2
$\begingroup$

I am working on an exercise to show that if $\{v_k\}_{k\ge 0}$ is an orthonormal basis of $L_2([0,1])$, then $v_{k,\ell}(s,t)=v_k(s)v_\ell(t)$ for all $k,\ell\ge 0$ is an orthonormal basis of $L_2([0,1]^2)$. I don't know why $\{v_{k,\ell}\}$ span $L_2([0,1]^2)$

In infinite-dimension Hilbert space $V$, $\{v_k\}$ span $V$ if any $u$ in $V$ can be arbitrarily approximated by finite linear combination of $v_k$.

My idea is that for a function $f\in L_2([0,1]^2)$, we can approximate it by sum of $\alpha_i\mathbf{1}_{[a_i,b_i]\times[c_i,d_i]}(s,t)=\alpha_i\mathbf{1}_{[a_i,b_i]}(s)\mathbf1_{[c_i,d_i]}(t)$. But it seems we could only approximate $\mathbf{1}_{[a_i,b_i]}$ by $\{v_k\}_{k\ge 0}$, but not represent it. So it may cause some problem.

$\endgroup$

2 Answers 2

1
$\begingroup$

First verify that $\{v_{k,l}\}_{k,l \ge 0}$ is an orthonormal set in $L^2([0,1]^2)$:

\begin{align} \langle v_{k,l}, v_{r,s}\rangle &= \iint_{[0,1]^2} v_k(x)v_l(y)\overline{v_r(x)v_s(y)}\,dx\,dy \\ &= \left(\int_{[0,1]} v_k(x)\overline{v_r(x)}\,dx\right)\left(\int_{[0,1]} v_l(y)\overline{v_s(y)}\,dy\right)\\ &= \langle v_k, v_r\rangle \langle v_l, v_s\rangle\\ &= \delta_{kr}\delta_{ls} \\ &= \delta_{(k,l),(r,s)} \end{align}

Now it suffices to show that $\{v_{k,l}\}_{k,l \ge 0}$ is a maximal set, i.e. if $h \perp v_{k,l}, \forall k,l \ge 0$, then $h = 0$.

Indeed, we have

$$0 = \langle h, v_{k,l}\rangle = \int_{[0,1]}\left(\int_{[0,1]}h(x,y)v_k(x)\,dx\right)v_l(y)\,dy = \left\langle y \mapsto \int_{[0,1]}h(x,y)v_k(x)\,dx, v_l\right\rangle, \forall l \ge 0$$

so $y \mapsto \int_{[0,1]}h(x,y)v_k(x)\,dx$ is equal to $0$ for almost every $y$, for every $k \ge 0$. Hence $$E_k = \left\{y \in [0,1] : \int_{[0,1]}h(x,y)v_k(x)\,dx \ne 0\right\}$$ has measure zero and so $E = \bigcup_{k \ge 0} E_k$ also has measure zero. It follows that for every $y \in [0,1] \setminus E$ we have $$\langle x \mapsto h(x,y), v_k(x) \rangle = \int_{[0,1]}h(x,y)v_k(x)\,dx = 0, \forall k \ge 0$$ and so $h(x,y) = 0$ for almost every $x \in [0,1]$.

Therefore $h= 0$ almost everywhere which completes the proof.

$\endgroup$
0
$\begingroup$

I know there is a purely measure theoretical answer for the more general case of a product of measure spaces, but this argument seems to work in this special case.

It suffices to show that $\operatorname{span}\{v_{k,\ell}\}$ is dense in $L_2([0,1]^2)$. Note that $C([0,1]^2)$ is dense in $L^2([0,1]^2)$, and that $A=\operatorname{span}\{(s,t)\mapsto f(s)g(t):f,g\in C([0,1])\}$ is dense in $C([0,1]^2)$. Thus $A$ is dense in $L_2([0,1]^2)$, and we can approximate functions in $A$ by elements of $\operatorname{span}\{v_{k,\ell}\}$. Combining all of this, we see that $\operatorname{span}\{v_{k,\ell}\}$ is dense in $L_2([0,1]^2)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .