Partial Differential Equations/The Fourier transform

In this chapter, we introduce the Fourier transform. The Fourier transform transforms functions into other functions. It can be used to solve certain types of linear differential equations.

Definition and calculation rules
We recall that $$f$$ is integrable $$\Leftrightarrow$$ $$|f|$$ is integrable.

Now we're ready to prove the next theorem:

Theorem 8.2: The Fourier transform of an integrable $$f$$ is well-defined.

Proof: Since $$f$$ is integrable, lemma 8.2 tells us that $$|f|$$ is integrable. But
 * $$\forall x, y \in \mathbb R^d : |f(x) e^{-2\pi i x \cdot y}| = |f(x)| \cdot \overbrace{|e^{-2\pi i x \cdot y}|}^{=1} = |f(x)|$$

, and therefore $$x \mapsto |f(x) e^{-2\pi i x \cdot y}|$$ is integrable. But then, $$x \mapsto f(x) e^{-2\pi i x \cdot y}$$ is integrable, which is why
 * $$\int_{\mathbb R^d} f(x) e^{-2\pi i x \cdot y} dx = \hat f(y)$$

has a unique complex value, by definition of integrability.

Theorem 8.3: Let $$f \in L^1(\mathbb R^d)$$. Then the Fourier transform of $$f$$, $$\hat f$$, is bounded.

Proof:


 * $$\begin{align}

\left| \int_{\mathbb R^d} f(x) e^{-2\pi i x \cdot y} dx \right| & \le \int_{\mathbb R^d} \left| f(x) e^{-2\pi i x  \cdot y} \right| dx & \text{triangle ineq. for the } \int \\ & = \int_{\mathbb R^d} |f(x)| dx & \left| e^{-2\pi i x \cdot y} \right| = 1 \\ & \in \mathbb R & f \in L^1(\mathbb R^d) \end{align}$$

Once we have calculated the Fourier transform $$\tilde f$$ of a function $$f$$, we can easily find the Fourier transforms of some functions similar to $$f$$. The following calculation rules show examples how you can do this. But just before we state the calculation rules, we recall a definition from chapter 2, namely the power of a vector to a multiindex, because it is needed in the last calculation rule.

Definition 2.6:

For a vector $$x = (x_1, \ldots, x_d) \in \R^d$$ and a $$d$$-dimensional multiindex $$\alpha \in \N_0^d$$ we define $$x^\alpha$$, $$x$$ to the power of $$\alpha$$, as follows:
 * $$x^\alpha := x_1^{\alpha_1} \cdots x_d^{\alpha_d}$$

Now we write down the calculation rules, using the following notation:

Notation 8.4:

We write
 * $$f(x) \rightarrow g(y)$$

to mean the sentence 'the function $$y \mapsto g(y)$$ is the Fourier transform of the function $$x \mapsto f(x)$$'.

Proof: To prove the first rule, we only need one of the rules for the exponential function (and the symmetry of the standard dot product):

1.


 * $$f(x)e^{-2 \pi i h \cdot x} \rightarrow \int_{\mathbb R^d} f(x) e^{-2 \pi i h \cdot x} e^{-2\pi i x \cdot y} dx = \int_{\mathbb R^d} f(x) e^{-2\pi i x \cdot (y + h)} dx = \hat f(y + h)$$

For the next two rules, we apply the general integration by substitution rule, using the diffeomorphisms $$x \mapsto x - h$$ and $$x \mapsto \delta^{-1} x$$, which are bijections from $$\mathbb R^d$$ to itself.

2.


 * $$f(x + h) \rightarrow \int_{\mathbb R^d} f(x + h) e^{-2\pi i x \cdot y} dx = \int_{\mathbb R^d} f(x) e^{-2\pi i (x - h) \cdot y} dx = e^{2\pi i h \cdot y} \int_{\mathbb R^d} f(x) e^{-2\pi i x \cdot y} dx = \hat f(y) e^{2\pi i h \cdot y}$$

3.


 * $$f(\delta x) \rightarrow \int_{\mathbb R^d} f(\delta x) e^{-2\pi i x \cdot y} dx = \int_{\mathbb R^d} \delta^{-d} f(x) e^{-2\pi i(\delta^{-1} x) \cdot y} dx = \delta^{-d} \int_{\mathbb R^d} f(x) e^{-2\pi i x \cdot (\delta^{-1} y)} dx = \delta^{-d} \hat f(\delta^{-1} y)$$

4.


 * $$\begin{align}

\int_{\mathbb R^d} \hat g(x) f(x) dx & = \int_{\mathbb R^d} \int_{\mathbb R^d} g(y) e^{2 \pi i x \cdot y} dy f(x) dx & \text{Def. of the Fourier transform} \\ & = \int_{\mathbb R^d} \int_{\mathbb R^d} f(x) g(y) e^{2 \pi i x \cdot y} dy dx & \text{putting a constant inside the integral} \\ & = \int_{\mathbb R^d} \int_{\mathbb R^d} f(x) g(y) e^{2 \pi i x \cdot y} dx dy & \text{Fubini} \\ & = \int_{\mathbb R^d} g(y) \int_{\mathbb R^d} f(x) e^{2 \pi i x \cdot y} dx dy & \text{pulling a constant out of the integral} \\ & = \int_{\mathbb R^d} g(y) \hat f(y) dy & \text{Def. of the Fourier transform} \\ \end{align}$$

The Fourier transform of Schwartz functions
In order to proceed with further rules for the Fourier transform which involve Schwartz functions, we first need some further properties of Schwartz functions.

Proof:

Let $$\varrho, \varsigma \in \mathbb N_0^d$$. Due to the general product rule, we have:
 * $$\partial_\varsigma x^\alpha \partial_\beta \phi(x) = \sum_{\varepsilon \in \mathbb N_0^d \atop \varepsilon \le \varsigma} \binom{\varsigma}{\varepsilon} \partial_\varepsilon(x^\alpha) \partial_{\varsigma - \varepsilon} \partial_\beta \phi(x)$$

We note that for all $$\alpha$$ and $$\varepsilon$$, $$\partial_\varepsilon(x^\alpha)$$ equals to $$x$$ to some multiindex power. Since $$\phi$$ is a Schwartz function, there exist constants $$c_\varepsilon$$ such that:
 * $$\|x^\varrho \partial_\varepsilon(x^\alpha) \partial_{\varsigma - \varepsilon} \partial_\beta \phi\|_\infty \le c_\varepsilon$$

Hence, the triangle inequality for $$\| \cdot \|_\infty$$ implies:
 * $$\|x^\varrho \partial_\varsigma x^\alpha \partial_\beta \phi\|_\infty \le \sum_{\varepsilon \in \mathbb N_0^d \atop \varepsilon \le \varsigma} \binom{\varsigma}{\varepsilon} c_\varepsilon$$

Proof:

We use that if the absolute value of a function is almost everywhere smaller than the value of an integrable function, then the first function is integrable.

Let $$\phi$$ be a Schwartz function. Then there exist $$b, c \in \mathbb R_{>0}$$ such that for all $$x \in \mathbb R^d$$:
 * $$|\phi(x)| \le \min \left\{ b \prod_{j=1}^d |x_j|^{-2}, c \right\}$$

The latter function is integrable, and integrability of $$\phi$$ follows.

Now we can prove all three of the following rules for the Fourier transform involving Schwartz functions.

Proof:

1.

For the first rule, we use induction over $$|\alpha|$$.

It is clear that the claim is true for $$|\alpha| = 0$$ (then the rule states that the Fourier transform of $$\phi$$ is the Fourier transform of $$\phi$$).

We proceed to the induction step: Let $$n \in \mathbb N_0$$, and assume that the claim is true for all $$\alpha \in \mathbb N_0^d$$ such that $$|\alpha| = n$$. Let $$\beta \in \mathbb N_0^d$$ such that $$|\beta| = n+1$$. We show that the claim is also true for $$\beta$$.

Remember that we have $$\partial_\beta \phi := \partial_{x_1}^{\beta_1} \cdots \partial_{x_d}^{\beta_d} \phi$$. We choose $$k \in \{1, \ldots, d\}$$ such that $$\beta_k > 0$$ (this is possible since otherwise $$|\beta| = 0$$), define
 * $$e_k := (0, \ldots, 0, \overbrace{1}^{k\text{th entry}}, 0, \ldots, 0)$$
 * $$\alpha := \beta - e_k$$

and obtain
 * $$\partial_\beta \phi = \partial_{x_k} \partial_\alpha \phi$$

by Schwarz' theorem, which implies that one may interchange the order of partial derivation arbitrarily.

Let $$R \in \mathbb R_{>0}$$ be an arbitrary positive real number. From Fubini's theorem and integration by parts, we obtain:
 * $$\begin{align}

\int_{[-R, R]^d} \partial_{x_k} \partial_\alpha \phi(x) e^{-2\pi i x \cdot y} dx & = \int_{[-R, R]^{d-1}} \int_{-R}^R \partial_{x_k} \partial_\alpha \phi(x) e^{-2\pi i x \cdot y} dx_k d(x_1, \ldots, x_{k-1}, x_{k+1}, \ldots, x_d) \\ & = \int_{[-R, R]^{d-1}} \left( \left( \partial_\alpha \phi(x) e^{-2\pi i x \cdot y} \right) \big|_{x_k = -R}^{x_k = R} - \int_{-R}^R \partial_\alpha \phi(x) (-2\pi i y_k) e^{-2\pi i x \cdot y} dx_k \right) d(x_1, \ldots, x_{k-1}, x_{k+1}, \ldots, x_d) \end{align}$$

Due to the dominated convergence theorem (with dominating function $$x \mapsto \partial_{x_k} \partial_\alpha \phi(x)$$), the integral on the left hand side of this equation converges to
 * $$\int_{\mathbb R^d} \partial_{x_k} \partial_\alpha \phi(x) e^{-2\pi i x \cdot y} dx$$

as $$R \to \infty$$. Further, since $$\phi$$ is a Schwartz function, there are $$b, c \in \mathbb R_{>0}$$ such that:
 * $$|\partial_\alpha \phi(x)| < \min\left\{ b \prod_{j=1 \atop j \neq k}^d |x_j|^{-2}, c\right\}$$

Hence, the function within the large parentheses in the right hand sinde of the last line of the last equation is dominated by the $$L^1(\mathbb R^{d-1})$$ function
 * $$(x_1, \ldots, x_{k-1}, x_{k+1}, \ldots, x_d) \mapsto \min\left\{ b \prod_{j=1 \atop j \neq k}^d |x_j|^{-2}, c\right\} + \int_{-\infty}^\infty |\partial_\alpha \phi(x) (-2\pi i y_k)| dx_k$$

and hence, by the dominated convergence theorem, the integral over that function converges, as $$R \to \infty$$, to:
 * $$\begin{align}

\int_{\mathbb R^{d-1}} \left( \int_{-\infty}^\infty \partial_\alpha \phi(x) (2\pi i y_k) e^{-2\pi i x \cdot y} dx_k \right) d(x_1, \ldots, x_{k-1}, x_{k+1}, \ldots, x_d) & = \int_{\mathbb R^d} \partial_\alpha \phi(x) (2\pi i y_k) e^{-2\pi i x \cdot y} dx & \text{Fubini} \\ & = \int_{\mathbb R^d} (2\pi i y^\beta) \phi(x) e^{-2\pi i x \cdot y} dx & \text{induction hypothesis} \end{align} $$ From the uniqueness of limits of real sequences we obtain 1.

2.

We use again induction on $$|\alpha|$$, note that the claim is trivially true for $$|\alpha| = 0$$, assume that the claim is true for all $$\alpha \in \mathbb N_0^d$$ such that $$|\alpha| = n$$, choose $$\beta \in \mathbb N_0^d$$ such that $$|\beta| = n+1$$ and $$k \in \{1, \ldots, d\}$$ such that $$\beta_k > 0$$ and define $$\alpha := \beta - e_k$$.

Theorems 8.6 and 8.7 imply that Further, Hence, Leibniz' integral rule implies:
 * for all $$y \in \mathbb R^d$$, $$\int_{\mathbb R^d} |\phi(x) e^{-2\pi i x \cdot y}| dx < \infty$$ and
 * for all $$y \in \mathbb R^d$$, $$\int_{\mathbb R^d} |\phi(x) \partial_{y_k} e^{-2\pi i x \cdot y}| dx < \infty$$.
 * $$\partial_{y_k} (\phi(x) e^{-2\pi i x \cdot y})$$ exists for all $$(x, y) \in \mathbb R^d \times \mathbb R^d$$.
 * $$\begin{align}

\partial_\beta \hat \phi (y) & = \partial_{x_k} \partial_\alpha \hat \phi(y) & \\ & = \partial_{y_k} \int_{\mathbb R^d} (2\pi i x)^\alpha \phi(x) e^{-2 \pi i x \cdot y} dx & \text{induction hypothesis} \\ & = \int_{\mathbb R^d} \partial_{y_k} (2\pi i x)^\alpha \phi(x) e^{-2 \pi i x \cdot y} dx & \\ & = \int_{\mathbb R^d} (2\pi i x)^\beta \phi(x) e^{-2 \pi i x \cdot y} dx \end{align}$$

3.


 * $$\begin{align}

\widehat{\phi * \theta}(y) & := \int_{\mathbb R^d} (\phi * \theta)(x) e^{-2 \pi i x \cdot y} dx & \text{Def. of Fourier transform} \\ & := \int_{\mathbb R^d} \int_{\mathbb R^d} \phi(z) \theta(x - z) dz e^{-2 \pi i x \cdot y} dx & \text{Def. of convolution} \\ & = \int_{\mathbb R^d} \int_{\mathbb R^d} e^{-2 \pi i x \cdot y} \phi(z) \theta(x - z) dz dx & \text{linearity of the integral} \\ & = \int_{\mathbb R^d} \int_{\mathbb R^d} e^{-2 \pi i x \cdot y} \phi(z) \theta(x - z) dx dz & \text{Fubini} \\ & = \int_{\mathbb R^d} \int_{\mathbb R^d} e^{-2 \pi i x \cdot y} e^{-2 \pi i z \cdot y} \phi(z) \theta(x) dx dz & \text{Integration by substitution using } x \mapsto x + z \text{ and } \forall b, c \in \mathbb R : e^{b+c} = e^b e^c \\ & = \int_{\mathbb R^d} \overbrace{\int_{\mathbb R^d} e^{-2 \pi i x \cdot y} \theta(x) dx}^{= \hat \theta(y)} \phi(z) e^{-2 \pi i z \cdot y} dz & \text{pulling a constant out of the integral} \\ & = \hat \theta(y) \overbrace{\int_{\mathbb R^d} \phi(z) e^{-2 \pi i z \cdot y} dz}^{= \hat \phi(y)} & \text{pulling a constant out of the integral} \end{align}$$

Proof:

Let $$\alpha, \beta \in \mathbb N_0^d$$ be two arbitrary $$d$$-dimensional multiindices, and let $$\phi \in \mathcal S(\mathbb R^d)$$. By theorem 8.6 $$\partial_\alpha ((-2\pi \mathrm i x)^\beta \phi)$$ is a Schwartz function as well. Theorem 8.8 implies:
 * $$x^\alpha \partial_\beta \hat \phi = \widehat{\partial_\alpha ((-2\pi \mathrm i x)^\beta \phi)}$$

By theorem 8.3, $$\widehat{\partial_\alpha ((-2\pi \mathrm i x)^\beta \phi)}$$ is bounded. Since $$\alpha, \beta \in \mathbb N_0^d$$ were arbitrary, this shows that $$\hat \phi \in \mathcal S(\mathbb R^d)$$.

Both the Fourier transform and the inverse Fourier transform are sequentially continuous:

Proof:

1. We prove $$\mathcal F(\phi_l) \to \mathcal F(\phi), l \to \infty$$.

Let $$\alpha, \beta \in \mathbb N_0^d$$. Due to theorem 8.8 1. and 2. and the linearity of derivatives, integrals and multiplication, we have
 * $$x^\alpha \partial_\beta(\mathcal F(\phi_l)(x) - \mathcal F(\phi)(x)) = \int_{\mathbb R^d} \partial_\alpha ((-2\pi i y)^\beta (\phi_l(y) - \phi(y))) e^{-2\pi i x \cdot y} dy$$.

As in the proof of theorem 8.3, we hence obtain
 * $$\left| x^\alpha \partial_\beta(\mathcal F(\phi_l)(x) - \mathcal F(\phi)(x)) \right| \le \|\partial_\alpha ((-2\pi i x)^\beta (\phi_l - \phi)))\|_{L^1}$$.

Due to the multi-dimensional product rule,
 * $$\partial_\alpha ((-2\pi i x)^\beta (\phi_l(x) - \phi(x))) = \sum_{\varsigma \in \mathbb N_0^d \atop \varsigma \le \alpha} \binom{\varsigma}{\alpha} \partial_\varsigma ((-2\pi i x)^\beta) \partial_{\alpha - \varsigma} (\phi_l(x) - \phi(x))$$.

Let now $$\epsilon > 0$$ be arbitrary. Since $$\phi_l \to \phi$$ as defined in definition 3.11, for each $$n \in \{1, \ldots, d\}$$ we may choose $$N_1 \in \mathbb N$$ such that
 * $$\forall k \ge N_1 : \sum_{\varsigma \in \mathbb N_0^d \atop \varsigma \le \alpha} \binom{\varsigma}{\alpha} \|x_n^2 \partial_\varsigma ((-2\pi i x)^\beta) \partial_{\alpha - \varsigma} (\phi_l(x) - \phi(x))\|_\infty < \epsilon$$.

Further, we may choose $$N_2 \in \mathbb N$$ such that
 * $$\forall k \ge N_2 : \sum_{\varsigma \in \mathbb N_0^d \atop \varsigma \le \alpha} \binom{\varsigma}{\alpha} \|\partial_\varsigma ((-2\pi i x)^\beta) \partial_{\alpha - \varsigma} (\phi_l(x) - \phi(x))\|_\infty < \epsilon$$.

Hence follows for $$k \ge N := \max\{N_1, N_2\}$$:
 * $$\begin{align}

\|\partial_\alpha ((-2\pi i x)^\beta (\phi_l - \phi)))\|_{L^1} & := \int_{\mathbb R^d} |\partial_\alpha ((-2\pi i x)^\beta (\phi_l(x) - \phi(x)))| dx \\ & \le \int_{\mathbb R^d} \epsilon \min\{x_1^{-2}, \ldots, x_d^{-2}, 1\} dx \\ & = \epsilon \int_{\mathbb R^d} \min\{x_1^{-2}, \ldots, x_d^{-2}, 1\} dx \end{align}$$

Since $$\epsilon > 0$$ was arbitrary, we obtain $$\mathcal F(\phi_l) \to \mathcal F(\phi), l \to \infty$$.

2. From 1., we deduce $$\mathcal F^{-1}(\phi_l) \to \mathcal F^{-1}(\phi), l \to \infty$$.

If $$\phi_l \to \phi$$ in the sense of Schwartz functions, then also $$\theta_l \to \theta$$ in the sense of Schwartz functions, where we define
 * $$\theta_l(x) := \phi_l(-x)$$ and $$\theta(x) := \phi(-x)$$.

Therefore, by 1. and integration by substitution using the diffeomorphism $$x \mapsto -x$$, $$\mathcal F^{-1}(\phi_l) = \mathcal F(\theta_l) \to \mathcal F(\theta) = \mathcal F^{-1}(\phi)$$.

In the next theorem, we prove that $$\mathcal F^{-1}$$ is the inverse function of the Fourier transform. But for the proof of that theorem (which will be a bit long, and hence to read it will be a very good exercise), we need another two lemmas:

Proof:

1. $$\mathcal F^{-1}(G) = G$$:

We define
 * $$\mu: \mathbb R^d \to \mathbb R, \mu(\xi) := e^{\pi \|\xi\|^2} \hat G(\xi)$$.

By the product rule, we have for all $$n \in \{1, \ldots, d\}$$
 * $$\partial_{\xi_n} \mu (\xi) = 2 \pi \xi_n e^{\pi \|\xi\|^2} \hat G(\xi) + e^{\pi \|\xi\|^2} \partial_{\xi_n} \hat G(\xi)$$.

Due to 1. of theorem 8.8, we have
 * $$2 \pi \xi_n \hat G(\xi) = - i \widehat{\partial_{x_n} G} (\xi) = - i \int_{\mathbb R^d} (2 \pi x_n) e^{-\pi \|x\|^2} e^{2 \pi i \xi \cdot x} dx$$;

from 2. of theorem 8.8 we further obtain
 * $$\partial_{\xi_n} \hat G(\xi) = \int_{\mathbb R^d} (-2\pi i x_n) e^{-\pi \|x\|^2} e^{2 \pi i \xi \cdot x} dx$$.

Hence, $$\mu$$ is constant. Further,
 * $$\begin{align}

\mu(0) & = \int_{\mathbb R^d} e^{-\pi\|x\|^2} dx & \\ & = \int_{\mathbb R^d} \frac{1}{\sqrt{2\pi}^d} e^{-\|x\|^2/2} dx & \text{substitution using } x \mapsto \frac{1}{\sqrt{2\pi}^d} x \\ & = 1 & \text{lemma 6.2} \end{align}$$.

2. $$\mathcal F^{-1}(G) = G$$:

By substitution using the diffeomorphism $$x \mapsto -x$$,
 * $$\forall x \in \mathbb R^d : \mathcal F^{-1}(G)(x) = \mathcal F(G)(x) = G(x)$$.

For the next lemma, we need example 3.4 again, which is why we restate it:

Example 3.4: The standard mollifier $$\eta$$, given by
 * $$\eta: \mathbb R^d \to \mathbb R, \eta(x) = \frac{1}{c}\begin{cases} e^{-\frac{1}{1-\|x\|^2}}& \text{ if } \|x\|_2 < 1\\

0& \text{ if } \|x\|_2 \geq 1 \end{cases}$$ , where $$c := \int_{B_1(0)} e^{-\frac{1}{1-\|x\|^2}} dx$$, is a bump function (see exercise 3.2).

Proof:

Let $$\alpha, \beta \in \mathbb N_0^d$$ be arbitrary. Due to the generalised product rule,
 * $$\partial_\beta(\phi_n - \phi) = \sum_{\varsigma \in \mathbb N_0^d \atop \varsigma \le \beta} \binom{\beta}{\varsigma} \frac{1}{n^{|\beta - \varsigma|}} \partial_{\beta - \varsigma} \eta(x/n) \partial_\varsigma \phi(x) - \partial_\beta \phi(x)$$.

By the triangle inequality, we may hence deduce
 * $$|x^\alpha \partial_\beta (\phi_n - \phi)| \le \frac{1}{n} \sum_{\varsigma \in \mathbb N_0^d \atop \varsigma < \beta} \binom{\beta}{\varsigma} \frac{1}{n^{|\beta - \varsigma| - 1}} |x^\alpha \partial_\varsigma \phi(x)| |\partial_{\beta - \varsigma} \eta(x/n)| + |x^\alpha \partial_\beta \phi(x)| |1 - \eta(x/n)|$$.

Since both $$\phi$$ and $$\eta$$ are Schwartz functions (see exercise 3.2 and theorem 3.9), for each $$\varrho \in \mathbb N_0^d$$ we may choose $$b_\varrho, c_\varrho \in \mathbb R$$ such that
 * $$\|\partial_{\beta - \varrho} \eta\|_\infty < b_\varrho$$ and $$\|x^\alpha \partial_\varrho \phi\|_\infty < c_\varrho$$.

Further, for each $$k \in \{1, \ldots, d\}$$, we may choose $$c_k \in \mathbb R^d$$ such that
 * $$\|x_k x^\alpha \partial_\beta \phi\|_\infty < c_k$$.

Let now $$\epsilon > 0$$ be arbitrary. We choose $$N_1 \in \mathbb N$$ such that for all $$n \ge N_1$$
 * $$\frac{1}{n} \sum_{\varsigma \in \mathbb N_0^d \atop \varsigma < \beta} \binom{\beta}{\varsigma} \frac{1}{n^{|\beta - \varsigma| - 1}} b_\varsigma c_\varsigma < \epsilon/2$$.

Further, we choose $$R \in \mathbb R_{>0}$$ such that
 * $$\|x\| > R \Rightarrow |x^\alpha \partial_\beta \phi(x)| < \epsilon/2$$.

This is possible since
 * $$|x^\alpha \partial_\beta \phi(x)| \le \min \left\{ \frac{c_1}{|x_1|}, \ldots, \frac{c_d}{|x_d|} \right\}$$

due to our choice of $$c_1, \ldots, c_d$$.

Then we choose $$N_2 \in \mathbb N$$ such that for all $$n \ge N_2$$
 * $$\forall x \in B_R(0) : |1 - \phi(x/n)| < 1/c_\beta$$.

Inserting all this in the above equation gives $$|x^\alpha \partial_\beta (\phi_n - \phi)| < \epsilon$$ for $$n \ge N := \max \{N_1, N_2\}$$. Since $$\alpha$$, $$\beta$$ and $$\epsilon$$ were arbitrary, this proves $$\phi_n \to \phi$$ in the sense of Schwartz functions.

Proof:

1. We prove that if $$\phi$$ is a Schwartz function vanishing at the origin (i. e. $$\phi(0) = 0$$), then $$\mathcal F^{-1}(\mathcal F(\phi))(0) = 0$$.

So let $$\phi$$ be a Schwartz function vanishing at the origin. By the fundamental theorem of calculus, the multi-dimensional chain rule and the linearity of the integral, we have
 * $$\phi(x) = \phi(x) - \phi(0) = \int_0^1 \frac{d}{dt} \phi(tx) dx = \sum_{j=1}^d x_j \int_0^1 \partial_{x_j} \phi(tx) dt$$.

Defining $$\phi_n(x) := \eta(x/n) \phi(x)$$,
 * $$\theta_{j, n}(x) := \eta(x/n) \int_0^1 \partial_{x_j} \phi(tx) dt$$,

and multiplying both sides of the above equation by $$\eta(x/n)$$, we obtain
 * $$\phi_n(x) = \sum_{j=1}^d x_j \theta_{j, n}(x)$$.

Since by repeated application of Leibniz' integral rule for all $$\alpha \in \mathbb N_0^d$$
 * $$\partial_\alpha \theta_{j, n}(x) = \sum_{\varsigma \le \alpha} \binom{\alpha}{\varsigma} \frac{1}{n^{|\varsigma|}} \partial_\varsigma \eta(x/n) \int_0^1 \partial_{\alpha - \varsigma} \partial_{x_j} \phi(tx) dt$$,

all the $$\theta_{j, n}$$ are bump functions (due to theorem 4.15 and exercise 3.?), and hence Schwartz functions (theorem 3.9). Hence, by theorem 8.8 and the linearity of the Fourier transform (which follows from the linearity of the integral),
 * $$\mathcal F(\phi_n) (y) = \sum_{j=1}^d \mathcal F(x_j \theta_{j, n})(y) = \frac{1}{-2\pi i}\sum_{j=1}^d \partial_{y_j} \mathcal F(\theta_{j, n})(y)$$.

Hence,
 * $$\mathcal F^{-1}(\mathcal F(\phi_n))(0) = \int_{\mathbb R^d} \mathcal F(\phi_n)(y) \overbrace{e^{-2\pi i 0 \cdot y}}^{=1} dy = \frac{1}{-2\pi i} \sum_{j=1}^d \int_{\mathbb R^d} \partial_{y_j} \mathcal F(\theta_{j, n})(y) dy$$.

Let $$k \in \{1, \ldots, d\}$$. By Fubini's theorem, the fundamental theorem of calculus and since $$\theta_{k, n}$$ is a bump function, we have
 * $$\int_{\mathbb R^d} \partial_{y_k} \mathcal F(\theta_{k, n})(y) dy = \int_{\mathbb R^{d-1}} \int_{-\infty}^\infty \partial_{y_k} \mathcal F(\theta_{k, n})(y) dy_k d(y_1, \ldots, y_{k-1}, y_{k+1}, \ldots, y_d) = 0$$.

If we let $$n \to \infty$$, theorem 8.11 and lemma 8.13 give the claim.

2. We deduce from 1. that if $$\phi$$ is an arbitrary Schwartz function, then $$\mathcal F^{-1}(\mathcal F(\phi))(0) = \phi(0)$$.

As in lemma 8.12, we define
 * $$G: \mathbb R^d \to \mathbb R, G(x) := e^{-\pi \|x\|^2}$$.

Let now $$\phi$$ be any Schwartz function. Then $$\phi - \phi(0) G$$ is also a Schwartz function (see exercise 3.?). Further, since $$G(0) = 1$$, it vanishes at the origin. Hence, by 1.,
 * $$\mathcal F^{-1}(\mathcal F(\phi - \phi(0) G))(0) = 0$$.

Further, due to lemma 8.12 and the linearity of the Fourier transform,
 * $$0 = \mathcal F^{-1}(\mathcal F(\phi - \phi(0) G))(0) = \mathcal F^{-1}(\mathcal F(\phi)) - \phi(0) G(0)$$.

3. We deduce from 2. that if $$\phi$$ is a Schwartz function and $$x \in \mathbb R^d$$ is arbitrary, then $$\mathcal F^{-1}(\mathcal F(\phi))(x) = \phi(x)$$ (i. e. $$\mathcal F^{-1}(\mathcal F(\phi)) = \phi$$.

Let $$\phi \in \mathcal S(\mathbb R^d)$$ and $$x \in \mathbb R^d$$ be arbitrary. Due to the definition of $$\mathcal F^{-1}$$,
 * $$\mathcal F^{-1}(\mathcal F(\phi))(x) = \int_{\mathbb R^d} \mathcal F(\phi)(y) e^{2 \pi i x \cdot y} dy$$.

Further, if we define $$\theta(z) := \phi(z + x)$$,
 * $$\mathcal F(\phi)(y) e^{2 \pi i x \cdot y} = \int_{\mathbb R^d} \phi(z) e^{-2 \pi i z \cdot y} e^{2 \pi i x \cdot y} dz = \int_{\mathbb R^d} \theta(z) dz e^{-2 \pi i z \cdot y} = \mathcal F(\theta)(y)$$.

Hence, by 2.,
 * $$\mathcal F^{-1}(\mathcal F(\phi))(x) = \int_{\mathbb R^d} \mathcal F(\theta)(y) dy = \mathcal F^{-1}(\mathcal F(\theta))(0) = \theta(0) = \phi(x)$$.

4. We deduce from 3. that for any Schwartz function $$\phi$$ we have $$\mathcal F(\mathcal F^{-1}(\phi)) = \phi$$.

Let $$\phi \in \mathcal S(\mathbb R^d)$$ and $$x \in \mathbb R^d$$ be arbitrary. Then we have
 * $$\mathcal F(\mathcal F^{-1}(\phi))(y) = \mathcal F^{-1}(\mathcal F^{-1}(\phi))(-y) = \int_{\mathbb R^d} \mathcal F^{-1}(\phi)(x) e^{-2 \pi i x \cdot y} dx = \int_{\mathbb R^d} \mathcal F(\phi)(-x) e^{-2 \pi i x \cdot y} dx = \mathcal F^{-1}(\mathcal F(\phi))(y) = \phi(y)$$.

The Fourier transform of tempered distributions
Proof:

1. Sequential continuity follows from the sequential continuity of $$\mathcal T$$ and $$\mathcal F$$ (theorem 8.11) and that the composition of two sequentially continuous functions is sequentially continuous again.

2. Linearity follows from the linearity of $$\mathcal T$$ and $$\mathcal F$$ and that the composition of two linear functions is linear again.