Norm 2, f Integrable function, show: ##||f-g||_2<\epsilon##

[physics1000]

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 :)

 

[mathman]

Simple method: nth step of Riemann sum approx. to integral - within each section, place a constant = approx. to function used above except at ends of length $\epsilon /(2n)$. Connect ends of remaining intervals by straight lines. This will give a continuous curve with integral ~ $\epsilon$ of approx. sum.

 

[physics1000]

Simple method: nth step of Riemann sum approx. to integral - within each section, place a constant = approx. to function used above except at ends of length $\epsilon /(2n)$. Connect ends of remaining intervals by straight lines. This will give a continuous curve with integral ~ $\epsilon$ of approx. sum.

Ahh forgot to mention, I know we can solve it using Riemann method, but I understand that concept real hard. we have not touched Riemann since calculus two ( integrable subject ).
Is there not any way with norms maybe? :(
I really do not know anything already with riemann

 

[pasmith]

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 :)

The point here is that $g$ must periodic and continuous, but $f$ is only defined on $[0,2\pi]$ and is merely Riemann integrable. However, if $f$ is $L_2$-equivalent to a continuous function $f_c$, in the sense that $\|f - f_c\|_2 = 0$, then you can clearly work with $f_c$ instead.

A starting point is to consider the periodic extension of $f$, and deal with the ways in which it might fail to be continuous. One obvious problem is that you are not given that $f(0) = f(2\pi)$. But you can deal with that by multiplying $f$ by a continuous function which is 1 everywhere except in small neighbourhoods of $0$ and $2\pi$, where it rapidly adjusts to zero; for example, the following family of functions for $\delta > 0$: $$h_\delta: [0, 2\pi] \to \mathbb{R}: x \mapsto \begin{cases} \frac x\delta & 0 \leq x < \delta \\<br /> 1 & \delta \leq x \leq 2\pi - \delta \\<br /> \frac{2\pi - x}{2\pi - \delta} & 2\pi - \delta < x \leq 2\pi. \end{cases}$$
Another problem is that although you know that $f$ is Riemann integrable, so its discontinuities are confined to a set of measure zero, you are not actually told that $f$ is continuous. If it isn't $L_2$-equivalent to a continuous function, then you will need to take a similar approach at each point of discontinuity.

It remains to show that this approach results in a continuous periodic function $g$ such that $\|g - f\|_2 < \epsilon$.

 

[physics1000]

The point here is that $g$ must periodic and continuous, but $f$ is only defined on $[0,2\pi]$ and is merely Riemann integrable. However, if $f$ is $L_2$-equivalent to a continuous function $f_c$, in the sense that $\|f - f_c\|_2 = 0$, then you can clearly work with $f_c$ instead.

A starting point is to consider the periodic extension of $f$, and deal with the ways in which it might fail to be continuous. One obvious problem is that you are not given that $f(0) = f(2\pi)$. But you can deal with that by multiplying $f$ by a continuous function which is 1 everywhere except in small neighbourhoods of $0$ and $2\pi$, where it rapidly adjusts to zero; for example, the following family of functions for $\delta > 0$: $$h_\delta: [0, 2\pi] \to \mathbb{R}: x \mapsto \begin{cases} \frac x\delta & 0 \leq x < \delta \\<br /> 1 & \delta \leq x \leq 2\pi - \delta \\<br /> \frac{2\pi - x}{2\pi - \delta} & 2\pi - \delta < x \leq 2\pi. \end{cases}$$
Another problem is that although you know that $f$ is Riemann integrable, so its discontinuities are confined to a set of measure zero, you are not actually told that $f$ is continuous. If it isn't $L_2$-equivalent to a continuous function, then you will need to take a similar approach at each point of discontinuity.

It remains to show that this approach results in a continuous periodic function $g$ such that $\|g - f\|_2 < \epsilon$.

Ahh, I thought I could skip the delta's part, I guess its impossible.
Thanks, I will try to work with it :)

 

[Office_Shredder]

What about this: $\int_0^xf(t)dt$ is continuous, so can be approximated by some polynomial $G(x)$. $G'(x)$ is a polynomial that approximates $f$ in the $L_2$ sense (I haven't checked this works but it feels like it must) then tweak the endpoints to make it periodic.

Edit: OK, this isn't going to work easily. For example $F(x)=x$ is uniformly approximatws by $G(x)=x+\epsilon \sin(x/\epsilon)$ to arbitrary precision as $\epsilon\to 0$ but $F'=1$ and $G'=1+\cos(x/\epsilon)$.

This isn't quite the same situation since $G$ isn't a polynomial and $G$ is the sequence that is a good approximator, but if you can find a sequence of good polynomials it's probably hard to describe them and prove they work.