If $ \lambda= \hat{f} (n)$, then $ f - \lambda \delta_0$ is not invertible with respect to the convolution product.
Please feel free to jump directly to the end for the explanation.
Let $\mathbb{T}= [0,1]$ be identified with the circle $\{z: \mathbb{C} : |z| =1 \} = \{ e^{2 \pi i t} : t \in [0,1]\} $. An integrable periodic function $f \in L^1 (\mathbb{T})$ can be associated with its decomposition as a Fourier series
Please feel free to jump directly to the end for the explanation.
Let $\mathbb{T}= [0,1]$ be identified with the circle $\{z: \mathbb{C} : |z| =1 \} = \{ e^{2 \pi i t} : t \in [0,1]\} $. An integrable periodic function $f \in L^1 (\mathbb{T})$ can be associated with its decomposition as a Fourier series
$$ f(t) \sim \sum_{n=-\infty}^\infty \hat{f} (n) e^{2 \pi i n t}, $$
where the Fourier coefficients are defined by
$$ \hat{f} (n) := \int_0^1 f(t) e^{-2 \pi i n t} dt . $$
Intuitively the Fourier coefficients $\hat{f} (n)$ tell us the weights of the components $e^{2 \pi i n t}$ within f. $ \hat{f}$ is the spectrum of the signal $f$.
In linear algebra, the spectrum of a square matrix (or more generally a bounded linear operator) $A$ is the set of eigenvalues $\{ \lambda \in \mathbb{C} : A- \lambda I $ is not invertible $\} $. Extending this, the spectrum of an element $a$ in a unital Banach algebra $A$ is $\{\lambda \in \mathbb{C}: a - \lambda 1$ is not invertible $\}$. Is there any way to connect this spectrum with the spectrum of a function?
Let us recall the definition of convolution of two functions:
$$f *g (x) := \int_0^1 f(x-t) g(t) dt .$$
It has the following property: $\widehat{f*g} (n) = \hat{f} (n) \hat{g} (n) $. Note that $L^1(\mathbb{T}) $ is an algebra with respect to this convolution product (not with respect to the usual multiplication)! However, $L^1(\mathbb{T})$ does not have an identity element (Is there a function such that f * g = f for all f?), we cannot talk about invertibility in this algebra. (Hence in harmonic analysis we need something called approximate identity as a substitute, sometimes appearing as summability kernel in classical case.) For convenience, let us move to a larger algebra $M(\mathbb{T})$ which contains $L^1(\mathbb{T})$.
$M(\mathbb{T})$ is the algebra of regular Borel measures on $\mathbb{T}$. A function $f \in L^1(\mathbb{T})$ is identified with the measure $f(t)dt \in M(\mathbb{T})$. The convolution product and Fourier coefficients of a measure is extended as follows:
$$
\begin{align*} f * \mu (x) :=& \int_0^1 f(x-t) d \mu(t) , \\ \mu * \nu (E) :=& \int_0^1 \int_0^1 1_E (x+y) d\mu(x) d\nu(y) \\ =& \int_0^1 \mu(E-y) d\nu(y) \end{align*}
$$
and
$$ \hat{\mu} (n) := \int_0^1 e^{-2 \pi i n t} d\mu(t) $$
where $1_E$ is the indicator function of $E$. $\widehat{\mu * \nu} (n) = \hat{\mu} (n) \hat{\nu} (n)$ is satisfied analogously.
Now we have an identity element in $M(\mathbb{T})$, namely the Dirac measure $\delta_0$:
Now we have an identity element in $M(\mathbb{T})$, namely the Dirac measure $\delta_0$:
$$ f* \delta_0 (x) = \int_0^1 f(x-t) d \delta_0(t) = f(x) .$$
Its Fourier transform is $\hat{\delta}_0 (n) = \int_0^1 e^{-2 \pi i n t} d\delta_0 (t) = 1$ for all $n$.
Finally, we can show our claim, by considering the contrapositive of the statement.
Suppose $f - \lambda \delta_0$ is invertible, then $$ (f - \lambda \delta_0) * h = \delta_0 $$ for some $h \in M(\mathbb{T}) $. Taking Fourier transform we have $$ (\hat{f} (n) - \lambda ) \hat{h} (n) = 1 $$ for all $n$. This implies that $ \lambda \neq \hat{f} (n) $ for all $n$.
Remark:
1. Originally I want to prove the converse as well, but it involves some theory of Gelfand transform.
2. To be frank, this may not be the best way of presenting $\hat{f} (n)$ as eigenvalues. A more proper explanation would be along the line of eigenspace decomposition of the regular representation
$$ \begin{align*} \pi : L^1 (\mathbb{T}) &\to B(L^2(\mathbb{T})) \\ \pi(f) g &:= f *g . \end{align*} $$ In this case we can really claim that $\lambda = \hat{f} (n)$ if and only if $\lambda$ is an eigenvalue of $\pi(f)$.
3. I hope this post is still an interesting observation though, and will arouse your interest in the linkage between Fourier analysis, spectral theorem and representation theory. The interplay of the group action with the differential operator is behind the scene. One may also ponder on this question: why is $e^{2 \pi i n t}$ so special?
Personally the motivation along this investigation is the line of different generalizations of spectrum, from eigenvalues to spectrum of a ring and a $C^*$-algebra!
Finally, we can show our claim, by considering the contrapositive of the statement.
Suppose $f - \lambda \delta_0$ is invertible, then $$ (f - \lambda \delta_0) * h = \delta_0 $$ for some $h \in M(\mathbb{T}) $. Taking Fourier transform we have $$ (\hat{f} (n) - \lambda ) \hat{h} (n) = 1 $$ for all $n$. This implies that $ \lambda \neq \hat{f} (n) $ for all $n$.
Remark:
1. Originally I want to prove the converse as well, but it involves some theory of Gelfand transform.
2. To be frank, this may not be the best way of presenting $\hat{f} (n)$ as eigenvalues. A more proper explanation would be along the line of eigenspace decomposition of the regular representation
$$ \begin{align*} \pi : L^1 (\mathbb{T}) &\to B(L^2(\mathbb{T})) \\ \pi(f) g &:= f *g . \end{align*} $$ In this case we can really claim that $\lambda = \hat{f} (n)$ if and only if $\lambda$ is an eigenvalue of $\pi(f)$.
3. I hope this post is still an interesting observation though, and will arouse your interest in the linkage between Fourier analysis, spectral theorem and representation theory. The interplay of the group action with the differential operator is behind the scene. One may also ponder on this question: why is $e^{2 \pi i n t}$ so special?
Personally the motivation along this investigation is the line of different generalizations of spectrum, from eigenvalues to spectrum of a ring and a $C^*$-algebra!
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