The Fourier Transform of exp(-cx^2) and Its Convolution

We develop two almost straightforward way to compute the Fourier transform of $\exp(-cx^2)$, in the sense that any contour integration and the calculus of residues are not required at all. The first cool approach enables us to think about these elementary concepts much deeper, so I highly recommend to study this approach as long as you are familiar with ODE of first order.
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The Banach Algebra of Borel Measures on Euclidean Space

This blog post is intended to deliver a quick explanation of the algebra of Borel measures on \(\mathbb{R}^n\). It will be broken into pieces. All complex-valued complex Borel measures \(M(\mathbb{R}^n)\) clearly form a vector space over \(\mathbb{C}\). The main goal of this post is to show that this is a Banach space and also a Banach algebra.

In fact, the \(\mathbb{R}^n\) case can be generalised into any locally compact abelian group (see any abstract harmonic analysis books), this is because what really matters here is being locally compact and abelian. But at this moment we stick to Euclidean spaces. Note since \(\mathbb{R}^n\) is \(\sigma\)-compact, all Borel measures are regular.

To read this post you need to be familiar with some basic properties of Banach algebra, complex Borel measures, and the most important, Fubini's theorem.

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Left Shift Semigroup and Its Infinitesimal Generator

Left shift operator

Throughout we consider the Hilbert space \(L^2=L^2(\mathbb{R})\), the space of all complex-valued functions with real variable such that \(f \in L^2\) if and only if \[ \lVert f \rVert_2^2=\int_{-\infty}^{\infty}|f(t)|^2dm(t)<\infty \] where \(m\) denotes the ordinary Lebesgue measure (in fact it's legitimate to consider Riemann integral in this context).

For each \(t \geq 0\), we assign an bounded linear operator \(Q(t)\) such that \[ (Q(t)f)(s)=f(s+t). \] This is indeed bounded since we have \(\lVert Q(t)f \rVert_2 = \lVert f \rVert_2\) as the Lebesgue measure is translate-invariant. This is a left translation operator with a single step \(t\).

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Quasi-analytic Vectors and Hamburger Moment Problem (Operator Theory)

Analytic and quasi-analytic vectors

Guided by researches in function theory, operator theorists gave the analogue to quasi-analytic classes. Let \(A\) be an operator in a Banach space \(X\). \(A\) is not necessarily bounded hence the domain \(D(A)\) is not necessarily to be the whole space. We say \(x \in X\) is a \(C^\infty\) vector if \(x \in \bigcap_{n \geq 1}D(A^n)\). This is quite intuitive if we consider the differential operator. A vector is analytic if the series \[ \sum_{n=0}^{\infty}\lVert{A^n x}\rVert\frac{t^n}{n!} \] has a positive radius of convergence. Finally, we say \(x\) is quasi-analytic for \(A\) provided that \[ \sum_{n=0}^{\infty}\left(\frac{1}{\lVert A^n x \rVert}\right)^{1/n} = \infty \] or equivalently its nondecreasing majorant. Interestingly, if \(A\) is symmetric, then \(\lVert{A^nx}\rVert\) is log convex.

Based on the density of quasi-analytic vectors, we have an interesting result.

(Theorem) Let \(A\) be a symmetric operator in a Hilbert space \(\mathscr{H}\). If the set of quasi-analytic vectors spans a dense subset, then \(A\) is essentially self-adjoint.

This theorem can be considered as a corollary to the fundamental theorem of quasi-analytic classes, by applying suitable Banach space techniques in lieu.

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