Good night, I'm trying to solve this problem about Hilbert space.

Let $\mathbb R_N [X]$ be the vector space of polynomials whose their degrees are less than or equal to $N$. On $\mathbb R_N [X]$, we define the inner product $\langle \cdot, \cdot \rangle$ by $$\langle p,q \rangle = \int_0^1 p(x)q(x) \, \mathrm{d}x, \quad p,q \in \mathbb R_N [X]$$ Prove that $\mathbb R_N [X]$ with $\langle \cdot, \cdot \rangle$ is a Hilbert space.

My attempt:

For $p \in \mathbb R_N [X]$, the induced norm is $\| \cdot \|$ such that $$\| p \| = \sqrt {\int_0^1 p^2(x) \, \mathrm{d}x }$$

Let $(p_n)_{n \in \mathbb N}$ is a Cauchy sequence in $\mathbb R_N [X]$ where $p_n = \sum_{k=0}^N p^{n}_k X^k$ for all $n \in \mathbb N$. It follows that $$\forall \epsilon >0, \exists M \in \mathbb N,\forall n,m> M:\| p_n - p_m \| = \sqrt {\int_0^1 \left (\sum_{k=0}^N (p^{n}_k - p^{m}_k) X^k \right)^2 \, \mathrm{d}x} < \epsilon$$

I guess that $(p_k^n)_{n \in \mathbb N}$ is Cauchy sequence in $\mathbb R$ for all $i = \overline{0,N}$, but I fail to get the desired result.

Could you please shed me some light? Thank you so much!

  • $\begingroup$ Out of curiosity: is it important that the term $p(x)q(x)$ can have a degree of at most $2N>N$? $\endgroup$ – mrtaurho Oct 4 at 21:03
  • $\begingroup$ Hi @mrtaurho, I'm sorry, but I'm unable to understand your comment. Because $\operatorname{deg} (p), \operatorname{deg} (q) \le N$, $\operatorname{deg} (pq) \le 2N$ :)). $\endgroup$ – Akira Oct 4 at 21:08
  • $\begingroup$ Yes, sure. I was just confused that the space is defined as having only polynomials of degree $\leqslant N$ but the defintion of the inner product enables one to get higher degree polynomials as an intermediate. I do not think this is of any importance while now thinking about it. (BTW, this has nothing to do with the solution to the problem). $\endgroup$ – mrtaurho Oct 4 at 21:11
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    $\begingroup$ More generally, every finite-dimensional inner product space is complete, hence a Hilbert space. Indeed, a $d$-dimensional inner product space is isometrically isomorphic to $\mathbb{R}^d$ with its standard Euclidean inner product, which of course is complete. So it is a little bit silly to go around trying to prove that a specific finite-dimensional inner product space is complete... $\endgroup$ – Nate Eldredge Oct 4 at 21:48

Let $\|P\|_{\infty}=\underset{x\in[0,N]}{\sup}{|P(x)|}$. First notice that if $P\in\mathbb{R}_N[X]$, then $$ P=\sum_{k=0}^N{P(k)L_k} $$ where $L_i(j)=\delta_{i,j}$ for all $0\leqslant i,j\leqslant N$. Thus if $\|P_n-P_m\|_{\infty}<\varepsilon$, in particular $|P_n(i)-P_m(i)|\leqslant \|P_n-P_m\|_{\infty}<\varepsilon$ and $(P_n(i))_{n\in\mathbb{N}}$ is a Cauchy sequence and thus converges toward $\ell_i$. Finally $$ \forall n\in\mathbb{N},\,\left\|P_n-\sum_{k=0}^N{\ell_k L_k}\right\|_{\infty}\leqslant \sum_{k=0}^N{|P_n(k)-\ell_k|\|L_k\|_{\infty}}\underset{n\rightarrow +\infty}{\longrightarrow}0 $$ and $$ \lim\limits_{n\rightarrow +\infty}P_n=\sum_{k=0}^N{\ell_k L_k}\in\mathbb{R}_N[X] $$ Since $\dim\mathbb{R}_N[X]<+\infty$, this works with $\|\cdot\|$ instead of $\|\cdot\|_{\infty}$ as well.

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    $\begingroup$ How does the interval $[0,N]$ enter the discussion? The polynomials are of degree at most $N$ and defined in $[0,1]$. Also, what is the point of presenting an argument for a different norm without proving that it suffices for the relevant norm? $\endgroup$ – uniquesolution Oct 4 at 21:29
  • $\begingroup$ The reason I chose the norm $\|\cdot\|_{\infty}$ is to have the inequality $|P_n(i)-P_m(i)|\leqslant\|P_n-P_m\|_{\infty}$ for all $i\in[\![0,N]\!]$ which is not necessary true with $\|\cdot\|$. Moreover, I defined $\|\cdot\|_{\infty}$ with the $\sup$ on $[0,N]$ because I want that the inequality $|P_n(i)-P_m(i)|\leqslant\|P_n-P_m\|_{\infty}$ holds for all $i\in[0,N]\cap\mathbb{N}$, any interval can define this norm, but this would not necessary give me the desired inequality. $\endgroup$ – Tuvasbien Oct 4 at 21:42

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