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Can somebody either point me to, or show me a proof, that the Legendre polynomials, or any set of eigenfunctions, are complete?

Andrews
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  • Related: http://physics.stackexchange.com/q/68822/2451 – Qmechanic Dec 03 '13 at 16:18
  • By the link on that answer, it seems like they suggest that given the finite dimensional spectral theorem, we should generalize to infinite dimension to state that we can find a "basis". I don't know why we can generalize, nor does that imply that the Legendre Polynomials are the basis, although they are orthogonal-just because there are infinitely many doesn't mean that they are a "basis" right? –  Dec 03 '13 at 16:31
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    For them to be a basis you need to be able to write your function as a linear combination of the legendre polynomials. –  Dec 03 '13 at 16:48
  • There is also a spectral theorem for normal operators in the infinite dimensional case. Note however, that the spectrum or a part of the spectrum might be continous. In this case you don't have a complete set of eigenfunctions. –  Dec 03 '13 at 18:52

4 Answers4

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We follow J. Weidmanns Linear Operators in Hilbert Spaces. We find in section $3.2$ Orthonormal systems and orthonormal bases:

Example 6: In $L_2(-1,1)$ the set $F=\{f_n:n\in \mathbb{N}_0\}$ with $f_n(x)=x^n$ is a linearly independent system.

The application of the Gram-Schmidt orthogonalization process provides an orthonormal system (ONS) $M=\{p_n:n\in\mathbb{N}_0\}$, where $p_n(x)=\sum_{j=0}^na_{n,j}x^j$ holds with $a_{n,n}>0$; i.e., $p_n$ is a polynomial of degree $n$ with a positive leading coefficient.

These polynomials are called the Legendre polynomials. As $F$ is total, the Legendre polynomials constitute an orthonormal basis (ONB) in $L_2(-1,1)$.

The polynomials $p_n$ can be given explicitly: \begin{align*} p_n(x)=\frac{1}{2^nn!}\left(\frac{2n+1}{2}\right)^{\frac{1}{2}}\frac{d^n}{dx^n}\left(x^2-1\right)^n,\qquad\qquad n\in\mathbb{N}_0\tag{1} \end{align*}

$L_2(M)$ denotes a Lebesgue space for a Lebesgue measurable subset $M$ of $\mathbb{R}^m$. If $B$ is a subset of a Hilbert space $H$, then it is said to be total if the linear hull $L(B)$ (the set of finite linear combinations of elements of $B$) is dense in $H$, i.e. ${\overline{L(B)}}=H$.

In order to prove formula (1) it is sufficient to show that the expression given for $p_n$ is a polynomial of degree $n$ whose leading coefficient is positive and that $\langle p_n,p_m\rangle=\delta_{n,m}$.

The first assertion is obvious. For $j<m$ we obtain by a $(j+1)$-fold integration by parts (the integrated terms vanish) that \begin{align*} \int_{-1}^1x^jp_m(x)\,dx&=C_m\int_{-1}^1x^j\frac{d^m}{dx^m}\left(x^2-1\right)^m\,dx\\ &=\ldots\\ &=C_m(-1)^{j+1}\int_{-1}^1\frac{d^{j+1}}{dx^{j+1}}(x^j)\frac{d^{m-j-1}}{dx^{m-j-1}}\left(x^2-1\right)^m\,dx\\ &=0 \end{align*} This implies that $\langle p_n,p_m\rangle=0$ for $n\ne m$. It remains to prove that $\|p_n\|=1$. By integration by parts we obtain that \begin{align*} \int_{-1}^1&\left[\frac{d^n}{dx^n}\left(x^2-1\right)^n\right]^2\,dx\\ &=(-1)^n\int_{-1}^1\left(x^2-1\right)^n\frac{d^{2n}}{dx^{2n}}\left(x^2-1\right)^n\,dx\\ &=(2n)!\int_{-1}^1(1-x)^n(1+x)^n\,dx\\ &=(2n)!\frac{n}{n+1}\int_{-1}^1(1-x)^{n-1}(1+x)^{n+1}\,dx\\ &=(2n)!\frac{n(n-1)}{(n+1)(n+2)}\int_{-1}^1(1-x)^{n-2}(1+x)^{n+2}\,dx\\ &=\ldots\\ &=(2n)!\frac{n(n-1)\cdots 1}{(n+1)(n+2)\cdots 2n}\int_{-1}^1(1+x)^{2n}\,dx\\ &=(2n)!\frac{1}{2n+1}\left[(1+x)^{2n+1}\right]_{-1}^1\\ &=(n!)^2(2n+1)^{-1}2^{2n+1} \end{align*}

From this it follows that $\|p_n\|=1$ and the Legendre polynomials constitute an ONB in $L_2(-1,1)$.

Markus Scheuer
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  • I do not understand the justification of $$(-1)^n\int_{-1}^{1}(x^2-1)\frac{d^{2n}}{dx^{2n}}(x^2-1)^{n}dx=(2n)!\int_{-1}^{1}(1-x)^{n}(1+x)^{n}dx$$ since by the binomial theorem $(x^2-1)^{n}=\sum_{k}^{n}\binom{n}{k}(-1)^{k}x^{2n-2k}$ then $\frac{d^{2n}}{dx^{2n}}(x^2-1)^{n}=(2n)!$ what implies $$(-1)^n\int_{-1}^{1}(x^2-1)\frac{d^{2n}}{dx^{2n}}(x^2-1)^{n}dx=(-1)^{n}(2n)!\int_{-1}^{1}(x^2-1)dx=(-1)^{n+1}(2n)!\frac{4}{3}.$$ Which is different from $(n!)^{2}(2n+1)^{-1}2^{2n+1}$. – Diego Fonseca Nov 14 '17 at 11:56
  • I already saw the problem, the expression $$(-1)^n\int_{-1}^{1}(x^2-1)\frac{d^{2n}}{dx^{2n}}(x^2-1)^{n}dx$$ is not correct, the correct thing is $$(-1)^n\int_{-1}^{1}(x^2-1)^{n}\frac{d^{2n}}{dx^{2n}}(x^2-1)^{n}dx.$$ – Diego Fonseca Nov 14 '17 at 15:03
  • @DiegoFonseca: Great! Thanks a lot for taking the time and finding the mistake. Typo corrected. Best, – Markus Scheuer Nov 14 '17 at 15:47
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I think that B_Scheiner's answer could be condensed to the statement that it depends on it being of Sturm–Liouville type.

Once you have a linear inner product and resolution of the identity, then $$\begin{align}f(x) &= \int\!dx'\, \delta(x-x') f(x')\\ & = \int\!dx'\, \left[ \sum_{n} u_n(x) u_n(x') w(x) \right] f(x')\\ & = \sum_n u_n(x) \left[\int\!dx'\, u_n(x') w(x) f(x') \right]\\ & = \sum_n u_n(x) f_n \end{align}$$

Thus all you really need to do is prove that the set of all Legendre polynomials satisfies $\sum_n p_n(x) p_n(y) = \delta(x-y)$ over the domain.

This properly belongs in math.se though, but I suppose it's something every physicist should know.

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Let $M_n = \int_a^b |f(x)-\sum_i a_i f_i(x)|^2 dx$ where $f_i$ is an orthonormal set of functions (such as the legendre polynomials). The set of $ f_i$ is complete if there is a set of coefficients $\{a_i\}$ such that $\lim_{n -> \infty} M_n=0$. If you can show that you can approximate a function on a closed interval in a way such that $M_n$ goes to zero as n goes to infinity then you are golden ( maybe look into the wierstrass approximation theorem).

I should mention that the legendre polynomials are part of what are termed Sturm -Louiville problems, and your question could be generalized to a much larger set of polynomials.

Anode
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  • You can use $$ to get centered equations (e.g. $$ \lim_{n\to\infty}M_n=0$$) – Kyle Kanos Dec 03 '13 at 16:21
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    A lot of the beauty of quantum mechanics comes from the properties of Strum-Liouville systems. I would recommend "Mathematics of Classical and Quantum Physics " by Byron and Fuller chapter 5. –  Dec 03 '13 at 16:21
  • Thanks I couldnt remember how to do that. –  Dec 03 '13 at 16:22
  • Can I do this for a general function though? I have yet to take any analysis courses-however I feel like you would have to put in some specific function to show it converges... –  Dec 03 '13 at 16:28
  • Forgive me because its been a while since my last analysis class, but I believe that as long as the function is continuous, and bounded on the interval you should be able to prove it in general.The chapter I recommended in byron and fuller really has a good discussion on these topics that should be readable even if you dont have any analysis background. –  Dec 03 '13 at 16:31
  • How though? I don't know how you would progress from having a $f(x)$ floating around. –  Dec 03 '13 at 16:32
  • I'll get back to you on that soon –  Dec 03 '13 at 16:44
  • An alternative reference about proving the completeness of Strum-Liouville eigenfunction is Hilbert and Courant, Methods of mathematical physics, vol I, section VI-3 – user26143 Dec 03 '13 at 16:48
  • Has this question been answered satisfactorily yet? I don't feel like it has, but I'm also a little lost in the analysis. –  Dec 04 '13 at 07:28
  • @Anonymous: Isn't the other answer complete and correct? –  Feb 22 '14 at 13:24
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The easiest proof that the Legendre polynomials are complete is with the Stone-Weierstrass theorem. You just have to prove that linear combinations of Legendre functions are a unital subalgebra of the continuous functions that separates points.

To prove a collection of eigenvectors forms a basis, take a look at Theorem 7 in Appendix D.6 in Lawrence Evans' Partial Differential Equations. This should get you started. It gives conditions that guarantee this: Just apply the theorem to the solution operator of a differential equation.