Completeness of an inner product space

Click For Summary

Discussion Overview

The discussion revolves around the completeness of the space of polynomials with complex coefficients, specifically in the context of an inner product defined on this space. Participants explore the implications of this inner product and the completeness of the space, referencing concepts from functional analysis and approximation theory.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant introduces the inner product defined on the space of polynomials and questions the completeness of this space.
  • Another participant references the Weierstrass approximation theorem, suggesting that since continuous functions can be approximated by polynomials, the space of polynomials is not complete.
  • A participant proposes using the function f(x)=e^{-x} and its polynomial approximations to demonstrate that the space is not complete, noting that the limit of these polynomials is not a polynomial.
  • Another participant reiterates the use of f(x)=e^{-x} and its Taylor series expansion, emphasizing that the convergence of polynomial approximations to a non-polynomial function illustrates the incompleteness.

Areas of Agreement / Disagreement

Participants generally agree on the idea that the space of polynomials is not complete, but there are multiple approaches and justifications presented, indicating a lack of consensus on the most rigorous method to demonstrate this.

Contextual Notes

The discussion includes references to approximation theorems and specific functions, but the application of these concepts to the inner product space of polynomials remains nuanced and not fully resolved.

Hjensen
Messages
22
Reaction score
0
I'm on a course which is currently introducing me to the concept of Hilbert spaces and the professor in charge was giving examples of such spaces. He ended by considering [tex]V[/tex], the space of polynomials with complex coefficients from [tex]\mathbb{R}[/tex] to [tex]\mathbb{C}[/tex]. He then, for [tex]f,g\in V[/tex], defined

[tex](f,g)=\int_{0}^{\infty}f(x)\bar{g(x)}e^{-x}dx[/tex]​

and claimed - without proof - that [tex]V[/tex] equipped with [tex](\cdot ,\cdot )[/tex] is an inner product space, but that [tex]V[/tex] is not complete. Could anyone come up with a clever way of showing that this is true?
 
Physics news on Phys.org
There is a classical theorem which applies here (from Wikipedia).
-----------------------------------------------------------
In mathematical analysis, the Weierstrass approximation theorem states that every continuous function defined on an interval [a,b] can be uniformly approximated as closely as desired by a polynomial function. Because polynomials are among the simplest functions, and because computers can directly evaluate polynomials, this theorem has both practical and theoretical relevance, especially in polynomial interpolation. The original version of this result was established by Karl Weierstrass in 1885.
------------------------------------------------------------------------

In other words, (not polynomial) continuous functions can be the limit of polynomials. Therefore the space of only polynomials is not complete.

You would need to do a little work to extend this to the space of functions you are concerned with, but the general idea would be to approximate non - polynomials with polynomial sequences using the given inner product norm, rather than uniform approximation (sup norm).
 
Thank you for your help, I'll get on it later today.
 
Last edited:
Edit: Rubbish.
 
Last edited:
Let f(x)=e-x and fn(x) be the polynomial defined by the first n terms. This will show it is not complete, since fn(x) converges to f(x), which is not a polynomial. Moreover, it is easy to show that any polynomial will have a finite norm.
 
mathman said:
Let f(x)=e-x and fn(x) be the polynomial defined by the first n terms.
of a Taylor's series expansion of [itex]e^{-x}[/itex].

This will show it is not complete, since fn(x) converges to f(x), which is not a polynomial. Moreover, it is easy to show that any polynomial will have a finite norm.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K