Shifted Legendre polynomials symmetry relation

Question

I have to prove that $p_n(1-x)=(-1)^np_n(x)$ ($x\in[0,1]$) for all $n\in\mathbb{N}$, where $(p_n)_n$ is the family of Legendre polynomials on $[0,1]$: given $(x^n)_{n=0,1,\ldots}$, $(p_n)_n$ is the family of orthogonal monic polynomials that one constructs using Gram-Schmidt process starting from $(x^n)_n$ on $L^2([0,1])$. Now, I have to prove that $\forall n=0,1,2,\ldots$, $p_n(1-x)=(-1)^np_n(x)$ for all $x\in[0,1]$, that would follow by proving that $\tilde{p}_n\perp p_j$ for all $j\neq p$ and $(p_n,\tilde{p}_n)_2=(-1)^n$. I denoted by $\tilde{p}_n(x)=p_n(1-x)$. By induction, one has that the cases $n=0,1$ are trivial. Then, I tried to use the relation given by Gram-Schmidt, so that (using the induction)
\begin{equation}
(\tilde{p}_{n+1},p_j)_2=((1-x)^{n+1},p_j)_2-\frac{(x^{n+1},p_n)_2}{\Vert p_n\Vert_2^2}(\tilde{p}_n,p_j)_2=((1-x)^{n+1},p_j)_2+(-1)^{n+1}\frac{(x^{n+1},p_n)_2}{\Vert p_n\Vert_2^2}(p_n,p_j)_2.
\end{equation}
If $j\neq n$ the second addend is $0$ since $(p_j)_j$ is an orthogonal family. In particular, the second addend vanishes for all $j\neq n$ and if $j=n+1$, this gives $(\tilde{p}_n,p_{n+1})=((1-x)^{n+1},p_{n+1})_2=(-1)^n$, since this is obviously the coefficient of $x^{n+1}$ respect to $p_{n+1}$. If $j=n$,
\begin{equation}
(\tilde{p}_{n+1},p_n)_2=((1-x)^{n+1},p_n)_2+(-1)^{n+1}(x^{n+1},p_n)_2=((1-x)^{n+1}+(-x)^{n+1},p_n)\overset{(?)}{=}0.
\end{equation}
Since it is obvious that $(Q_j,p_k)=0$ whenever $j<k$ and $Q_j$ is a polynomial of degree $j$, it follows that $(\tilde{p}_{n+1},p_j)_2=0$ for all $j>n+1$ (we don't even need the recursive definition of $p_{n+1}$ here). If $j<n$,
\begin{equation}
(\tilde{p}_{n+1},p_j)_2=((1-x)^{n+1},p_j)_2\overset{(?)}{=}0.
\end{equation}
The question is, how do I prove the (?)? I am not supposed to know anything of the shifted Legendre polynomials except for their definition I gave at the beginning and the recursive formula $p_{n+1}(x)=(x+A_n)p_{n}(x)-B_np_{n-1}(x)$, but I cannot use the fact that (one can prove) $A_n=-1/2$ for all $n$ (actually, I have to prove this fact starting from the just given recursive formula and the symmetry of $p_n$'s – and I already did it).

Answer 1

I think I found the answer, but the induction is useless here. The proof I posted before proves that $(\tilde{p}_n,p_n)_2=(-1)^n$ for all $n$. Now, it's enough to observe that $(\tilde{p}_k,p_j)_2=(p_k,\tilde{p}_j)_2$ for all $k,j$ and use the $(Q_j,p_n)_2=0$ for all $n>j$ thing: this implies directly that $(\tilde{p}_n,p_j)_2=0$ for all $j>n$, but also $(\tilde{p}_n,p_j)_2=(p_n\tilde{p}_j)_2=0$ if $j<n$.