- #1

- 104

- 4

Let ##F:[0,2\pi] --> Complex##

##F## is integrable riemman.

show for all ##\epsilon>0## you can find a ##g##, continuous and periodic ##2\pi## s,t: ##||f-g||_2<\epsilon##

What I tried ( in short ), which is nothing almost, but all I know:

because g in continuous and periodic, according to Weirsterass sentence, we can find polynom ##P(X)## such that ##||g-p||_2<\epsilon##

and I know that, because f is integrable, ##||S_n(f)(x)-f(x)||_2<\epsilon##

But I dont know how to use the two of them to reach the answer, since I can not connect between p and f or to reach ##S_n(f)(x)## someway.

any tip will be welcomed!

EDIT:

Oww, I just noticed there is actually a fourier server here, I didnt see it because it said analysis and not fourier.

if needed, move it there please, if not, then you can keep it here :)

##F## is integrable riemman.

show for all ##\epsilon>0## you can find a ##g##, continuous and periodic ##2\pi## s,t: ##||f-g||_2<\epsilon##

What I tried ( in short ), which is nothing almost, but all I know:

because g in continuous and periodic, according to Weirsterass sentence, we can find polynom ##P(X)## such that ##||g-p||_2<\epsilon##

and I know that, because f is integrable, ##||S_n(f)(x)-f(x)||_2<\epsilon##

But I dont know how to use the two of them to reach the answer, since I can not connect between p and f or to reach ##S_n(f)(x)## someway.

any tip will be welcomed!

EDIT:

Oww, I just noticed there is actually a fourier server here, I didnt see it because it said analysis and not fourier.

if needed, move it there please, if not, then you can keep it here :)

Last edited: