Functions in Normed Linear Space

bugatti79

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

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

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...etween+0+and+1

Fredrik

[tex]\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3[/tex] 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

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

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.

bugatti79

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....



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

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 [itex]sup_{t \in [0,1]} | f-z|=max\{t_1,t_2\}[/itex] or something like this?

Fredrik

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 [itex](f-z)(t)=t^3-t^2[/itex] for all [itex]t\in[0,1][/itex], then [itex](f-z)'(t)=3t^2-2t[/itex] for all [itex]t\in [0,1][/itex], so [itex](f-z)'(t)=0[/itex] implies [itex]t=2/3[/itex].

Yes, that's right.

bugatti79

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

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

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

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

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

bugatti79

Brilliant, thanks Frekrik. As you can see my ability isnt good but at least I can keep learning. :-)

bugatti79

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...?