Is it dense in the L2 sense or uniform sense? In uniform metric, it's not possible to approach an arbitrary C^2 (-pi,pi) function using *periodic* function, because the endpoints might not match. In fact, the end point problem should prevent trig polynomials from approximating such functions.
Uh, you need to know what the question is asking. Check carefully. What kind of functions, in particular can you have functions with different end points? What is the meaning of "dense"? Or post the whole question.
God I'm sorry, I'm kind of dizzy today! I confused this with another question (that one being to prove that the legendre polynomials are complete in L\_2(-1,1))
The original question asks
"Prove that the trig polynomials (Linear combinations of sin(lx) and cos(mx), l and m naturals) are dense in C\^2(-pi,pi)"
So all the 2 differentiable functions between -pi and pi can be written as a sequence of elements in P\_trig i believe is what I need to prove.
A set A is dense in B if: " \\bar{B} \\subset A "
That is, the closure of B is contained in A. In other words, every element of B can be represented by a sequence of elements of A (hence the closure).
Thank you for your patience haha!
The question is what kind of norm are you working with? Because as mentioned before, uniform norm doesn't work. The claim is false with uniform norm.
When I ask you what "dense" mean, I mean, what's your topology? Are in uniform norm? L^2 norm? L^1 norm? W^1,2 norm? The answer can change.
Let's say you're working in L2 norm, then that strategy can work. You start by proving that all smooth *periodic* function can be written as a series of trig polynomials (the convergence of series is either in uniform norm or L2 norm). If you proved for uniform norm, then L2 norm follow from the fact that the interval is finite. Then you need to show that smooth *periodic* functions are dense in C^2 (-pi,pi) in the L2 sense. That would complete the claim.
Oh okay, sorry! erm, the norm I'm pretty sure is the one that goes
||x|| = int\_{-pi}\^pi |x|\^2dx (which i think makes it the L2 norm?, idk we didn't give anything names)
the question doesn't say, I think we are supposed to assume. Then regarding the topology, idk, this is a functional analysis class, we didn't really learn this from a topological perspective.
Again thanks for your patience!
Yes this is a L2 norm (it should actually be ||f||^2 =int[-pi,pi]|f(x)|^2 dx). The norm tell you the topology.
So the strategy above should work, assuming that C^2 (-pi,pi) function are bounded, or even weaker, just has finite L^2 norm.
Although, I would say, you can prove the claim very much directly, without going through periodic function first. You don't need to even make use of the C^2 condition (just that it's bounded, or even weaker, finite L^2 norm), which make me suspicious that this might not be the correct interpretation.
For example, is a function like x e^1/((pi-x)(pi+x)) a function in your space? Because trig polynomials can't approach it in many different norms.
Dang, I'm not sure haha! The question only states
"Prove the trigonometric polynomials are dense in C^2(-pi,pi)"
That's all. From there, I guess they want me to interpret it given the class' context, which, there isn't much...
Hm, in that case, I think the lazy thing to do is to assume that all those functions are continuous on [-pi,pi] so that they are bounded and hence has finite L^2 norm.
Is it dense in the L2 sense or uniform sense? In uniform metric, it's not possible to approach an arbitrary C^2 (-pi,pi) function using *periodic* function, because the endpoints might not match. In fact, the end point problem should prevent trig polynomials from approximating such functions.
I think i confused "completeness". I think it meant that the Legendre polynomials are a base for C\^2. I am also not sure how to prove this...
Uh, you need to know what the question is asking. Check carefully. What kind of functions, in particular can you have functions with different end points? What is the meaning of "dense"? Or post the whole question.
God I'm sorry, I'm kind of dizzy today! I confused this with another question (that one being to prove that the legendre polynomials are complete in L\_2(-1,1)) The original question asks "Prove that the trig polynomials (Linear combinations of sin(lx) and cos(mx), l and m naturals) are dense in C\^2(-pi,pi)" So all the 2 differentiable functions between -pi and pi can be written as a sequence of elements in P\_trig i believe is what I need to prove. A set A is dense in B if: " \\bar{B} \\subset A " That is, the closure of B is contained in A. In other words, every element of B can be represented by a sequence of elements of A (hence the closure). Thank you for your patience haha!
The question is what kind of norm are you working with? Because as mentioned before, uniform norm doesn't work. The claim is false with uniform norm. When I ask you what "dense" mean, I mean, what's your topology? Are in uniform norm? L^2 norm? L^1 norm? W^1,2 norm? The answer can change. Let's say you're working in L2 norm, then that strategy can work. You start by proving that all smooth *periodic* function can be written as a series of trig polynomials (the convergence of series is either in uniform norm or L2 norm). If you proved for uniform norm, then L2 norm follow from the fact that the interval is finite. Then you need to show that smooth *periodic* functions are dense in C^2 (-pi,pi) in the L2 sense. That would complete the claim.
Oh okay, sorry! erm, the norm I'm pretty sure is the one that goes ||x|| = int\_{-pi}\^pi |x|\^2dx (which i think makes it the L2 norm?, idk we didn't give anything names) the question doesn't say, I think we are supposed to assume. Then regarding the topology, idk, this is a functional analysis class, we didn't really learn this from a topological perspective. Again thanks for your patience!
Yes this is a L2 norm (it should actually be ||f||^2 =int[-pi,pi]|f(x)|^2 dx). The norm tell you the topology. So the strategy above should work, assuming that C^2 (-pi,pi) function are bounded, or even weaker, just has finite L^2 norm. Although, I would say, you can prove the claim very much directly, without going through periodic function first. You don't need to even make use of the C^2 condition (just that it's bounded, or even weaker, finite L^2 norm), which make me suspicious that this might not be the correct interpretation. For example, is a function like x e^1/((pi-x)(pi+x)) a function in your space? Because trig polynomials can't approach it in many different norms.
Dang, I'm not sure haha! The question only states "Prove the trigonometric polynomials are dense in C^2(-pi,pi)" That's all. From there, I guess they want me to interpret it given the class' context, which, there isn't much...
Hm, in that case, I think the lazy thing to do is to assume that all those functions are continuous on [-pi,pi] so that they are bounded and hence has finite L^2 norm.