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Question first, motivation afterwards.

Let $P$ be the vector space of all real polynomials (with no restriction on the degree). Is there an inner product $\langle p\!\mid\!q\rangle$ on $P$, and a fixed polynomial $q\in P$, such that $$\langle p\!\mid\!q\rangle=p(0)$$ for all $p\in P$?

Motivation. This kind of thing is always possible on a finite-dimensional space $V$ over the field $F=\Bbb R$ or $F=\Bbb C$: moreover, even if the inner product is specified, and a linear functional $\phi:V\to F$ is specified (replacing the map $p\mapsto p(0)$), one can always find a suitable $q$.

It can also sometimes be done in infinite dimensional spaces: for example, if $$\langle p\!\mid\!q\rangle=\int_{-1}^1 p(x)q(x)\,dx\quad\hbox{and}\quad \phi(p)=\int_{-1}^1 p(x)\,dx\ ,$$ then obviously $\langle p\!\mid\!1\rangle=\phi(p)$ for all $p$.

If the functional is $p\mapsto p(0)$ then the "obvious attempt" is something like $$\langle p\!\mid\!q\rangle=p(0)q(0)+p(1)q(1)+\cdots\ ,$$ but this will not converge.

I was given an answer to this question which went something like "blah blah Hilbert space blah blah" and which I didn't understand at all, so it would be great if that could be avoided.

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To generalize Ian's answer: if you have any real vector space $V$ and any functional $\alpha:V\to\mathbb{R}$, you can put an inner product on $V$ such that there is an element $q\in V$ such that $\alpha(p)=\langle p\mid q\rangle$. Indeed, you may assume $\alpha\neq 0$, and choose a vector $q\in V$ such that $\alpha(q)=1$. Now let $B$ be a basis for $\ker(\alpha)$; then $C=B\cup\{q\}$ is a basis for $V$. Take the inner product on $V$ induced by this basis (i.e., to compute the inner product, expand each vector with respect to the basis and then take the dot product of their coefficients). Since $\alpha$ vanishes on every basis vector except $q$ and $\alpha(q)=1$, $\alpha(p)$ is just the coefficient of $q$ in the basis expansion of $p$ for all $p\in V$. That is, $\alpha(p)=\langle p\mid q\rangle$.

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  • $\begingroup$ Extremely clear, thank you very much! $\endgroup$ – David Mar 14 '16 at 3:34
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Each polynomial's coefficients form a member of $\ell^2$ when extended with zeros. Then you use the standard $\ell^2$ inner product with $q $ isomorphic to $(1,0,0,\dots) $, i.e. $q=1$.

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    $\begingroup$ To put it a little differently: the critical thing about your polynomials is that they're zero after finitely many coefficients. If you'd instead allowed the possibility of "infinite polynomials" (like Taylor series), then this approach might not work. This also raises another question: is there such a $q$ for any inner product rather than for the standard inner product on $\ell^2$? I ought to have an answer to that, but I don't. :( $\endgroup$ – John Hughes Mar 13 '16 at 23:25
  • $\begingroup$ Thanks @Ian. What is the standard $\ell^2$ inner product? $\endgroup$ – David Mar 13 '16 at 23:33
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    $\begingroup$ @David: For this case, we can express it as follows: given $p(x) = \sum_{i=0}^n a_i x^i$ and $q(x) = \sum_{i=0}^m b_i x^i$, set $\langle p \mid q \rangle := \sum_{i=1}^{\min(n,m)} a_i b_i$. $\endgroup$ – Nate Eldredge Mar 13 '16 at 23:49
  • $\begingroup$ @JohnHughes: No, there needn't exist such a $q$ for a general inner product; only if your evaluation functional is bounded with respect to that inner product. For instance, consider the $L^2([0,1])$ inner product $\langle p \mid q \rangle = \int_{0}^1 p(x)q(x)\,dx$. If such $q$ were to exist, Cauchy-Schwartz would tell us $|p(0)| = |\langle p \mid q \rangle | \le \|p\| \|q\|$. But by taking $p(x) = (1-x)^n$ for very large $n$, we see that we can have $p(0)=1$ while $\|p\| = \sqrt{\int_0^1 p(x)^2\,dx}$ is arbitrarily small. $\endgroup$ – Nate Eldredge Mar 13 '16 at 23:55
  • $\begingroup$ Thanks, @NateEldredge -- that makes sense. Clear and simple (once you see it, of course!) $\endgroup$ – John Hughes Mar 14 '16 at 3:14

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