0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.