Example 2: Let G_σ denote the Gaussian distribution with variance σ²: \[ G_{\sigma} (x) = \frac{1}{\sigma \sqrt{2\pi}}\, e^{- x^2/2\sigma^2} . \] This is a probability distribution because \( \displaystyle \quad \int_{\mathbb{R}} G_{\sigma} (x)\,{\text d}x = 1 . \)

We can show this as follows. Let \[ I = \int_{-\infty}^{+\infty} e^{-x^2 /2} {\text d} x . \] Then square I and use polar coordinates, we get \[ I^2 = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-x^2 /2} e^{-y^2 /2}{\text d} x\,{\text d} y = \int_0^{2\pi} {\text d}\theta \int_0^{\infty} e^{-r^2 /2} r\,{\text d} r = 2\pi . \] Its Fourier transform is \[ ℱ\left[ G_{\sigma} \right]_{x \to \xi} = \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-x^2 /2\sigma} e^{-{\bf j}x\xi} {\text d}x = \sigma\sqrt{2\pi}\, e^{-\xi^2 \sigma^2 /2} = 2\pi\,G_{1/\sigma} (\xi ) . \] This formula follows from the identity \[ - \frac{x^2}{2} - {\bf j}\,x\sigma \xi = -\frac{1}{2} \left( x + {\bf j}\,\sigma\xi \right)^2 - \frac{\sigma^2 \xi^2}{2} . \]    ■

1. For any function f from 𝒮 = 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 ∈ 𝒮, \[ ℱ_{x\to\xi}\left[ -{\bf j}x \,f \right] = \frac{\text d}{{\text d}x} \left[ f^F \right] , \] 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 ∈ 𝒮 also belongs to 𝒮.
2. Now let f be a function with compact support and regard it as an infinitely 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 ↑ ∞

Example 3:    ■

Because of the rapid decay of f and g, we can write either side of the identity above as a double integral, and integrate in either order. Then each side of this identity equals the double integral \[ \int_{-\infty}^{+\infty} {\text d} \xi \,\int_{-\infty}^{+\infty} {\text d} x \,f(x)\,g(\xi )\, e^{-{\bf j}x\xo} . \]

Example 4:    ■

Since f is differentiable and f′ is bounded, the mean-value theorem of calculus implies the following inequality \[ \left\vert f(b) - f(a) \right\vert \le \sup_t \left\vert f' (t) \right\vert |b-a| = M \left\vert b-a \right\vert . \] We have \( \displaystyle \quad f(x) = f' (t)\,\int_{-\infty}^{+\infty} g(y) \,{\tyext d} y = \int_{-\infty}^{+\infty} f(x)\,g(y) \,{\tyext d} y \quad \) because of the first property of function g. Using mean value theorem, we get \[ \left\vert \int_{-\infty}^{+\infty} f(x + \epsilon y)\,g(y)\,{\text d}y - f(x) \right\vert = \left\vert \int_{-\infty}^{+\infty} \left( f(x + \epsilon y) - f(x) \right) g(y) \,{\text d} y \right\vert \] \[ \le M \int_{-\infty}^{+\infty} \left\vert \epsilon\,y \right\vert |g(y)|\,{\text d}y . \] Since |yg(y)| is integrable, the expression in the right-hand side is bounded by a constant times ϵ. The desired conclusion then follows by the definition of a limit.

Example 5:    ■

Given e^t > 0, we must show that \[ \ledt\vert \int_{-\infty}^{+\infty} h(t) \left( 1 - e^{-\epsilon^2 t^2 /2} \right) {\text d}t \right\vert < \epsilon' \] for sufficiently small ϵ. Since h decays rapidly at ∞, there is an R such that \[ \left\vert \int_{|t| > R} h(t) \left( 1 - e^{-\epsilon^2 t^2 /2} \right) {\text d}t \right\vert \le \int_{|t| > R} |h(t)|\,{\text d}t \lt; \frac{\epsilon'}{2} . \] Once this R is determined, we can choose ϵ sufficiently small such that \[ \left\vert \int_{-R}^R h(t) \left( 1 - e^{-\epsilon^2 t^2 /2} \right) {\text d}t \right\vert \le 2R\,\sup |h| \left( 1 - e^{-\epsilon^2 t^2 /2} \right) < \frac{\epsilon'}{2} . \] The needed inequality follows.

Example 6:    ■

We use the Gaussian (with σ = 1) as an approximate identity and apply Lemma 2. Let \[ g(y) = \frac{1}{\sqrt{2\pi}} \,e^{-y^2 /2} . \] Since this function satisfies the hypotheses of Lemma 2, we get \[ f(x) = \lim_{\epsilon\to 0} \frac{1}{\sqrt{2\pi}} \,\int_{-\infty}^{+\infty} f(x + \epsilon y)\,g(y)\,{\text d}y = \lim_{\epsilon\to 0} \frac{1}{\sqrt{2\pi}} \,\int_{-\infty}^{+\infty} f(x + \epsilon y)\,e^{-y^2 /2} \,{\text d}y . \] We know from Example 2 that the Fourier transform of the normal distribution is again the Gaussian distribution. So \[ f(x) = \lim_{\epsilon\to 0} \frac{1}{2\pi} \int_{-\infty}^{+\infty} f(x + \epsilon y) \,{\text d}y \,\frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{-\eta^2 /2} e^{-{\bf j}\,y\eta} {\text d}\eta . \tag{E.1} \] In (E.1) we make the change of variables t = x + ϵy, obtaining \[ f(x) = \lim_{\epsilon\to 0} \frac{1}{\sqrt{2\pi}} \,\int_{-\infty}^{+\infty} {\text d}t\,f(t)\, \frac{1}{\sqrt{2\pi}} \,\int_{-\infty}^{+\infty} {\text d}\eta\,e^{-\eta^2 /2} e^{-{\bf j}\left( t-x \right) \frac{\eta}{\epsilon}} \] Now we change variables by setting η = ϵξ; doing so introduces the factor \( \displaystyle \quad e^{-\epsilon^2 \xi^2 /2} \quad \) and enables us to interchange the order of integration. The result gives \[ f(x) = \lim_{\epsilon\to 0} \frac{1}{\sqrt{2\pi}} \,\int_{-\infty}^{+\infty} {\text d}\xi\,e^{-{\bf j}t\xi} \frac{1}{\sqrt{2\pi}} \,\int_{-\infty}^{+\infty} {\text d}t\,f(t)\, e^{-\epsilon^2 \eta^2 /2 e^{{\bf j} x\xi} . \] The inner integral is simply the Fourier transform of f. Hence we obtain \[ f(x) = \lim_{\epsilon\to 0} \frac{1}{\sqrt{2\pi}} \,\int_{-\infty}^{+\infty} \hat{f} (\xi ) \, e^{{\bf j}x\xi} e^{- \epsilon^2 \xi^2 /2} {\text d}\xi . \tag{E.2} \] To finally obtain the inversion formula, we use Lemma 3 to interchange the limit and integral in (E.2)

Example 7:    ■

1. Prove that e^x&rup2; belongs to 𝒮.
2. Let f(x, y) = |x|^|y| for (x, y) ≠ (0, 0). Show that \[ \lim_{x\to 0} \lim_{y\to 0} f(x,y) \ne \lim_{y\to 0} \lim_{x\to 0} f(x,y) . \]

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4. Reed, M. and Simon, B., Methods of Modern Mathematical Physics, II: Fourier Analysis, Self–Adjointness, Academic Press, 1975
5. Rudin, W., Functional Analysis, McGraw-Hill,