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Consider the space of all Fourier transforms of $L^{1}(\mathbb R),$ that is, $$\mathcal{F}L^{1}=\mathcal{F}L^{1}(\mathbb R):= \{f\in L^{\infty}(\mathbb R):\hat{f}\in L^{1}(\mathbb R)\},$$ with the norm, $\|f\|_{\mathcal{F}(L^{1})}=\|\hat{f}\|_{L^{1}(\mathbb R)}.$ (By uniqueness theorem and using the fact that Fourier transform is a linear, one can deduce that, this is actually a norm)

My Questions:(1) Can we expect $\mathcal{F}L^{1}$ is a complete with respect to the norm $\|\cdot\|_{\mathcal{F}L^{1}}$ ? (2) Put, $A=\{f\in \mathcal{F}L^{1}: |f|\in \mathcal{F}L^{1}\}.$ Can we expect $A$ is closed in $\mathcal{F}L^{1}$ ? (3) (Bit Vague question) Can you suggests some proper closed subsets of $\mathcal{F}L^{1}$; can we expect its characterization ?

My attempt: Suppose $\{f_{n}\}_{n\in \mathbb N}$ is a Cauchy sequence in $\mathcal{F}L^{1},$ that is, there is $N\in \mathbb N$, such that, $\|\hat{f_{n}}-\hat{f_{m}}\|_{L^{1}(\mathbb R)}\to 0$, for every $n, m \geq N$; and since $(L^{1}(\mathbb R), \|\cdot\|_{L^{1}(\mathbb R)})$ is complete, there is a $g\in L^{1}(\mathbb R)$, such that, $\|\hat{f_{n}}-g\|_{L^{1}(\mathbb R)}\to 0$ as $n\to \infty.$ We must find, $h\in \mathcal{F}L^{1}$ such that $\|f_{n}-h\|_{\mathcal{F}L^{1}}\to 0$ as $n\to \infty.$ Now my guess work is that, we should take, $h:=\check{g}$; but then the problem is: can we expect, $\hat{h}=g$ (I know this one can expect, if both $f$ and $\hat{f}$ both are in $L^{1}(\mathbb R)$, by inversion formula); or am I missing some thing ?

Thanks,

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Question 3 is far too broad. How good are you at describing/characterizing all the closed subsets of ${\bf R}$, for instance? –  Yemon Choi Mar 31 at 23:14
    
And as @JohannesHahn points out below, Question 1 is trivial. –  Yemon Choi Mar 31 at 23:14
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I suspect the answer to Question 2 is negative, but I don't yet have a counterexample –  Yemon Choi Mar 31 at 23:16

1 Answer 1

up vote 4 down vote accepted

(1) The space is obviously complete since you defined the norm in such a way that $\mathcal{F}$ is an isometry $L^1 \cong \mathcal{F}L^1$ (maybe an isometry up to constants depending on your definition of the fourier transform). Consequently the answer to (3) is that the closed sets of $\mathcal{F}L^1$ are exactly the fourier transforms of closed sets of $L^1$.

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@JH;thanks; but I am sorry, I could not follow: see, we consider $\mathbb R$ with usual metric, namely modulus, $|\cdot|$, which is complete; I think, $d:(0, 1)\to \mathbb R $ such that, $x\mapsto |x|$; preserves distance; but (0, 1) is not a complete metric space; Or am I missing something ? –  Inquisitive Mar 31 at 16:26
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@DivyangBhimani: Yes, you're missing that $L^1$ is complete and $(0,1)$ isn't. Completeness is obviously invariant w.r.t. to isometries that is: If $X_1, X_2$ are isometric metric spaces then $X_1$ is complete iff $X_2$ is. In your case $\mathcal{F}: \mathcal{F}L^1\to L^1$ is by your definition of the norm on $\mathcal{F}L^1$ an isometry. –  Johannes Hahn Mar 31 at 16:47
    
@JH;I think, It is clear to me that, $\mathcal{F}:\mathcal{F}L^{1}\to L^{1}$ is an isometry; but why is it onto map ? (I think, this is needed to show, two spaces are isometric by definition;); Thanks a lot; –  Inquisitive Apr 2 at 4:25
    
@DivyangBhimani: Because $\mathcal{F}^2(f(x))=f(-x)$ (up to constants depending on your definition of $\mathcal{F}$) –  Johannes Hahn Apr 2 at 11:09
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Yes it is true for all $f\in L^1$. What do you mean by "if $\mathcal{F}f=\hat{f}$ ? $\hat{\cdot}$ and $\mathcal{F}$ are two notations for the same thing, the fourier transform... –  Johannes Hahn Apr 2 at 11:50

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