Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading Applied Partial Differential Equations by DuChateu and Zachmann, and the first couple of chapters contain quite a bit of review of Fourier series, as well as theory about L2 integrable functions and orthogonal/orthonormal basis sets of functions.

Some of the exercises require showing that a particular family of functions is not a complete orthogonal or orthonormal family over a certain interval. The definition given by the book which I am able to remember is that a family of functions is not complete in L2 if $\exists$ a non-zero function $g \in L^2: (g, u_k) = 0$, $\{u_k: k = 1, 2,\ldots\}$ The full definition of a complete family of functions is here.

There don't appear to be any exercises on how to prove a family of functions is complete; it would seem this is a more difficult task. For example, the family of functions $(\frac{2}{\pi})^{\frac{1}{2}}\sin(kx)$ is stated to be complete and orthonormal on the interval $L^2(0, \pi)$; IIRC the family of function $\sin(\frac{n\pi x}{L})$ is complete and orthogonal (but not orthonormal) on the interval $(0, L)$.

I don't think the definition in 2 would be much help, certainly one can't evaluate every possible piecewise continuous function on the interval in question? The (perhaps simpleminded) approach I had in mind (assuming the converse of the the above definition of a function not being complete is true) would be to go from there with a proof by contradiction, or else somehow show that if there is some function g that purports to make the inner product $(g, u_k)$ zero on the interval, that this function by necessity would have to converge to $u_j, j \neq k$ at all points on the interval to make the inner product integral go to zero. I haven't gotten very far in my attempts, though, and any further advice on how to go about this would be appreciated.

share|cite|improve this question
I think you must had functions $\sqrt{\frac{2}{\pi}} \cos kx $ for your familie to be complete in the sense you mentionned. Then theory of Fourier series gives you that any $\mathcal{C}^{\infty}$ function which is orthogonal to every $u_k$ is zero. But since $\mathcal{C}^{\infty}$ is dense in $L^2$ , it's true for every function in $L^2$. – Selim Ghazouani Mar 17 '12 at 10:28
up vote 6 down vote accepted

The usual way to prove that families of functions are complete in this sense is via the Stone-Weierstrass theorem, a proof of which can be found here. Essentially this states that a family of real-valued functions $A$ over a compact Hausdorff space $S$ is complete if the following conditions hold:

  • $A$ is an algebra, i.e.
    • $\forall f, g \in A: fg \in A$
    • $\forall f, g \in A: f + g \in A$
    • $\forall \lambda \in \mathbb{R}, f \in A: \lambda f \in A$
    • (all these operations are defined pointwise, i.e. $fg(x) = f(x)g(x)$, etc)
  • $A$ contains all constant functions, i.e.
    • $\forall \lambda \in \mathbb{R} : \exists f \in A \quad f(x) = \lambda \forall x$
  • $A$ separates points, i.e.
    • $\forall x, y \in S : \exists f \in A \quad f(x) \ne f(y)$

These conditions hold for a large number of commonly used families of functions, including polynomials and trigonometric functions.

share|cite|improve this answer
Thanks. I haven't studied enough real analysis to completely follow all this, but it certainly gives me something to work towards. – Bitrex Mar 27 '12 at 10:32
The nice thing about Stone-Weierstrass is that all of the complicated functional analysis is encapsulated in the proof of the theorem, so if you're willing to take the proof on trust, you can use the results without having to do any analysis at all. – Martin O'Leary Mar 27 '12 at 17:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.