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Introduction to Linear Algebra with Mathematica
Glossary
Preface
The Schwartz space ๐ฎ(ℝ) was introduced by the French mathematician Laurent Schwartz (1915--2002), a different person from a German mathematician Hermann Schwarz, whose name is associated with the Cauchy-Bunyakovsky-Schwarz inequality. Actually, -Schwarz's contribution to this inequality is quite subtle---his 1888 paper contains no new information except excellent translation into German.
In opposite to to the Banach space 𝔏¹(ℝ) and Hilbert space 𝔏²(ℝ), the Fourier transform and its inverse are both well-defined in the Schwartz space ๐ฎ(ℂ) = S(ℂ). We always use the notation ∫ f for \( \displaystyle \int_{-\infty}^{+\infty} f(x)\.{\text d}x \) throughout this chapter; especially, the limits of integration are always ±∞ if nothing is said to the contrary.
Fourier integral in ๐ฎ
Example 1: First we define a function h on ℝ by \[ h(t) = \begin{cases} 0, & \quad\mbox{for} t \le 9 , \\ e^{-1/t} &\quad\mbox{for} t > 0 . \end{cases} \] This function is infinitely differentiable on all of ℝ, and all of its derivatives vanish at 0. Put g(t) = h(t) h(1 โ t). Then g is also infinitely differentiable. Furthermore, g(t) > 0 for 0 < t < 1 and g(t) = 0 otherwise. Thus, function g is a smooth function with compact support. ■
Example 2: ■
The definition of ๐ฎ regards differentiation and multiplication on an equal footing. Let \( \displaystyle \quad \mathtt{D} = {\text d}/{\text d}x \quad \) denote differentiation and let M = Mjฮพ denote multiplication operator by jξ. Working on ๐ฎ is convenient for several reasons; in particular, the Fourier transform exchanges these operations. Furthermore, as we will show in the following Theorem, the Fourier transform maps ๐ฎ to itself bijectively. We can interpret the last two items of the following Proposition as saying that the Fourier transform gives the spectral decomposition for the derivative operator.
- The Fourier transform ℱ maps the Schwartz space onto itself.
- The inverse transform ℱ−1 does what it should: \[ ℱ^{-1} \left( f^F \right) = f . \]
- \( \displaystyle \quad \hat{f}_h (\xi ) = e^{{\bf j}h\xi} \hat{f} (\xi ) , \quad \) where fh(x) = f(x + h).
- \( \displaystyle \quad \texttt{D}\,\hat{f}(\xi ) = \frac{\text d}{{\text d}\xi}\,\hat{f}(\xi ) = - {\bf j}\,ℱ\left( M_x f \right) , \quad \) taht is, \( \displaystyle \quad \texttt{D}_{{\bf j}\xi} \,ℱ = ℱ\,M_x . \)
- \( \displaystyle \quad ℱ\,\texttt{D}_x = M_{{\bf j}\xi}ℱ. \)
- \( \displaystyle \quad \texttt{D} = ℱ^{-1} M \,ℱ \quad \) and \( \displaystyle \quad M = ℱ\,\texttt{D}\,ℱ^{-1} . \)
- For any function f from ๐ฎ = S, we have using integration by pars that \[ ℱ_{x\to\xi}\left[ f' \right] = \int f' (x)\,e^{-{\bf j}x\xi} {\text d}x = -\int f(x) \left( e^{-{\bf j}x\xi} \right)' {\text d} x = {\bf j}\xi \int f(x) \, e^{-{\bf j}x\xi} {\text d}x . \] Also, by the rapid decrease of f ∈ ๐ฎ, \[ ℱ_{x\to\xi}\left[ -{\bf j}x \,f \right] = \frac{\text d}{{\text d}x} \left[ f^F \right] , \] and so, by induction, \[ ℱ_{x\to\xi}\left[ \texttt{D}^p \left( -{\bf j} x \right)^q f \right] = \left( {\bf j} \xi \right)^p \texttt{D}^p f^F , \] for any nonnegative integers p and q. Therefore, \[ \left\vert \xi \right\vert^p \left\vert \texttt{D}^q f^F \right\vert \le \left\| \texttt{D}^p x^q f \right|_1 < \infty . \] Hence, the Fourier transform of f ∈ ๐ฎ also belongs to ๐ฎ.
- Now let f be a function with compact support and regard it as an in๏ฌnitely differentiable function on the circle −T/2 ≤ x ≤ T/2, as is periodic and extended by zero outside this interval [−T/2, T/2]. Then you can express f for |x| < T/2 as a rapidly convergent Fourier series of period T: \begin{align*} f(x) &= \sum_{n=-\infty}^{\infty} e^{{\bf j}nx/T} \frac{1}{T} \int_{-T/2}^{T/2} f(y)\,e^{-{\bf j}ny/T} {\text d}y \\ &= \sum_{n=-\infty}^{\infty} e^{{\bf j}nx/T} \frac{1}{T} \, f^F \left( \frac{n}{T} \right) . \end{align*} But this is just a Riemann sum approximating to the integral \[ ℱ_{y\to x}\left[ f^F (y) \right] = \int_{-\infty}^{\infty} f^F (y) \,e^{{\bf j}yx} {\text d}y . \] In order to prove that \[ ℱ^{-1} \left[ f^F \right] = f , \] for functions with compact support, you have only to check that the sum converges to the integral as T ↑ ∞
- Similar reasoning leads to the formula \[ \| f \|_2 = \| f^F \|_2^2 = \int_{-T/2}^{T/2} | f |^2 = \sum_{n=-\infty}^{+\infty} \frac{1}{T} \, \left\vert f^F \left( \frac{n}{T} \right)\right\vert^2 \] from which we derive the Plancherel identity: \[ \| f \|_2 = \frac{1}{2\pi} \,\| f^F \|_2 = \frac{1}{2\pi} \left( \int_{-\infty}^{\infty} | f^F (y) |^2 {\text d}y \right)^{1/2} . \]
Example 3: ■
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