# Functions in Normed Linear Space

by bugatti79
Tags: functions, linear, normed, space
P: 660
 Quote by Fredrik 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
 Mentor P: 21,305 |a| <= 6 is equivalent to -6 <= a <= 6.
P: 660
 Quote by 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.
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
 Emeritus Sci Advisor PF Gold P: 9,415 $$\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3$$ 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.
P: 660
 Quote by 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.
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 :-)
 Emeritus Sci Advisor PF Gold P: 9,415 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.
Mentor
P: 21,305
 Quote by 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?
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.
 Quote by bugatti79 Is knowing what the resulting max value not enough, to determine whether f-z is in A? Thanks Fredrik :-)
P: 660
 Quote by 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.
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....

 Quote by 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.
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?
Emeritus
PF Gold
P: 9,415
 Quote by 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'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 $(f-z)'(t)=3t^2-2t$ for all $t\in [0,1]$, so $(f-z)'(t)=0$ implies $t=2/3$.

 Quote by bugatti79 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.
P: 660
 Quote by 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 $(f-z)(t)=t^3-t^2$ for all $t\in[0,1]$, then $(f-z)'(t)=3t^2-2t$ for all $t\in [0,1]$, so $(f-z)'(t)=0$ 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?
Emeritus
PF Gold
P: 9,415
 Quote by bugatti79 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.

 Quote by bugatti79 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.
P: 660
 Quote by 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...etween+0+and+1
 Quote by 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.
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?
Emeritus
PF Gold
P: 9,415
 Quote by 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?
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.
P: 660
 Quote by Fredrik 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 isnt good but at least I can keep learning. :-)
P: 660
 Quote by Fredrik 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...?
 HW Helper Thanks PF Gold P: 7,662 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?
Emeritus
PF Gold
P: 9,415
 Quote by 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...?
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).
P: 660
 Quote by 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?
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

 Quote by Fredrik 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

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