Hilbert space: Find an orthonormal basis of $V$.

Question

I'm working with small exercises about Hilbert spaces and I found this problem, which reminds me about the course Linear Algebra that I took for some years ago. The exercise that I have trouble with,

Exercise: Let $H=L^2([0,1],\mu)$ be a Hilbert space, here $\mu$ denotes the Lebesgue measure. We are given three vectors (functions defined on the interval [0,1]), $f_0=1$, $f_1=x$ and $f_2=x^2$.

1. Let $V$ be a subspace of $H$ spanned by $f_0$ and $f_1$. Find an orthonormal basis of $V$.
2. Let $P_V:H\rightarrow V$ be the orthogonal projection from $H$ onto $V$. Find $P_V(f_2)$.
3. Find the distance between $f_2$ and $V$.

Of course I know how to find basis from the core course in Linear Algebra, but I've never seen this type of exercise on this level and therefore I hope I can get some hints and ideas.

Answer 1

Hints: I am also new to Hilbert Space Theory, but I have some ideas as how to proceed. I give you just hints as per your request.

1. Apply Gram Schmidt to find the orthonormal basis.

2. We need to compute

$$P_V(f_2)=\text{arg}\min\limits_{v\in V}||v-x^2||$$

As every $v \in V$ is a linear combination of basis vectors $p_1,p_2$ found in the previous exercise, we can write $v= \lambda p_1+(1-\lambda)p_2$. Thus

$$||v-x^2||=||\lambda p_1+(1-\lambda)p_2-x^2||$$

This norm could be computed and taking derivative with respect to $\lambda$ and setting to $0$,we get the necessary $\lambda$ and hence the required projection vector can be computed.

3. Distance is $0$ as $f_2 \in V$.

Partial attempts. I hope this helps.

Answer 2

For an orthonormal basis you don't have to invoke Gram-Schmidt. (And I believe that neither Gram nor Schmidt would have used Gram-Schmidt to determine that basis.)

Take as first element $f_0=1$, which obviously has length $1$ and as second element any straight line for which the area under and over the ordinate is equal (that makes it perpendicular to $f_0$). An Ansatz would be $a(x-1/2)$. As the integral of $(x-1/2)^2$ is $1/12$ we get $\pm2\sqrt3(x-1/2)$.

You don't need Gram-Schmidt for the projection either. Let $u$, $v$, and $w$ vectors of a Hilbert space. To find the orthogonal projection of $w$ on the subspace generated by $u$ and $v$ split $w$ in two vectors $w=w^\top+w^\perp$ where $w^\top$ ist the projection and $w^\perp$ is perpendicular to the subspace. We have $$ w^\perp=w-w^\top=w-(\lambda u+\mu v). $$ As $\langle w^\perp,u\rangle =\langle w^\perp,w\rangle=0$ we want to solve the system $$ \begin{eqnarray} \lambda\langle u, u\rangle +\mu\langle u, v\rangle&=\langle u, w\rangle\\ \lambda\langle v, u\rangle +\mu\langle v, v\rangle&= \langle v, w\rangle. \end{eqnarray} $$ In case $u$ and $v$ are linearly independent, the Gram-matrix $$ G=\begin{pmatrix} \langle u, u\rangle &\langle u, v\rangle\\ \langle v, u\rangle &\langle v, v\rangle \end{pmatrix} $$ is regular and $$\begin{pmatrix} \lambda\\ \mu \end{pmatrix}=G^{-1} \begin{pmatrix} \langle u,w\rangle\\ \langle v,w\rangle \end{pmatrix}. $$ In our case $u=f_0=1$, $v=f_1=x$, and $w=f_2=x^2$. We easily compute $$\langle u,u\rangle=1,\langle v,v\rangle=1/3,\langle u,v\rangle=1/2,\langle u,w\rangle=1/3, \langle v,w\rangle=1/4, $$ hence $$G=\begin{pmatrix} 1 & 1/2\\ 1/2 & 1/3 \end{pmatrix} $$ from where $$ \begin{pmatrix} \lambda\\ \mu \end{pmatrix}=\frac{1}{1\cdot1/3-1/4} \begin{pmatrix} 1/3 & -1/2\\ -1/2 & 1 \end{pmatrix} \begin{pmatrix} 1/3\\1/4 \end{pmatrix}= \begin{pmatrix} -1/6\\1 \end{pmatrix}, $$ that is, $P_V(f_2)=-1/6+x$.

Furthermore, the distance from $f_2$ to $V$ is the length of $f_2-P_V(f_2)=x^2-x+1/6$ which is easily calculated to be $\sqrt{1/180}$.