Derivation of Legendre Polynomials from only orthogonality

Question

I recently stumbled on the idea of Legendre polynomials, from the perspective of using them to approximate functions over a region, and I discovered all of these other ways to express them, using differential equations, recurrence relations, and other forms. I can follow how these imply the orthogonality conditions, but I'm curious about going the other direction. Specifically:

Is it possible, starting from only

1. The orthogonality condition: If $m\not=n$, $\int_{-1}^1P_mP_ndx=0$, and

2. The normalization condition: $P_n(1) = 1$

To derive the recursion relation, the differential equation, and the Rodriguez form for the Legendre Polynomials?

Answer 1

I think I figured it out. I was inspired by the technique in this answer. They used the inner product $$(f,g) = \int_{-1}^1fgdx$$ Then $(xf,g)=(f,xg)$, and using this, they applied this to the Legendre polynomials: $$(xP_n, P_m) = (P_n, xP_m)$$

Since $P_n$ are an orthogonal basis of all polynomials, you can express $xP_n$ as a sum of Legendre polynomials, but since $xP_m$ only has degree up to m+1, the above identity equates to zero if $m+1<n$. Therefore, since $xP_n$ has degree $n+1$, it must be a linear combination of $P_{n-1}$ and $P_{n+1}$ (not $P_n$ as it has the wrong parity of powers of x).

However, that is true of any orthogonal sequence, regardless of the specific orthogonality condition. To get further, we need to use the boundary [-1,1]. This is where we use integration by parts:

$(f',g)=-(f,g')$ if $fg=0$ at $x=1$ or $-1$. Thus, if $f$ or $g$ has $1-x^2$ as a factor, then we can move derivatives from one to the other.

There must be two derivatives to cancel out the two powers of $x$ in $1-x^2$, and $1-x^2$ must be a factor in exactly one of $f$ and $g$. Therefore, using: $$(((1-x^2)P_n')',P_m)$$ $$=-((1-x^2)P_n',P_m')$$ $$=-(P_n',(1-x^2)P_m')$$ $$=(P_n,((1-x^2)P_m')')$$

The expression: $((1-x^2)P_n')'$ has degree $n$, and therefore can be expressed as a sum of Legendre polynomials of degree up to $n$. However, if m<n, the final inner product must be zero, as $((1-x^2)P_m')'$ has degree $m<n$, which $P_n$ is orthogonal to.

Therefore, the only Legendre polynomial that is not orthogonal to $((1-x^2)P_n')'$ is $P_n$ itself. Thus, $((1-x^2)P_n')'=\lambda P_n$. From this, you can derive all the other properties of the Legendre functions (as other answers show).