Orthogonality of functions related to Legendre polynomials

Question

If $q\in P^{0}_{k}(I)$, i.e $q$ is a polynomial of degree $\leq k$ that vanishes at two end points of the interval $I=(0,1)$ and $$\psi(x)=\frac{(-1)^{k-1}}{k!}\frac{1}{x(1-x)}\frac{d^{k-1}}{dx^{k-1}}[x^{k+1}(1-x)^{k}].$$ How I will prove that $\int^{1}_{0}\psi(x)q(x)=0,~\forall q\in P^{0}_{k}(I).$ Please help. I will be grateful.

Answer 1

One might think that the most efficient answer would be to apply a straightforward change of variables to the Legendre polynomials defined on $(-1, 1)$, but the paper by Crouzeix and Thomée (which was referred to in a comment, now deleted) imposes the atypical requirement $\psi(0) = 0$, which also helps to explain the atypical exponents in the expression $x^{k+1}(1-x)^{k}$; so it seems to be necessary to start from scratch. (Unless, that is, someone can explain what the authors have in mind when they write that the requirements $\psi(0) = 0$, $\psi(1) = 1$, $\int^{1}_{0} \psi(x)q(x)\,dx = 0,~\forall q\in P^{0}_{k}(I)$ "imply easily" the stated formula for $\psi$!)

One has to assume $k > 0$ (or else the problem doesn't make sense); and if $k = 1$, the only solution is $\psi(x) = x$, which agrees with the stated formula; so we can assume from now on that $k \geqslant 2$.

For $k \geqslant 2$, the main condition reduces to $\int^{1}_{0} x(1 - x)\psi(x)p(x)\,dx = 0,~\forall p\in P_{k-2}$, where $P_{k-2}$ is the set of all polynomials of degree $\leqslant k - 2$.

By a standard argument, as in $\S12.3$ of Powell, Approximation Theory and Methods (1981), this condition on its own determines $\psi$ as a multiple of a certain polynomial of degree $k - 1$.

The condition $\psi(1) = 1$ then normalises $\psi$. Finally, the condition $\psi(0) = 0$ imposes an extra factor of $x$, making $\psi$ a polynomial of degree $k$, as expected.

The simplest way to make this last rather fuzzy statement precise is probably to write $\psi(x) = x\varphi_{k-1}(x)$, from the outset, where $\varphi_{k-1}$ is some continuous function on $(0, 1)$, so that the main condition is now: $$ \int^{1}_{0} w(x)\varphi_{k-1}(x)p(x)\,dx = 0,~\forall p\in P_{k-2}, $$ which incorporates a "weight function", $w(x) = x^2(1 - x)$.

Powell's Theorem 12.5 (I'm afraid that at the moment I lack the time to consult other books for the same result) tells us that this is satisfied if we can find a $(k - 1)$-times differentiable function $u$ on $(0, 1)$ such that: \begin{gather*} u^{(k - 1)}(x) = w(x)\varphi_{k-1}(x) \qquad (0 \leqslant x \leqslant 1), \\ u^{(i)}(0) = u^{(i)}(1) = 0 \qquad (i = 0, 1, \ldots k - 2). \end{gather*} The function $u$ is easy to guess, but we don't even need to guess, because it has been stated for us: $$ u(x) = Cx^{k + 1}(1 - x)^k \qquad (0 \leqslant x \leqslant 1), $$ where $C$ is a constant, to be determined by the remaining condition $\varphi_{k-1}(1) = 1$.

Since $u^{(k - 1)}(x)/w(x)$ is a sum of terms which all vanish as $x \to 1$, except for $(-1)^{k - 1}(k!)Cx^{k - 1}$, this condition becomes $C = (-1)^{k - 1}/k!$, and we are done.

[All this has done is to derive the required special result from a standard theorem. Should I also write out a proof of that theorem? It's not long, but it may be redundant.]