A 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 &gt; 0: <br /> h_\delta: [0, 2\pi] \to \mathbb{R}: x \mapsto \begin{cases} \frac x\delta &amp; 0 \leq x &lt; \delta \\<br /> 1 &amp; \delta \leq x \leq 2\pi - \delta \\<br /> \frac{2\pi - x}{2\pi - \delta} &amp; 2\pi - \delta &lt; 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 &lt; \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 &gt; 0: <br /> h_\delta: [0, 2\pi] \to \mathbb{R}: x \mapsto \begin{cases} \frac x\delta &amp; 0 \leq x &lt; \delta \\<br /> 1 &amp; \delta \leq x \leq 2\pi - \delta \\<br /> \frac{2\pi - x}{2\pi - \delta} &amp; 2\pi - \delta &lt; 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 &lt; \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.