Compactness of the Fourier transform
Let $X = L^2(RR)$ be the Hilbert $CC$-space of the complex-valued functions that are a.e. defined and Lebesgue square-integrable on $RR$, equipped with the standard inner product $X$ x $X \to CC: (f,g) \to \int_{-\infty}^{+\infty} f(x) g^\star (x) dx$, where $z^\star = $Re$(z) - i\cdot$Im$(z)$, for any $z \in CC$. Then consider the linear operator $F(\cdot): X \to X: f(\cdot) \to \int_{-\infty}^\infty e^{-2i\pi \omega x} f(x) dx$.
THE PROBLEM: is $F(\cdot)$ compact? Can you describe its point spectrum?
THE PROBLEM: is $F(\cdot)$ compact? Can you describe its point spectrum?
Risposte
I read your posts and comments and I admit - as you said - my definition of an isometry is too weak. This not standing, it is a definition. Anyway, since I like generalities, there's no doubt I will share yours for the times incoming. Then thank you so much for your precious contributions.
I think that your definition of isometry is somewhat "too weak" since it only maintains the distance with respect to zero.
Things being like that, there is no need to go to infinitely dimensional spaces to find nonlinear maps having that property (take any
map $g:S\to S$, where $S$ is the unit sphere, and define $f(x)=||x|| g(x/||x||)$). Even continuity (except at zero) is lost.
But please take your time to think about it.
By the way I think I can show an example of a nonlinear isometry (in my sense).
Take $X_1=RR$ with the usual norm $||x||_1=|x|$, and $X_2=RR^2$ with the infinity norm:
$||(x, y)||_2="max"(|x|,|y|)$.
Define $T(x):=(x,\atan(x))$.
Since $|\atan(x)-\atan(y)|<|x-y|$ you get $||T(x)-T(y)||_2="max"(|x-y|,|\atan(x)-\atan(y)|)=|x-y|=||x-y||_1$
So $T$ is an isometry but it clearly is a nonlinear map.
Anyway thanks for the stimulating topic.
Things being like that, there is no need to go to infinitely dimensional spaces to find nonlinear maps having that property (take any
map $g:S\to S$, where $S$ is the unit sphere, and define $f(x)=||x|| g(x/||x||)$). Even continuity (except at zero) is lost.
But please take your time to think about it.
By the way I think I can show an example of a nonlinear isometry (in my sense).
Take $X_1=RR$ with the usual norm $||x||_1=|x|$, and $X_2=RR^2$ with the infinity norm:
$||(x, y)||_2="max"(|x|,|y|)$.
Define $T(x):=(x,\atan(x))$.
Since $|\atan(x)-\atan(y)|<|x-y|$ you get $||T(x)-T(y)||_2="max"(|x-y|,|\atan(x)-\atan(y)|)=|x-y|=||x-y||_1$
So $T$ is an isometry but it clearly is a nonlinear map.
Anyway thanks for the stimulating topic.
Take for sure I will read what you wrote carefully, but probably tomorrow. In my understandings, a map $T:X_1 \to X_2$ between two normed spaces $(X_1, ||\cdot||_1)$ and $(X_2, ||\cdot||_2)$ over the real (resp., complex) field is an isometry iff $||Tx||_2 = ||x||_1$. So $||T0||_2 = ||0||_1 = 0$, hence $||T0||_2 = 0$. But in any normed space $(X,||\cdot||)$: $||x|| = 0$ iff $x = 0$. Yet I see that our definitions don't agree (your sounds to be a generalization) - and maybe this is because we don't agree too.
...is an isometry, then $||T0||_2=||0||_1=0$, ...
In my opinion an isometry between two metric spaces $X_1$ and $X_2$ is a map $T:X_1\to X_2$ such that
$"dist"_2(T(x_1),T(x_2))="dist"_1(x_1,x_2)$ (the case of normed spaces being a particular one). In this setting
I can't see how you can derive $T(0)=0$. Actually $T(x):=x+1$ is an isometry between $RR$ and $RR$.
Moreover the map $T:l^p\to l^p$ defined by
$(Tx)_n=x_n$ if $n\geq2$, $(Tx)_1=|x_1|$ is NOT an isometry, since taking $x=(1,0,0,...)$ and $x'=(-1,0,0,...)$
yields $||Tx=Tx'||=0\ne||x-x'||$.
Sorry, if $H_1$, $H_2$ are real Hilbert spaces, any isometry $T:H_1\to H_2$ with $T0=0$ is in fact linear
(I haven't tried the generalization to the complex case, but it should hold too). This can be see as follows:
(denote by $<\cdot,\cdot>_{1}$, $<\cdot,\cdot>_{2}$the inner products)
(1) $
To see this use $||Tx-Ty||_2^2=||x-y||_1^2=||x||1^2+||y||_1^2-2
$||Tx-Ty||_2^2=||Tx||_2^2+||Ty||_2^2-2
the conclusion easily follows.
(2) $T(x-y)=Tx-Ty$ for all pairs $x,y$ in $H_1$.
We have:
$||T(x-y)-(Tx-Ty)||_2^2=||T(x-y)||_2^2+||Tx-Ty||_2^2-2
$||x-y||_1^2+||x-y||_1^2-2
(3) $T(\lambda x)=\lambda Tx$ for $x$ in $H_1$ and $\lambda$ in $RR$
Again:
$||T(\lambda x)-\lambda Tx||_2^2=||T(\lambda x)||_2^2+||\lambda Tx||_2^2-2
$||\lambda x||_1^2+\lambda^2||x||_1^2-2\lambda<\lambda x,x>_1=2\lambda^2||x||_1^2-2\lambda^2
So if a counterexample exists it has to be looked for in some "ugly" linear space, probably a non uniformly convex one (may be $l^\infty$ ??).
But I have no precise idea in this respect.
"ViciousGoblinEnters":
[...] is an isometry $T$ a linear operator??
(well assuming that $T0=0$ otherwise a simple translation would be a counterexample).
Your assumption is unnecessary. In fact, if $(X_1, ||\cdot||_1)$ and $(X_2,||\cdot||_2)$ are real (resp., complex) normed spaces and $T: X \to Y$ is an isometry, then $||T0||_2 = ||0||_1 = 0$, so necessarily $T0 = 0$. Anyway, your question seems interesting.
And some minutes later ...
Consider the space $X = l^p(CC)$ of all complex sequences $x=\{x_n\}_{n \in NN}$ for which $||x||_p := (\sum_{k=1}^\infty |x_k|^p)^{1/p} < \infty$, where $NN = \{1, 2, \ldots\})$ and $p \ge 1$ is a real. The map $||\cdot||_p: X \to RR: x \to ||x||_p$ is, in fact, a norm. Moreover, $(X,||\cdot||_p)$ is made a complex infinite dimensional Banach space (in the standard way). For any $\{x_n\}_{n \in NN} \in l^p$, define the sequence $\{y_n\}_{n \in NN}$ taking $y_1 = |x_1|$ and $y_n = x_n$, whenever $n \ge 2$. Apparently, $\{y_n\}_{n \in NN}\in l^p$. So we have got an operator $T: X \to X$. Much more important, $T$ isn't linear. This not standing, it's an isometry.
"ViciousGoblinEnters":
It seems true in a Hilbert space, while it could be more tricky to understand what happens in a normed space.
Not really! - take $p = 2$ above.
How annoying ...
Let $T:X\to Y$ verify $||Tx_1-Tx_2||_Y=||x_1-x_2||_X$. Let $(x_n)$ be a sequence in $X$ such that
the points $y_n:=Tx_n$ converge to some point $y$ in $T$. Then $(y_n)$ is a Cauchy sequence in $Y$. Since
$||x_m-x_m||_X=||y_n-y_m||_Y$
we get that $(x_n)$ is a Cauchy sequece in $X$, which therefore converges to a point $x$ in $X$. Hence
$y_n=Tx_n\to Tx$ so $y=Tx$. We have thus proved that any limit point of a sequence of points in the range
of $T$ must lie in the range, that is the range is closed.
Actually linearity plays no role in this proof -- mmmh, this suggests a new question: is an isometry $T$ a linear operator??
(well assuming that $T0=0$ otherwise a simple translation would be a counterexample).
It seems true in a Hilbert space, while it could be more tricky to understand what happens in a normed space.
Let $T:X\to Y$ verify $||Tx_1-Tx_2||_Y=||x_1-x_2||_X$. Let $(x_n)$ be a sequence in $X$ such that
the points $y_n:=Tx_n$ converge to some point $y$ in $T$. Then $(y_n)$ is a Cauchy sequence in $Y$. Since
$||x_m-x_m||_X=||y_n-y_m||_Y$
we get that $(x_n)$ is a Cauchy sequece in $X$, which therefore converges to a point $x$ in $X$. Hence
$y_n=Tx_n\to Tx$ so $y=Tx$. We have thus proved that any limit point of a sequence of points in the range
of $T$ must lie in the range, that is the range is closed.
Actually linearity plays no role in this proof -- mmmh, this suggests a new question: is an isometry $T$ a linear operator??
(well assuming that $T0=0$ otherwise a simple translation would be a counterexample).
It seems true in a Hilbert space, while it could be more tricky to understand what happens in a normed space.
"ViciousGoblinEnters":
Anyway I guess that an isometry defined on a infinite dimensional Banach space should have closed range [...]
If you think so, prove it.
You are right!!- actually I should have noted that the the Fourier transform is an INVERTIBLE isometry,
$F^{-1}$ being given by $F\circ j$, where $(j u)(t)=u(-t)$.
Anyway I guess that an isometry defined on a infinite dimensional Banach space should have closed range,
hence it cannot be compact, by the same argument.
$F^{-1}$ being given by $F\circ j$, where $(j u)(t)=u(-t)$.
Anyway I guess that an isometry defined on a infinite dimensional Banach space should have closed range,
hence it cannot be compact, by the same argument.
"ViciousGoblinEnters":
The fact that $F(\cdot)$ is an isometry clearly implies that it is NOT a compact operator. Indeed it maps, say, the
unit ball (which is bounded) onto itself [...]
I see your point of view, but I have just realized your previous arguments need to be revised - guess that I'm going to be more explicit. Given a real (resp., complex) infinite dimensional Banach space $(X,||\cdot||)$, we all agree that proving the identity $I: X \to X: x \to x$ isn't compact is trivial - as you showed. Anyway, how can you demonstrate any isometry $A: X \to X$ fulfills that $A(B) = B$, provided $B = \{x \in X: ||x|| < 1\}$? This is false!
In fact, if $\{x_n\}_{n \in NN}$ is any sequence of complex numbers, let $||x|| = \sum_{n=1}^\infty |x_n|$. Then define $X = l^1 = \{\{x_n\}_{n \in NN} \subseteq CC: ||x|| < \infty\}$. As known, $(X,||\cdot||)$ is a complex (infinite dimensional) Banach space. So consider the forward shift $A: X \to X: (x_1, x_2, \ldots) \to (0, x_1, x_2, \ldots)$. Clearly, $A$ is an isometry. This not standing, $A(B)$ is a proper subspace of $B$.
... as no compact continuous linear operator between infinite dimensional real (resp., complex) Banach spaces has continuous inverse.
And why is this true? Because otherwise the identity map would be compact - and the identity map is not compact in an infinite dimensional space
because in such a space the ball is not compact.
This is my point of view, at least..
Ok, this is a way. Alternatively, consider that $F(\cdot)$ is invertible and its inverse is still bounded in the operator norm. Then $F(\cdot)$ can't be compact, as no compact continuous linear operator between infinite dimensional real (resp., complex) Banach spaces has continuous inverse.
The fact that $F(\cdot)$ is an isometry clearly implies that it is NOT a compact operator. Indeed it maps, say, the
unit ball (which is bounded) onto itself, and the unit ball is not a compact set in an infinite dimensional space.
unit ball (which is bounded) onto itself, and the unit ball is not a compact set in an infinite dimensional space.
It looks like your arguments are circular, but maybe I'm wrong. In fact, suppose $(X,||\cdot||)$ is a real (resp., complex) Banach space. Then say $E(X)$ is the real (resp., complex) vector space of all linear maps $A: X \to X$ and $L(X)$ is the subspace of all $A \in E(X)$ such that $||A||_{op} = $ inf$\{c \in RR: ||Ax|| \le c ||x||,\forall x \in X\} < \infty$. Then $L(X)$ is known to be a real (resp., complex) Banach space in the norm $L(X) \to RR: A \to ||A||_{op}$.
Now, if $A \in L(X)$ is an isometry (this meaning $||Ax|| = ||x||$, for any $x \in X$), then $||A||_{op} = 1$, hence $A$ is bounded in the operator norm - so continuous. On the other hand, how would you establish that the image of a compact set $U \subseteq X$ under an isometry $A: X \to X$ is compact without implicitly using the continuity of $A$?
EDIT: too messy!
Now, if $A \in L(X)$ is an isometry (this meaning $||Ax|| = ||x||$, for any $x \in X$), then $||A||_{op} = 1$, hence $A$ is bounded in the operator norm - so continuous. On the other hand, how would you establish that the image of a compact set $U \subseteq X$ under an isometry $A: X \to X$ is compact without implicitly using the continuity of $A$?
EDIT: too messy!
Ok, but mine is only a humble attempt.
The image of a compact set in $X$ under an isometry is a compact set in $X$. Compactness implies total boundedness, so the image of a bounded set under the Fourier transform is toally bounded. This implies (?) that the Fourier transform is a bounded operator.
The image of a compact set in $X$ under an isometry is a compact set in $X$. Compactness implies total boundedness, so the image of a bounded set under the Fourier transform is toally bounded. This implies (?) that the Fourier transform is a bounded operator.
Yes, the Fourier transform is an isometry of $X$. So? I'm asking for a complete proof, not just a sketch. Yet I can say your way is the right one - the question is almost trivial, if you know some theory.
I believe the Parseval - Plancherel theorem is the key to the first question: $langle f , g rangle = int_(-oo)^(oo)f(x) g^(**)(x)dx=int_(-oo)^(oo)F(omega)G^(**)(omega)d omega= langle F , G rangle$, where $F(cdot)$ and $G(cdot)$ are the Fourier transforms of $f(cdot)$ and $g(cdot)$, respectively. In other words, the Fourier transform is an isometry of $X=cc L^2(RR)$, since $||f||_2^2=||F||_2^2$, with $||cdot||_2$ the $ccL^2$ norm.
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