## Functions in Normed Linear Space

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

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

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

 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 :-)
 Mentor 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
 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 :-)

 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?

Mentor
 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.

 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?

Mentor
 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.

 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?

Mentor
 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.

 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. :-)

 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...?