2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $$\|f*k-f*k_j\|_1 \le \|f\|_1 \|k-k_j\|_1\to 0\tag3 $$

Completeness

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...

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.