# Welcome.

## January 7, 2011

### Harmonic Analysis Lecture Notes 3

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:54 pm

Fourier Transform in ${{\mathbb R}^n}$

Definition For ${f \in L^1({\mathbb R}^d)}$, we define

$\displaystyle \hat f(\xi) = \int_{{\mathbb R}^n} e^{-2\pi i x \cdot \xi} f(x) \; dx.$

Note that ${||\hat f||_{L^\infty} \leq ||f||_{L^1}}$ (aka Minkowski’s inequality).

Definition A ${C^\infty}$ smooth function is said to be a Schwartz function (this is Laurent Schwartz, not Hermann Schwarz of the C-S inequality) if

$\displaystyle x^\beta \partial_x^\alpha f \in L^\infty ({\mathbb R}^d) \text{ for all multi-indices }\alpha, \beta,$

where ${x^\beta := x_1^{\beta_1} x_2^{\beta_2} \cdots x_d^{\beta_d}}$ and ${\partial_x^\alpha = \partial_{x_1}^{\alpha_1}\partial_{x_2}^{\alpha_2} \cdots \partial_{x_d}^{\alpha_d}}$. The space of Schwartz function is denoted ${\mathcal{S}({\mathbb R}^d)}$.

Proposition Suppose ${f \in \mathcal{S}({\mathbb R}^d)}$, then

1. If ${g(x) = f(x - y)}$, then ${\hat g(\xi) = e^{-2\pi i y \cdot \xi} \hat f(\xi) }$.
2. If ${g(x) = e^{2\pi i \eta \cdot x}f(x)}$ then ${\hat g(\xi) = \hat f(\xi - \eta) }$.
3. If ${g(x) = f(Tx)}$, where ${T \in GL({\mathbb R}^d)}$, then ${\hat g(\xi) = |\det(T)|^{-1} \hat f((T^{-1})^t \xi)}$.
4. If ${g = \bar f}$ then ${\hat g(\xi) = \overline{\hat f(-\xi)}}$.
5. If ${g = k *f = \int_{{\mathbb R}^d} k(x - y) f(y) \; dy}$ with ${k \in L^1}$, then ${g \in L^1}$ (again, Minkowski’s inequality) and ${\hat g(\xi) = \hat k(\xi) \hat f(\xi)}$.
6. If ${g = \partial^{\alpha} f}$ then ${\hat g(\xi) = (2\pi i \xi)^{\alpha} \hat f(\xi)}$.
7. If ${g(x) = x^\beta f(x)}$ then ${\left( \frac{i}{2\pi} \partial_\xi\right)^{\beta} \hat f(\xi)}$.

Proof: For assertion (3), using the change of variables ${z = Tx}$,

$\displaystyle \begin{array}{rcl} \hat g(\xi) &=& \displaystyle \int_{{\mathbb R}^d} e^{-2\pi i x \cdot \xi} f(Tx) \; dx \\ \\ &=& \displaystyle |\det(T)|^{-1} \int_{{\mathbb R}^d} e^{-2\pi i \xi \cdot T^{-1} z} f(z) \; dz\\ \\ &=& \displaystyle |\det(T)|^{-1} \int_{{\mathbb R}^d} e^{-2\pi i (T^{-1})^t\xi \cdot z} f(z) \; dz\\ \\ &=&\displaystyle |\det(T)|^{-1} \hat f((T^{-1})^t \xi). \end{array}$

For assertion (5), using the change of variables ${x = z+y}$ and Fubini

$\displaystyle \begin{array}{rcl} \hat g(\xi) &=&\displaystyle \int_{{\mathbb R}^d} \int_{{\mathbb R}^d} e^{-2\pi i x \cdot \xi} k(x-y) f(y) \; dy \; dx \\ \\ &=&\displaystyle \int_{{\mathbb R}^d} \int_{{\mathbb R}^d} e^{-2\pi i (y+z) \cdot \xi} k(z) f(y) \; dy \; dx \\ \\ &=&\displaystyle \hat k(\xi) \hat f(\xi). \end{array}$

For the assertion (7), note that ${x^\beta e^{-2\pi i x \cdot \xi} = \left( \frac{i}{2\pi} \partial_\xi\right)^{\beta} e^{-2\pi i x\cdot \xi}}$. The remaining assertions are left as an exercise to the reader. $\Box$

Remark

1. Using ${|| \hat f||_{L^\infty} \leq ||f||_{L^1}}$, we can extend assertions (1) through (5) to general ${f \in L^1}$.
2. From assertions (6) and (7) we see that if ${f \in \mathcal{S}({\mathbb R}^d)}$, then ${\hat f \in \mathcal{S} ({\mathbb R}^d)}$.
3. From assertion (3) we see that if ${T \in O({\mathbb R}^d)}$, then ${g = f \circ T \implies \hat g = \hat f \circ T}$. In particular, any reflection/rotation symmetry of ${f}$ is inherited by ${\hat f}$.

Lemma Let ${A = A^t}$ be a complex matrix with positive definite real part. Then

$\displaystyle \int_{{\mathbb R}^d} e^{-x^t A x} e^{-2\pi i x \cdot \xi} \; dx = \pi^{d/2} \det(A)^{-\frac{1}{2}} e^{-\pi^2 \xi^t A^{-1} \xi}.$

Note: The fact that ${A}$ is indeed invertible is left as an exercise to the reader. Proof: By analytic continuation it suffices to consider the case where ${A}$ is real (this method also reveals which branch of ${\sqrt{\det(A)}}$ one should use). Let ${y = \pi A^{-1}\xi}$ then

$\displaystyle (x+ i y)^{t} A(x + iy) = x^t A x + 2\pi i x \cdot \xi - \pi^2 \xi^t A^{-1} \xi$

Thus, the whole problem is to show

$\displaystyle \int e^{-(x + iy)^t A(x+ iy)} \; dx = \pi^{d/2} \det(A)^{-\frac{1}{2}}.$

Note that

$\displaystyle \nabla_y (LHS) = i \int_{{\mathbb R}^d} \nabla_x e^{-(x+iy)^t A(x+iy)} = 0,$

with the last equality by the fundamental theorem of calculus, and thus we are left to deal with the case ${y = 0}$. Breaking ${A}$ into eigenvalues and eigenvectors, we are just left to show that

$\displaystyle \int_{\mathbb R} e^{-\lambda x^2} \; dx = \sqrt{\pi / \lambda}$

which is a classical computation from multivariable calculus. $\Box$

Remark From this result it follows that the Fourier transform of ${e^{-\pi x^2}}$ is itself.

Theorem (Fourier Inversion) If ${f \in \mathcal{S}({\mathbb R}^d)}$, then

$\displaystyle f(x) = \int_{{\mathbb R}^d} e^{2\pi i x \cdot \xi} \hat f(\xi) \; d\xi$

(note that the integral on the right hand side is absolutely convergent since ${\hat f \in \mathcal{S}({\mathbb R}^d)}$). Equivalently,

$\displaystyle (\hat f)\hat \; (x) = f(-x).$

We call this “inverse Fourier transform” ${\check f(\xi) = \int_{{\mathbb R}^d} e^{2\pi i x \cdot \xi} f(x) \; dx}$.

Proof: Let ${I_\epsilon(x) = \int_{{\mathbb R}^d} e^{-\pi \epsilon^2 \xi^2} \hat f(\xi) e^{2\pi i x \cdot \xi} \; d\xi}$. By dominated convergence theorem, ${\lim_{\epsilon \downarrow 0} I_\epsilon(x)}$ is equal to the right hand side above. On the other hand, Fubini’s theorem gives

$\displaystyle \begin{array}{rcl} I_\epsilon(x) &=& \displaystyle \int_{{\mathbb R}^d}\int_{{\mathbb R}^d} \underbrace{e^{-\epsilon^2 \pi \xi^2} e^{2\pi i (x-y) \cdot \xi} f(y)}_{\text{in }L^1({\mathbb R}^d \times {\mathbb R}^d) \text{ because } \epsilon>0} \; dy \; d\xi\\ \\ &=& \displaystyle \int_{{\mathbb R}^d} \epsilon^{-d} e^{-\pi (x-y)^2/\epsilon^2} \; f(y) \; dy \end{array}$

which converges to ${f(x)}$ as ${\epsilon \downarrow 0}$ because ${f}$ is continuous. $\Box$