Countable Orthonormal Basis for Hilbert Space

Actually, we have a non-constructive proof.Lemma : Let $mathcalH$ be a Hilbert space. Then it is impossible

that $mathcalH$ has a countable dense subset $D$ and an uncountable

orthonormal set $e_imid iin I$.Proof: Prove by contradiction. Suppose the contrary that $mathcalH$

has a countable dense subset $D$ and an uncountable orthonormal set

$e_imid iin I.$ Let $rfrac14$ and for each $iin I$,

let $B_iB(e_i,r)$, then open ball centered at $e_i$ with

radius $r$. Note that $B_icap B_jemptyset$ whenever $ineq j$.

(For, if there exists $xin B_icap B_j$, then $||e_i-e_j||leq||e_i-x||||x-e_j||e_i$, where the sum is independent

of ordering. (I leave the proof to you)./////////////////////////////////////////////////////////////I expand the last line, state it as a theorem and give a self-contained proof. This is known as the Fourier expansion in Hilbert space.////////////////////////////////////////////////////////////Theorem: Let $mathcalH$ be a Hilbert space (over $mathbbR$

or $mathbbC$) and let $e_imid iin I$ be a maximal (with

respect to $subseteq$) orthonormal set (where the index $I$ may

be uncountable). Then for each $xinmathcalH$, we have $xsum_iin Ilangle x,e_irangle e_i$,

where the series converges in unorder sense (explained in Claim (4)

below).Proof: Claim (1) (Bessel's inequality): For any finite subset $I_1subseteq I$

and any $xinmathcalH$, $sum_iin I_1|langle x,e_irangle|^2leq||x||^2$. Proof of Claim (1): Denote $Pxsum_iin I_1langle x,e_irangle e_i$.

Observe that $Pxbot x-Px$, so $||x||^2||Px||^2||x-Px||^2geq||Px||^2sum_iin I_1|langle x,e_irangle|^2$.Claim (2): For any $xinmathcalH$, $iin Imidlangle x,e_irangleneq0$

is a countable set.Proof of Claim (2): Let $I'iin Imidlangle x,e_irangleneq0$.

Prove by contradiction. Suppose the contrary that $I'$ is uncountable.

For each $ninmathbbN$, let $I_niin Imid|langle x,e_irangle|geqfrac1n$.

Observe that $I'cup_nI_n$, so there exists $n$ such that $I_n$

is uncountable. Choose $k$ such that $frackn^2>||x||^2$.

Choose a finite subset $I'_nsubseteq I_n$ that contains $k$

elements, then $sum_iin I'_n|langle x,e_irangle|^2geqfrackn^2>||x||^2$,

contradicting to claim (1).Claim (3): Given $xinmathcalH$ and let $I_xiin Imidlangle x,e_irangleneq0$.

Fix an enumeration for $I_x$, say $I_xi_1,i_2,ldots$

(finite or infinite), then the series $sum_klangle x,e_i_krangle e_i_k$

is convergent.Proof of Claim (3): If $I_x$ is a finite set, we are done. Suppose

that $I_x$ is a countably infinite set. Let $s_nsum_k1^nlangle x,e_i_krangle e_i_k$.

By Claim (1), for each $n$, $sum_k1^n|langle x,e_i_krangle|^2leq||x||^2$,

so the series $sum_k1^infty|langle x,e_i_krangle|^2$

is convergent. We show that $s_n$ is a Cauchy sequence in $mathcalH$.

Let $varepsilon>0$. Choose $N$ such that $sum_kN^infty|langle x,e_i_krangle|^20$

be arbitrary. Let $U_yB(y,varepsilon)$ be the open ball centered

at $y$ with radius $varepsilon$. Choose $N$ such that $||sum_k1^Nlangle x,e_i_krangle e_i_k-y||leqfracvarepsilon4$.

(If $I_x$ in Claim (3) is a finite set, let $N$ be the number

of elements in $I_x$.) We adopt the notation in Claim (3) and continue

to work with the enumeration $I_xi_1,i_2,ldots$. Let

$I_1i_1,i_2,ldots,i_N$. Clearly $I_1inmathcalC$.

Consider the case that $I_x$ is an infinite set (the finite case

is trivial). By continuity of norm (i.e., the continuity of the map

$xmapsto||x||$), we have

$$

lim_n||sum_k1^Nlangle x,e_i_krangle e_i_k-sum_k1^nlangle x,e_i_krangle e_i_k||||sum_k1^Nlangle x,e_i_krangle e_i_k-y||leqfracvarepsilon4.

$$

On the other hand, for any $n>N$, we have $||sum_k1^Nlangle x,e_i_krangle e_i_k-sum_k1^nlangle x,e_i_krangle e_i_k||^2sum_kN1^n|langle x,e_i_krangle|^2$.

Hence $sum_kN1^infty|langle x,e_i_krangle|^2leqleft(fracvarepsilon4right)^2$.

Let $I_2inmathcalC$ be arbitrary such that $I_1preceq I_2$.

Then

$$

||theta(I_2)-y||leq||sum_k1^Nlangle x,e_i_krangle e_i_k-y||||sum_iin I_2setminus I_1langle x,e_irangle e_i||leqfracvarepsilon4||sum_iin I_2setminus I_1langle x,e_irangle e_i||.

$$

Observe that

$$

||sum_iin I_2setminus I_1langle x,e_irangle e_i||^2sum_iin I_2setminus I_1|langle x,e_irangle|^2leqsum_kN1^infty|langle x,e_i_krangle|^2leqleft(fracvarepsilon4right)^2

$$ because for any $iin I_2setminus I_1$, if $inotini_N1,i_N2,ldots$,

then $langle x,e_irangle0$.Now we have: $||theta(I_2)-y||leqfracvarepsilon2$. This

shows that $sum_iin Ilangle x,e_irangle e_iy$ in unordered

sense.Claim (5): The $y$ defined in Claim (3) and Claim (4) is $x$. That

is $xsum_iin Ilangle x,e_irangle e_i$.Proof of Claim (5): Let $zx-y$. Prove by contradiction. Suppose

the contrary that $zneq0$. We adopt the notation in Claim (3) and

Claim (4). Recall that for each $ainmathcalH$, the map $xmapstolangle x,arangle$

is continuous. Let $iin I$ be arbitrary. Conside the case that $I_x$

is infinite (The finite case is trivial.). We have

$$

langle z,e_iranglelim_nrightarrowinftylangle x-sum_k1^nlangle x,e_i_krangle e_i_k,e_irangle.

$$ If $inotin I_x$, we have $langle x,e_irangle0$ and $langle e_i_k,e_irangle0$

for all $k$, and hence $langle z,e_irangle0$. Suppose that

$iin I_0$, say $ii_k'$ for some $k'inmathbbN$. Then for

any $ngeq k'$, we have

$$

langle x-sum_k1^nlangle x,e_i_krangle e_i_k,e_iranglelangle x,e_irangle-langlesum_k1^nlangle x,e_i_krangle e_i_k,e_iranglelangle x,e_i_k'rangle-langle x,e_i_k'rangle0.

$$ Therefore $langle z,e_irangle0$ in all cases. Define $tildezz/||z||$,

then $e_imid iin Icuptildez$ is an orthonormal set,

containing $e_imid iin I$ as a proper subset. This contradicts

to the maximality of $e_imid iin I$.

If $Dx_n : nin mathbbN$ is a countable set dense in a Hilbert space $mathcalH$, how can I show that Gram-Schmidt algorithm applied to $D$ (or a subset of $D$) produces an orthonormal numerable basis for $mathcalH$?

So far I have been able to prove that every ortonormal basis of $mathcalH$ has to be numerable.

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