Functions in Normed Linear Space

[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