A problem on orthogonality of two functions in function space

Question

The Problem is: For any $n \geq 1$, define $$P_n(x)= \frac{d^n}{dx^n}(x^n(1-x)^n)$$ Show that $$m \neq n \Rightarrow \int_0^1 \!\!{P_n(x) P_m(x)dx} =0$$

My approach: Actually, I was thinking that the problem is linked to orthogonality of polynomial functions $P(x)$ in the inner product space $\mathbb R[x]$ equipped with the inner product $\langle f(x),g(x)\rangle= \int_0^1{f(x)g(x)dx}$ . But I am unable to take the approach properly. I also think that there is an application of Fourier series into it, but I don't know much about Fourier series. A small hint is warmly appreciated.

Answer 1

Let me use the notation $f^{(n)}(x)$ instead of $\frac{\mathrm d^n}{\mathrm dx^n} f(x)$ and define $p_n(x) = x^n (1-x)^n$ so that $P_n(x) = p_n^{(n)}(x)$.

Note that $p_n$ has roots of multiplicity $n$ at both $x=0$ and $x=1$ so for the derivatives it holds that $$ p_n^{(k)}(x) = 0 \quad\text{for $x=0,1$ and $k<n$}. $$ Using integration by parts we have \begin{align*} \int_0^1 p_n^{(n)}(x) p_m^{(m)}(x) &= \left[ p_n^{(n)}(x) \underbrace{p_m^{(m-1)}(x)}_0\right]_0^1 - \int_0^1 p_n^{(n+1)}(x) p_m^{(m-1)}(x) \\&= - \int_0^1 p_n^{(n+1)}(x) p_m^{(m-1)}(x). \end{align*} Repeating this $m$ times, we obtain \begin{align*} \int_0^1 p_n^{(n)}(x) p_m^{(m)}(x) &= (-1)^m \int_0^1 p_n^{(n+m)}(x) p_m(x). \end{align*} When $m>n$ we note that $n+m>2n=\deg(p_n)$ so that $p_n^{(n+m)}$ is the zero polynomial and hence the integral becomes zero.

Answer 2

Do it with integration by parts.

You'll need to show that for $k<n$ you have $$ \frac{d^k}{dx^k}(x^n(1-x)^n) \Big|_{x=0} = \frac{d^k}{dx^k}(x^n(1-x)^n) \Big|_{x=1}= 0 $$ From that you'll be able to show that the boundary terms from the integration by parts vanish, and you'll get

$$ \int_{0}^1 \frac{d^n}{dx^n}(x^n(1-x)^n) \frac{d^m}{dx^m}(x^m(1-x)^m)dx = (-1)^n \int_{0}^1 x^n(1-x)^n \frac{d^{m+n}}{dx^{m+n}}(x^m(1-x)^m)dx $$

Assuming $n>m$ it's easy to show that the integrand on right side vanishes.