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

In summary: 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.For example, if you take the function h_0: [0,2\pi] \to \mathbb{R}: x \mapsto 0, then \|h_0\|_2 = 0 for all x in [0,
  • #1
physics1000
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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 :)
 
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  • #2
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.
 
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  • #3
mathman said:
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
 
  • #4
physics1000 said:
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 [itex]g[/itex] must periodic and continuous, but [itex]f[/itex] is only defined on [itex][0,2\pi][/itex] and is merely Riemann integrable. However, if [itex]f[/itex] is [itex]L_2[/itex]-equivalent to a continuous function [itex]f_c[/itex], in the sense that [itex]\|f - f_c\|_2 = 0[/itex], then you can clearly work with [itex]f_c[/itex] instead.

A starting point is to consider the periodic extension of [itex]f[/itex], and deal with the ways in which it might fail to be continuous. One obvious problem is that you are not given that [itex]f(0) = f(2\pi)[/itex]. But you can deal with that by multiplying [itex]f[/itex] by a continuous function which is 1 everywhere except in small neighbourhoods of [itex]0[/itex] and [itex]2\pi[/itex], where it rapidly adjusts to zero; for example, the following family of functions for [itex]\delta > 0[/itex]: [tex]
h_\delta: [0, 2\pi] \to \mathbb{R}: x \mapsto \begin{cases} \frac x\delta & 0 \leq x < \delta \\
1 & \delta \leq x \leq 2\pi - \delta \\
\frac{2\pi - x}{2\pi - \delta} & 2\pi - \delta < x \leq 2\pi. \end{cases}[/tex]
Another problem is that although you know that [itex]f[/itex] is Riemann integrable, so its discontinuities are confined to a set of measure zero, you are not actually told that [itex]f[/itex] is continuous. If it isn't [itex]L_2[/itex]-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 [itex]g[/itex] such that [itex]\|g - f\|_2 < \epsilon[/itex].
 
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  • #5
pasmith said:
The point here is that [itex]g[/itex] must periodic and continuous, but [itex]f[/itex] is only defined on [itex][0,2\pi][/itex] and is merely Riemann integrable. However, if [itex]f[/itex] is [itex]L_2[/itex]-equivalent to a continuous function [itex]f_c[/itex], in the sense that [itex]\|f - f_c\|_2 = 0[/itex], then you can clearly work with [itex]f_c[/itex] instead.

A starting point is to consider the periodic extension of [itex]f[/itex], and deal with the ways in which it might fail to be continuous. One obvious problem is that you are not given that [itex]f(0) = f(2\pi)[/itex]. But you can deal with that by multiplying [itex]f[/itex] by a continuous function which is 1 everywhere except in small neighbourhoods of [itex]0[/itex] and [itex]2\pi[/itex], where it rapidly adjusts to zero; for example, the following family of functions for [itex]\delta > 0[/itex]: [tex]
h_\delta: [0, 2\pi] \to \mathbb{R}: x \mapsto \begin{cases} \frac x\delta & 0 \leq x < \delta \\
1 & \delta \leq x \leq 2\pi - \delta \\
\frac{2\pi - x}{2\pi - \delta} & 2\pi - \delta < x \leq 2\pi. \end{cases}[/tex]
Another problem is that although you know that [itex]f[/itex] is Riemann integrable, so its discontinuities are confined to a set of measure zero, you are not actually told that [itex]f[/itex] is continuous. If it isn't [itex]L_2[/itex]-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 [itex]g[/itex] such that [itex]\|g - f\|_2 < \epsilon[/itex].
Ahh, I thought I could skip the delta's part, I guess its impossible.
Thanks, I will try to work with it :)
 
  • #6
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.
 
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What does "Norm 2" refer to in this statement?

"Norm 2" refers to the L2 norm, also known as the Euclidean norm, which is a measure of the length or magnitude of a vector in a vector space.

What is an integrable function?

An integrable function is a function that can be represented by a definite integral, meaning it can be integrated over a specific interval to find the area under the curve.

How do you show that ||f-g||_2<\epsilon?

To show that ||f-g||_2<\epsilon, you would need to use the definition of the L2 norm, which is the square root of the integral of the squared difference between two functions, and then manipulate the equation to show that it is less than the given epsilon value.

Why is it important for ||f-g||_2 to be less than epsilon?

It is important for ||f-g||_2 to be less than epsilon because it indicates that the difference between the two functions, f and g, is small and therefore they are close to each other in terms of their L2 norm. This is useful in many applications, such as signal processing and data analysis, where minimizing the difference between two functions is desirable.

What are some real-world examples of where this statement would be applicable?

This statement would be applicable in many areas of science and engineering, such as image and audio processing, where minimizing the difference between two signals is important for accurate analysis and reconstruction. It is also used in statistics and machine learning for measuring the similarity between two datasets or models. Additionally, it is used in physics and engineering for analyzing the convergence of numerical methods and approximations.

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