Studying introductory harmonic analysis, I hit upon the following question:

Let $f\in L^1(\mathbb{R})$. Define $E_f=\overline{\text{span}}\{f_y\}_{y\in\mathbb{R}}$.

  1. Assume $k \in L^1(\mathbb{R})$ is compactly supported. show $k*f \in E_f$.
  2. Show the condition on $k$ is not necessary.
  3. Show $\{f_y\}_{y\in \mathbb{R}}$ is complete in $L^1(\mathbb{R})$ iff $\widehat{f}(x) \ne 0 $ for all $x\in \mathbb{R}$.

1 easily implies 2, as for any $k\in L^1(\mathbb{R})$ the sequence $k_N = k\cdot {1}_{[-N,N]}$ converges to $k$, and $f*k_N \in E_f$.

I cannot work out $1$. Have tried using simple functions for both $f$ and $k$, have tried to show it for $k= 1_K$ but no luck. I noticed the image of $E_f$ under $\widehat{(\cdot)}$ operator is $\overline{\text{span}}\{e^{iy(\cdot)}\widehat{f}(\cdot)\}_{y\in\mathbb{R}}$, and $\widehat{k*f}=\widehat{k}\cdot\hat{f}$. So, in a way, it is ``enough'' to show $\widehat{k}$ can be approximated by such exponents. The problem is, I do not understand how such approximation coudl help. Inverting the transform is not possible.

As for 3, I cannot work both directions. If we assume $\widehat{f} \ne 0 $, the obvious approach would be to solve the following: for any $g\in L^1$, find $k\in L^1$ s.t: $$g = f*k$$ this happens iff $\widehat{k} = \widehat{g}/ \hat{f}$ but can such $k$ be guaranteed?


You need a lemma: translation is continuous in $L^1$, meaning for every $\epsilon>0$ there exists $\delta>0$ such that $\|f-f_y\|_1 <\epsilon$ when $|y|<\delta$. Notice that this also implies $\|f_y-f_z\|_1<\epsilon$ when $|y-z|<\delta$.

So, if $k$ is a characteristic function $\chi_{[a,b]}$ of an interval $[a,b]$ of length less that $\delta$, then $$ \|f*k - (b-a)f_a\|_1 = \left\|\int_a^b (f_y -f_a)\,dy \right\|_1 < \epsilon(b-a) \tag{1} $$

Next consider $k=\chi_{[a,b]}$ with arbitrary $a,b$. Partition $[a,b]$ into subintervals of length $<\delta$ and use (1) to conclude that $$ \left\|f * k - \frac{b-a}{n} \sum f_{y_j}\right\|_1 < \epsilon(b-a) \tag2 $$ where $y_j$ is a point of the $j$th subinterval. Since $\epsilon$ is arbitrary, (2) shows $f*k\in E_f$.

By linearity, we get the same when $k$ is a step function. By density (since $E_f$ is closed) the same holds for arbitrary $k\in L^1$. Indeed, if $k$ can be approximated by some $k_j$ in $L^1$ norm, then $$\|f*k-f*k_j\|_1 \le \|f\|_1 \|k-k_j\|_1\to 0\tag3 $$


The Fourier transform is a continuous map from $L^1$ to $C_0$ (continuous functions that tend to $0$ at infinity, with uniform norm). If $\hat f(\xi)=0$, then $\hat f_y(\xi)=0$ for all $y$, since translation becomes modulation on the Fourier side. By the aforementioned continuity of the Fourier transform, we get $\hat g(\xi)=0$ for all $g\in E_f$. But there are $L^1$ functions, like the Gaussian, whose Fourier transform is everywhere nonzero; so they are not in $E_f$, so $E_f\ne L^1$.

The converse, $\hat f\ne 0\implies E_f = L^1$, is much deeper; that's Wiener's Tauberian theorem for $L^1$. I don't think it's fair to assign it as an exercise... Wiener's paper was 100 pages in Annals of Mathematics...


I would like to post a suggestion towards a solution for the other direction: $\hat{f}\ne0 \implies E_f = L^1$.

It is clear, due to the first two questions, that for any $h\in L^1$, it is enough to find $g\in L^1$ such that $ h = f*g$. If we find such $g$, then $h\in E_f$. If such equality should hold, then we must have $\hat{h}/\hat{f} = \hat{g}$ (note $\hat{f} \ne 0$) but such $\hat{g}$ cannot necessarily be inverted to give a $g\in L^1$.

Define $(k_t)_{t>0}$ to be an approximative unit with compactly supported fourier transform. E.g., take $K(x)=\frac{1}{\pi}\frac{\sin^2(x)}{x^2}$. And define $k_t(x)=K(x/t) /t$. It can be verified that $\hat{K}$ is a triangular function with supported at [-2,2]. We also have $\hat{k_t}(\lambda) = \hat{K}(\lambda t)$, and so it is supported in $[-2/t , 2/t]$ which exapnds to $\mathbb{R}$ as $t\rightarrow 0$ (which is the limit inwhich the approximative unit becomes a unit).

Overall we have $k_t *h \xrightarrow{L^1} h$ at $t\rightarrow 0$ (Forall $h\in L^1$). We wish to solve $$k_t *h = f*g_t$$ for some $g_t \in L^1$ (which we will find). We denote (without assuming anything about the nature of $g$), $$\hat{k_t}\hat{h}/\hat{f} = \hat{g_t}$$ So $\hat{g_t}$ is compactly supported and must be in $L^1$ (it is not necessarily a fourier transform of some function!). It has an ``inverse'' transform $\check{\hat{g_t}}$ (this is not necessarily the wanted $g_t$, since it is possible it is not integrable).

Can this be mended?

  • 1
    $\begingroup$ It is enough to find $g_n \in L^1$ such that $f \ast g_n \to h$ in $L^1$ but in general $g_\infty \not \in L^1$ (its Fourier transform is $\frac{\hat{h}}{\hat{f}}$). Assume for $n >0$, $G_n(\xi) = \frac{e^{-\pi \xi^2 n^2}}{\hat{f}(\xi)} \in L^1$ and look at $h \ast e^{-\pi x^2/n^2} = f \ast g_n(x)$ where $\hat{g_n} = G_n \hat{h}$. In general you'll apply the same argument, replacing the Gaussian with a Schwartz function $\hat{u}$ decreasing fast enough so that $\frac{\hat{u}(n \xi)}{\hat{f}(\xi)} \in L^1$. $\endgroup$ – reuns Jul 13 '17 at 12:12
  • $\begingroup$ @user1952009 ok this is what I was trying to do, but having defined $g_n$ that way, what guarantees that we have $h*e ^{-\pi x^2/n^2} = f*g_n$? $\endgroup$ – Ranc Jul 13 '17 at 12:49
  • 1
    $\begingroup$ I meant $G_n(\xi ) = \frac{e^{-\pi \xi^2 / n^2}}{\hat{f}(\xi)}, \hat{g_n} = \hat{G}_n \hat{h}$ so that $h \ast n e^{-\pi n^2 x^2} = f \ast g_n$ by the convolution theorem $\endgroup$ – reuns Jul 13 '17 at 13:01
  • $\begingroup$ $G_n$ is integrable, but why should \hat{G_n} be? I know such a theorem but it a-priori requires the integrability of both $\hat{g_n}$ and $g_n$. $\endgroup$ – Ranc Jul 13 '17 at 13:23
  • 1
    $\begingroup$ If you are afraid then look at $\hat{g_n} \ast n e^{- \pi n^2 \xi^2 / n^2}$ that is Schwartz $\endgroup$ – reuns Jul 13 '17 at 13:34

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