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


  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


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

Blog at WordPress.com.

%d bloggers like this: