Convergence of C[0,2*pi] with f(x)=sin(x) and sup|f(x)|=1

[bugatti79]

Folks,

For C[0,2*pi] and given a function f(x)=sin(x) the supremum |f(x)|=max|f(x)| for x in [a,b]

I calculate the sup|f(x)| to be = 0 but my notes say 1. The latter answer would be the case if f(x) was cos(x)...right?

 

[micromass]

With both sin(x) and cos(x), the supremum is 1. Can you explain how you found 0 as answer??

 

[bugatti79]

With both sin(x) and cos(x), the supremum is 1. Can you explain how you found 0 as answer??

I thought you take the highest evaluation resulting from either sin(0) and sin(2*pi)? Both are 0...?

 

[micromass]

No, you take |f(x)| for all values in $[0,2\pi]$. And you take the maximum for all those values x. So the maximum can also occur at 1/2 or 1 or whatever.

 

[bugatti79]

No, you take |f(x)| for all values in $[0,2\pi]$. And you take the maximum for all those values x. So the maximum can also occur at 1/2 or 1 or whatever.

ahh right...you consider all values from 0 to 2pi on the real line and it is sin(pi/2) in this case?

Thanks

 

[micromass]

ahh right...you consider all values from 0 to 2pi on the real line and it is sin(pi/2) in this case?

Thanks

Yes, both at $\pi/2$ as at $3\pi/2$ is the maximum reached.

 

[bugatti79]

Thanks,

Warning noted.