Norm of differentiation operator with respect to Gaussian norm

Question

Here is a problem from Luenberger's optimization by vector space methods. I would appreciate steps to proceed.

Let $\mathcal{P}_n\subset\mathbb{R}[x]$ be polynomials of degree at most $n\ge0$. Compute the norm of the differentiation operator $D:\mathcal{P}_n\rightarrow\mathcal{P}_n$, $(Dp)(x)=p'(x)$ relative to the norm $$\|p\|^2=\int_{\mathbb{R}}|p(x)|^2e^{-x^2/2}\text{ d}t.$$

Could this be done with Lagrange multipliers, trying to maximize $\int_{\mathbb{R}}(Dp)^2\text{ d}\mu$ where $\int_{\mathbb{R}}p^2\text{ d}\mu=1$? I tried this, and got to $\lambda=\frac{\int_{\mathbb{R}}D^2p\text{ d}\mu}{\int_{\mathbb{R}}p\text{ d}\mu}$, but did not know where to go from here.

Answer 1

Let's first consider $D$ on $L^2(\mathbb R, e^{-x^2/2}\, dx)$. An integration by parts shows that $D^*=-D+x$. The (probabilist's) Hermite polynomials $H_n$ provide an orthogonal basis of this space, and $$ \|H_n\|^2=\sqrt{2\pi}n! , \quad (-D+x)H_n=H_{n+1}. $$ So if we pass to the normalized versions $h_n$, then $(-D+x)h_n=\sqrt{n+1}h_{n+1}$.

If we now consider $D, D^*$ on $\mathcal P_n=L(h_0,\ldots, h_n)$, most of this remains true, except that we of course cannot have $D_n^*H_n=H_{n+1}$, which is not in the space. Rather, $\int H_n Dp e^{-x^2/2}\, dx=0$ by orthogonality, since $Dp$ is of degree $\le n-1$ now, so $D_n^*h_n=0$. Hence $\|D_n\|=\|D_n^*\|=\sqrt{n}$.