Let $n$ be a given natural number, and let $X$ denote the vector space consisting of the zero polynomial and of all polynomials of degree at most $n$, with real or complex numbers as co-efficients, and defined on a given closed interval $[a,b]$ of the real line, with the inner product defined by $$\langle x, y \rangle \ \colon= \ \int_a^b \ x(t) \ \overline{y(t) } \ \mathrm{d} t \ \ \ \mbox{ for all } \ x, y \in X.$$

Then is $X$ complete with respect to the norm induced by the above inner product?

I know that the space of all continuous functions is not complete in the above norm.

Let $x_m \colon= \sum_{j=0}^n \alpha_{jm} t^j$, where $t \in [a,b]$ and $m \in \mathbb{N}$, be a Cauchy sequence in $X$. Then, given $\epsilon > 0$, there is a natural number $N = N(\epsilon)$ such that $$\Vert x_m - x_k \Vert < \epsilon \ \ \ \mbox{ for all } \ m, k \in \mathbb{N} \ \mbox{ such that } \ m > N \ \mbox{ and } \ k > N.$$ Or, $$\sqrt{\int_a^b \ \vert x_m(t) - x_k(t) \vert^2 \ \mathrm{d} t } < \epsilon \ \ \ \mbox{ for all } \ m, k \in \mathbb{N} \ \mbox{ such that } \ m > N \ \mbox{ and } \ k > N.$$ That is, $$\sqrt{\int_a^b \ \left\vert \sum_{j=0}^n \left( \alpha_{jm} - \alpha_{jk} \right) \ t^j \ \right\vert^2 \ \mathrm{d} t } < \epsilon \ \ \ \mbox{ for all } \ m, k \in \mathbb{N} \ \mbox{ such that } \ m > N \ \mbox{ and } \ k > N.$$

What next?

Our aim should be to achieve the "Cauchy-ness" of the sequence $\alpha_{jm}$ of real or complex numbers, for each $j = 0, 1, \ldots, n$.

Am I right? If so, how to achieve this goal?


marked as duplicate by Jonas Meyer real-analysis Jul 26 '15 at 0:19

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  • 1
    $\begingroup$ Maybe I'm wrong but I think one way to do the proof would be to see that your vector space is finite dimensional over a complete field and such vector spaces are complete (All norms are equivalent, so you it suffices to see this for the euclidean norm on $\mathbb{C}$ or $\mathbb{R}$) $\endgroup$ – jorst Jul 25 '15 at 12:00
  • $\begingroup$ @jorst yes I do know the following results: (1) On a finite-dimensional normed space, every norm is equivalent to every other norm. (2) Every finite-dimensional normed space is a Banach space. So this space is a Banach space, relative to every norm including the norm induced by the above inner product. Am I right? $\endgroup$ – Saaqib Mahmood Jul 25 '15 at 12:57
  • $\begingroup$ yes, exactly. So this answers your question in the other comment as well: the inner product induced by the above may not be the euclidean one but the norms are equivalent and that is all you need. $\endgroup$ – jorst Jul 25 '15 at 16:45

The vector space of all polynomials of degree at most $n$ is isomorphic with $\mathbb{R^{n+1}}$ and is complete for every natural number.

  • $\begingroup$ yes, these two spaces are isomorphic as vector spaces. But, are they also isomorphic as inner product spaces? That's the key question, you see. $\endgroup$ – Saaqib Mahmood Jul 25 '15 at 12:47

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