An incomplete orthonormal system in $L^2$

[gugo82]

There are several useful complete orthonormal systems in $L^2([0,1])$, e.g. the trigonometric system or the shifted Legendre polynomial system; here we show an orthonormal system of $L^2$ functions which is not complete.

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Just few prerequisites:

It's well known that the functional space $L^2([0,1])$ is an Hilbert space with inner product $\langle f,g\rangle :=\int_0^1 f*g " d"x$ (the integral is taken in the Lebesgue sense) and norm $|f|:=\sqrt(\langle f,f \rangle)$.
Two functions $f,g \in L^2([0,1])$ are orthogonal iff $\langle f,g \rangle =0$.
A family $S =\{ u_a\}_(a \in A) \subseteq L^2([0,1])$ is called orthonormal system iff:

$AA a,b\in A, \langle u_a,u_b\rangle =\{(0, " if " a!=b),(1, " if " a=b):}$

An orthonormal system $S$ is said to be complete in $L^2([0,1])$ iff the linear space generated by $S$ is everywhere dense in $L^2([0,1])$, i.e. iff $L^2([0,1])=\bar("span " S)$.
An orthonormal system $\S$ is said to be maximal in $L^2([0,1])$ iff, $AA u \in L^2([0,1])\setminus S$ with $|u|=1$, the family $S':=S \cup \{ u\}$ is not an orthonormal system in $L^2([0,1])$ (in other words, $S$ is maximal iff do not exist orthonormal systems $S'$ s.t. $S \subset S'$).

A very useful result, to be always kept in mind, is the following:

Let $S \subseteq L^2([0,1])$ an orthonormal system.
The following are equivalent:

i) $S$ is complete in $L^2([0,1])$;
ii) $S$ is maximal in $L^2([0,1])$;
iii) the null function $0$ is the only function in $L^2([0,1])$ which is orthogonal to each element of $S$.

(These are prerequisites; no need to proof.)

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Problem:

For each $n\in NN$ let:

$u_n(x):= "sign " sin (2^(n+1) pi x) \quad$ (here $"sign " y:=\{(1, " if " y>0),(0 , " if " y=0),(-1, " if " y<0):}$)

and put $S:=\{ u_n\}_(n \in NN)$.

Show that:

1. $S$ is an orthonormal system in $L^2([0,1])$;

2. $S$ is not complete in $L^2([0,1])$.

Hints: 1. Make an explicit computation of the product $\langle u_n ,u_m\rangle$.
2.There is a non-null constant function which is orthogonal to each $u_n$: find it!

The orthonormal system $S$ is usually called Rademacher system (named after Hans Rademacher, a German mathematician).

[gugo82]

"Gugo82":There are several useful complete orthonormal systems in $L^2([0,1])$, e.g. the trigonometric system or the shifted Legendre polynomial system; here we show an orthonormal system of $L^2$ functions which is not complete.

***
Problem:

For each $n\in NN$ let:

$u_n(x):= "sign " sin (2^(n+1) pi x) \quad$ (here $"sign " y:=\{(1, " if " y>0),(0 , " if " y=0),(-1, " if " y<0):}$)

and put $S:=\{ u_n\}_(n \in NN)$.

Show that:

1) $S$ is an orthonormal system in $L^2([0,1])$;

For $n \in NN$ we have $u_n^2=1$ a.e. in $[0,1]$, so that:

(1) $\quad \langle u_n,u_n \rangle= \int_0^1 u_n^2" d"x=1\quad$.

For $n>m in NN$, in each maximal subinterval $I_k\subseteq [0,1]$ where $u_m$ is a constant, the function $u_n$ changes sign exactly $2^(n-m) -1$ times (check it!); let $I_(j k) \subseteq I_k$, $j=1,\ldots ,2^(n-m)$ be the maximal subinterval of $I_k$ where $u_n$ is a constant: we have:

$\int_(I_k) u_m*u_n" d"x=\sum_(j=1)^(2^(n-m)) \int_(I_(j k)) u_m*u_n" d"x=$
$\quad \quad =\{(\sum_(j=1)^(2^(n-m)) (-1)^j*m(I_(j k)), ", if " u_m=1 " in " I_k), (-\sum_(j=1)^(2^(n-m)) (-1)^j*m(I_(j k)), ", if " u_m=-1 " in " I_k) :}$

Since for all $j$s $m(I_(j k))=1/2^n$, the last member of the previous chain is $0$; therefore:

(2) $\quad \langle u_m,u_n \rangle:=\int_0^1 u_m*u_n" d"x=\sum_(k=1)^(2^m) \int_(I_k) u_m*u_n" d"x=0 \quad$.

In view of 1-2), $S=\{ u_n\}_(n \in NN)$ is an orthonormal system in $L^2([0,1])$.

"Gugo82":2) $S$ is not complete in $L^2([0,1])$. (Hint: There is a non-null constant function which is orthogonal to each $u_n$: find it!)

The function $u(x):=1$ is non null and orthogonal to all the $u_n$s; therefore $S$ is not a complete orthogonal system.

Note that $S':=S\cup \{ u\}$ is actually a complete orthonormal system of $L^2$ functions.