the modulus is always positive?

Sorry, ignore this post. Thanks

 

[Mark44]

|a| <= 6 is equivalent to -6 <= a <= 6.

 

[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.

Ok, thanks

How would you verify whether another new function like z(x)=5x is an element in A?

According to Wolfram, it looks like this function is with the region...

http://www.wolframalpha.com/input/?i=plot+x%5E3%2C+x%5E3%2B6%2C+x%5E3%2B9%2C+x%5E3-3%2C+5x+between+0+and+1

 

[Fredrik]

\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3 The last step is easy since the derivative of f-z doesn't have any zeroes in [0,1]. This means that f-z is either increasing in the entire interval, or decreasing in the entire interval. Since (f-z)(0)=3 and (f-z)(1)=-1, it must be decreasing, and the maximum value is 3.

 

[bugatti79]

\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3 The last step is easy since the derivative of f-z doesn't have any zeroes in [0,1]. This means that f-z is either increasing in the entire interval, or decreasing in the entire interval. Since (f-z)(0)=3 and (f-z)(1)=-1, it must be decreasing, and the maximum value is 3.

This is becoming interesting...so the slope if you like is negative ie (0,3) and (1,-1)...but what is the importance of determining its slope?

Is knowing what the resulting max value not enough, to determine whether f-z is in A?

Thanks Fredrik :-)

 

[Fredrik]

What you need to determine is the max value in the interval. Since the slope is never zero inside the interval, we know that the function has its max value at one of the endpoints of the interval.

If the slope had been zero somewhere in the interval, then that might have been the point where the function has its max value.

 

[Mark44]

This is becoming interesting...so the slope if you like is negative ie (0,3) and (1,-1)...but what is the importance of determining its slope?

There seem to be some gaps in you knowledge...

If a differentiable function f is increasing on an interval [a, b], its smallest value is f(a) and its largest is f(b). OTOH, if f is decreasing on [a, b], its largest value is f(a) and its smallest is f(b). These are very simple ideas.

Is knowing what the resulting max value not enough, to determine whether f-z is in A?

Thanks Fredrik :-)

 

[bugatti79]

What you need to determine is the max value in the interval. Since the slope is never zero inside the interval, we know that the function has its max value at one of the endpoints of the interval.

If the slope had been zero somewhere in the interval, then that might have been the point where the function has its max value.

So if I had different function like f-z = t^3-t^2 say, then the derivative of this wrt t and subbing in 0 we would get 0 which would impy the max is at t=0?

I am just looking for an example that demonstrates the slope being 0 somewhere in the interval instead of at the ends...

\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3 The last step is easy since the derivative of f-z doesn't have any zeroes in [0,1]. This means that f-z is either increasing in the entire interval, or decreasing in the entire interval. Since (f-z)(0)=3 and (f-z)(1)=-1, it must be decreasing, and the maximum value is 3.

Just one little query regarding the \|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3

You put in 0 and you are left with 3. If you put in 1 we get -1. So we take the highest of those two because sup_{t \in [0,1]} | f-z|=max\{t_1,t_2\} or something like this?

 

[Fredrik]

So if I had different function like f-z = t^3-t^2 say, then the derivative of this wrt t and subbing in 0 we would get 0 which would impy the max is at t=0?

I'm not a fan of the notation f-z=t^3-t^2, because it looks like a function equal to a number. If f-z satisfies (f-z)(t)=t^3-t^2 for all t\in[0,1], then (f-z)&#039;(t)=3t^2-2t for all t\in [0,1], so (f-z)&#039;(t)=0 implies t=2/3.

Just one little query regarding the \|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3

You put in 0 and you are left with 3. If you put in 1 we get -1. So we take the highest of those two because sup_{t \in [0,1]} | f-z|=max\{t_1,t_2\} or something like this?

Yes, that's right.

 

[bugatti79]

I'm not a fan of the notation f-z=t^3-t^2, because it looks like a function equal to a number. If f-z satisfies (f-z)(t)=t^3-t^2 for all t\in[0,1], then (f-z)&#039;(t)=3t^2-2t for all t\in [0,1], so (f-z)&#039;(t)=0 implies t=2/3.

but how would we determine whether this new function is in A? Ok, you have identified that the maximum value lies somewhere at a point corresponding to t=2/3 because this is where the slope is 0. How do we establish whether the function lies above x^3+9 say?

 

[Fredrik]

but how would we determine whether this new function is in A?

I think you called the set D earlier. What we have to do is of course to determine \|f-g\|, and this is fairly straightforward. I'll leave that to you.

Ok, you have identified that the maximum value lies somewhere at a point corresponding to t=2/3 because this is where the slope is 0. How do we establish whether the function lies above x^3+9 say?

I'm not sure why you would want to know, but in general, if you have two functions F and G, and want to know if one of them is "above" the other, you need to find out if the equation F(x)=G(x) has any solutions. If it doesn't, the graphs never intersect.

 

[bugatti79]

Ok, thanks

How would you verify whether another new function like z(x)=5x is an element in D?

According to Wolfram, it looks like this function is with the region...

http://www.wolframalpha.com/input/?i=plot+x%5E3%2C+x%5E3%2B6%2C+x%5E3%2B9%2C+x%5E3-3%2C+5x+between+0+and+1

\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3 The last step is easy since the derivative of f-z doesn't have any zeroes in [0,1]. This means that f-z is either increasing in the entire interval, or decreasing in the entire interval. Since (f-z)(0)=3 and (f-z)(1)=-1, it must be decreasing, and the maximum value is 3.

Just a question on these 2 post. Do we conclude that the function z(x)=5x is in D because the maximum value of 3 is less than our highest intercept of 9 and greater than -3?

 

[Fredrik]

Just a question on these 2 post. Do we conclude that the function z(x)=5x is in D because the maximum value of 3 is less than our highest intercept of 9 and greater than -3?

A function g\in C[0,1] (where C[0,1] now denotes real-valued continuous functions on [0,1]) is in D if and only if 3\leq\|f-g\|\leq 6. So to check if z is in D, you have to find \|f-z\|, and I did. \|f-z\|=3. Since 3\leq 3\leq 6, z is in D.

 

[bugatti79]

A function g\in C[0,1] (where C[0,1] now denotes real-valued continuous functions on [0,1]) is in D if and only if 3\leq\|f-g\|\leq 6. So to check if z is in D, you have to find \|f-z\|, and I did. \|f-z\|=3. Since 3\leq 3\leq 6, z is in D.

Brilliant, thanks Frekrik. As you can see my ability isn't good but at least I can keep learning. :-)

 

[bugatti79]

A function g\in C[0,1] (where C[0,1] now denotes real-valued continuous functions on [0,1]) is in D if and only if 3\leq\|f-g\|\leq 6. So to check if z is in D, you have to find \|f-z\|, and I did. \|f-z\|=3. Since 3\leq 3\leq 6, z is in D.

So we can verify this by checking that the derivative of f-z is not 0. If is WAS 0, then there would be a possibility that the max value could be higher than 6 hence not in D...?

 

[LCKurtz]

I have been reading this thread since my first post which was #2. I am curious bugatti79 about two questions:
1. What course are you currently studying where this topic arose?
2. Have you taken a course in calculus?

 

[Fredrik]

So we can verify this by checking that the derivative of f-z is not 0. If is WAS 0, then there would be a possibility that the max value could be higher than 6 hence not in D...?

f-z has a local maximum at t if and only if (f-z)'(t)=0 and (f-z)''(t)<0. So if (f-z)'(t)≠0 for all t, then we can be sure that f-z doesn't have any local maxima in the interval.

It looks like you need to refresh your memory about some basic calculus (as a couple of people have already suggested).

 

[bugatti79]

I have been reading this thread since my first post which was #2. I am curious bugatti79 about two questions:
1. What course are you currently studying where this topic arose?
2. Have you taken a course in calculus?

1) I am attempting to study analysis as a module which I find difficult as you can see. I will give it a go and see what happens.
2) No I have not taken a course in calculus but I come from an engineering background so I have some exposure to basic calculus which I guess I need to brush up on!

I am finish with thread now

f-z has a local maximum at t if and only if (f-z)'(t)=0 and (f-z)''(t)<0. So if (f-z)'(t)≠0 for all t, then we can be sure that f-z doesn't have any local maxima in the interval.

Thanks