Show that there does not exist $f\in L^2(R)$ such that $\overline{{\rm span}\{f(\cdot-n):n\in Z\}}=L^2(R)$

Question

Show that there does not exist $f\in L^2(R)$ such that $$\overline{{\rm span}\{f(\cdot-n):n\in Z\}}=L^2(R).$$

In other words, for any square function $f$, the space of the span of all shifts of $f$ is at most a proper subspace of $L^2(R)$.

Recall a well-known fact, the werner density theorem, that $f\in L^2(R)$ implies that $$\overline{{\rm span}\{f(\cdot-x):x\in R\}}=L^2(R)$$ if and only if $\hat{f}\ne0$ a.e. Thanks for any hints.

Answer 1

You can see this by looking at the Fourier side:

By the wiener density theorem that you already stated, we get $\widehat{f} \neq 0$ almost everywhere (note the $\neq$ instead of $=$).

Now choose $g \in L^2(\Bbb{R})$ with $\widehat{g} = \chi_{(0,1)}$.

Assume $g \in \overline{\rm{span}\{f(\cdot -n) \,|\, n \in \mathbb{Z}\}}$. Then there is a sequence $g_n = \sum_{i=-m_n}^{m_n} \alpha_{i}^{(n)} f(\cdot -i)$ such that $g_n \rightarrow g$ in $L^2$. As the Fourier transform is continuous as a map $L^2 \rightarrow L^2$, we get $\widehat{g_n} \rightarrow \widehat{g}$ in $L^2$.

By switching to a subsequence, we can assume $\widehat{g_n} \rightarrow \widehat{g}$ almost everywhere.

But now note that

$$\widehat{g_n}(\xi) = \widehat{f}(\xi) \cdot \sum_{j=-m_n}^{m_n} \alpha_i^{(n)} e^{\pm 2\pi i j \xi},$$

where you might get a slightly different exponent depending on the version of the Fourier transform you are using. The $\pm$ comes from the fact that I can remember the correct exponent :)

Note that the function given by the sum in the above equation is periodic of period $1$. Call it $h_n$. Because of $\widehat{f} \neq 0$ almost everywhere, we get

$$h_n = \frac{\widehat{g_n}(\xi)}{\widehat{f}(\xi)} \rightarrow \frac{\widehat{g}(\xi)}{\widehat{f}(\xi)}$$

almost everywhere.

But the right hand side vanishes on $\Bbb{R}\setminus (0,1)$ and the left hand side is periodic with period $1$, so that the limit must also be periodic with period $1$ (up to "almost everywhere problems").

We derive $\widehat{g} = $ almost everywhere, a contradiction.

Answer 2

$\newcommand{\span}{\operatorname{span}}$It is equivalent to construct a nontrivial function in $L^2(\Bbb{R})$ such that it as a linear functional on $L^2(\Bbb{R})$ vanishes on the set $\span \{\phi(\cdot-n):n\in\Bbb{Z}\}$. Assume that $h\in L^2(\Bbb{R})$ is orthogonal to each $\phi(\cdot-n)$, $n\in\Bbb{Z}$, namely, $$ \langle h,\phi(\cdot-n) \rangle=\int_\Bbb{R} \phi(t-n)h(t)dt=0, \mbox{ for all } n\in\Bbb{Z}. $$ This is equivalent to $$ \int_\Bbb{R} \overline{\hat{h}}(\xi)\hat{\phi}(\xi)e^{-i2\pi n\xi}d\xi=0, \mbox{ for all } n\in\Bbb{Z}. $$ We decompose $\Bbb{R}$, the domain of integration, into unions of the interval $[k,k+1]$, $k\in\Bbb{Z}$. Hence, $$ \sum_{k\in\Bbb{Z}}\int_k^{k+1} \overline{\hat{h}}(\xi)\hat{\phi}(\xi)e^{-i2\pi n\xi}d\xi=0, \mbox{ for all } n\in\Bbb{Z}. $$ It reduces to $$ \int_0^1 \Big(\sum_{k}\overline{\hat{h}}(\xi+k)\hat{\phi}(\xi+k)\Big)e^{-i2\pi n\xi}d\xi=0, \mbox{ for all } n\in\Bbb{Z}. $$ We shall point out that $\sum_{k\in\Bbb{Z}}\overline{\hat{h}}(\xi+k)\hat{\phi}(\xi+k)\in L^1([0,1])$ since $\phi, h\in L^2(\Bbb{R})$. Using the fact that the Fourier transform is injective from $L^1([0,1])$ to $l^\infty(\Bbb{Z})$, we obtain $$ \sum_{k\in\Bbb{Z}}\overline{\hat{h}}(\xi+k)\hat{\phi}(\xi+k)=0,\mbox{ almost everywhere } \xi\in [0,1]. $$ For $\phi\in L^2(\Bbb{R})$, we have $\hat{f}\ne 0$ (in $L^2(\Bbb{R})$) on $[j,j+1]$ for certain $j\in\Bbb{Z}$. If not, $\hat{f}=0$, by the Plancherel theorem, we have $f=0$ and then $\span \{\phi(\cdot-n):n\in\Bbb{Z}\}=\{0\}$ obviously fails to dense in $L^2(\Bbb{R})$. Let $j'\in\Bbb{Z}$ with $j'\ne j$. Introduce $$ \hat{h}(\xi):=\left\{\begin{array}{ll} \overline{\hat{\phi}}(\xi-j+j'),& j\le \xi\le j+1,\\ -\overline{\hat{\phi}}(\xi-j'+j),& j'\le \xi\le j'+1,\\ 0,&\mbox{otherwise}. \end{array} \right. $$ Substituting $\hat{h}$ into () yields for $\xi\in [0,1]$ $$ \sum_{k\in\Bbb{Z}}\overline{\hat{h}}(\xi+k)\hat{\phi}(\xi+k) =\overline{\hat{h}}(\xi+j)\hat{\phi}(\xi+j)+\overline{\hat{h}}(\xi+j')\hat{\phi}(\xi+j') =\hat{\phi}(\xi+j')\hat{\phi}(\xi+j)-\hat{\phi}(\xi+j)\hat{\phi}(\xi+j')=0. $$ Note that the construction of $\hat{h}$ implies $h\in L^2(\Bbb{R})$ and $h\ne0$, which completes the proof.