# Inner Product and Orthogonal Polynomials

Consider the linear space of all real valued polynomials $P$, equipped with the inner product $\langle f, g\rangle=\int^{\infty}_{-\infty}f(t)g(t)e^{-t^2}dt$

(a) Verify that this is an inner product indeed. (You must also explain why the integral converges.) (b) The standard basis on $P$ consists of all monomials $1, t, t^2,...,t^n...$ Use a Gram-Schmidt process to obtain the first 4 orthogonal polynomials: $p_0$, $p_1$, $p_2$ and $p_4$. The polynomials $p_0, p_1,\cdots, p_n\cdots$ obtained by the Gram-Schmidt process are called Hermite polynomials.

• This integral doesn't converge. You're missing a minus sign. Oct 19, 2012 at 2:56
• Yes! You are definitely right! Thank you for pointing that out. Do you know how to show it is convergent? Oct 19, 2012 at 2:58
• You can edit your question to include the $-$ on $e^{-t^2}$. Oct 19, 2012 at 3:24

All this is awfully close to asking for a full solution of your homework... but one can at least explain why the integrals $\langle P,Q\rangle=\displaystyle\int^{+\infty}_{-\infty}P(t)Q(t)\mathrm e^{-t^2}\mathrm dt$ indeed converge.
The first idea is that every polynomial $P$ is bounded on bounded sets and does not grow too quickly at infinity, in the sense that there exists some positive $C$ and $n$ such that $|P(t)|\leqslant C(1+|t|^n)$ for every $t$ (can you prove this?).
Using this for $PQ$, one sees that the task is to prove that the integral $\displaystyle\int^{+\infty}_{-\infty}|t|^n\mathrm e^{-t^2}\mathrm dt$ converges, for every $n\geqslant0$. To do that, note that, for $|t|\leqslant1$, $|t|^n\mathrm e^{-t^2}\leqslant1$, which is integrable on $|t|\leqslant1$. For $|t|\geqslant1$, one can proceed as follows. For every $t$, $\mathrm e^{t^2}\geqslant t^{2n+2}/c_n$ with $c_n=(n+1)!$ hence $|t|^n\mathrm e^{-t^2}\leqslant c_n/|t|^{n+2}\leqslant c_n/t^{2}$. Since $t\mapsto1/t^2$ is integrable on $|t|\geqslant1$, the proof is complete.