January 10, 2011

Harmonic Analysis Lecture Notes 4

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:53 pm

We have seen that the Fourier transform is a bijection on Schwartz space. In fact, it is a homeomorphism if we give {\mathcal{S}({\mathbb R}^d)} its natural topology: the sub-basic sets are

\displaystyle  	\mathcal{O}_{g,\epsilon, \alpha, \beta} = \{ f \in \mathcal{S}({\mathbb R}^d) : ||x^\beta \partial^\alpha (f-g)||_{L^\infty} < \epsilon\}

Under this topology, {\mathcal{S}({\mathbb R}^d)} is a completely metrizable LCS (locally convex space).

This LCS has many continuous linear functionals, they are called the tempered distributions {\mathcal{S}'({\mathbb R}^d)}. Note that any {f \in L^1}, {f \in L^\infty}, or {f \in {\mathbb C}[x_1, \ldots, x_d]} can be identified with a distribution via

\displaystyle  	f \leftrightarrow \left( \varphi \mapsto \int_{{\mathbb R}^d} f \varphi \; dx \right)

Other examples:

  • {\varphi \mapsto \varphi(0)} is called the Dirac delta function.
  • {\varphi \mapsto \lim_{\epsilon \downarrow 0} \int_{|x| > \epsilon} \frac{x_j}{|x|^{d+1}} \varphi(x) \; dx} where {1 \leq j \leq d} is called the Cauchy principal value.

Any continuous linear transformation {T} on {\mathcal{S}({\mathbb R}^d)} then defines a continuous linear map {T'} (the transpose) via {(T'f)(\varphi) = f(T\varphi)}, where {f \in \mathcal{S}'({\mathbb R}^d)}.

Example As {\partial_j} is continuous on {\mathcal{S}}, its transpose “-{\partial_j}” is continuous on {\mathcal{S}'({\mathbb R}^d)}.

Example Similarly, the Fourier transform extends to a homeomorphism of {\mathcal{S}'} because it is its own transpose (cf. below).

Lastly, note that

  1. {\mathcal{S}({\mathbb R}^d)} is dense in {\mathcal{S}'({\mathbb R}^d)}.
  2. “Distributionally” means in the sense of distributions, more sociologically, it just means integrate against {\varphi \in \mathcal{S}({\mathbb R}^d)} and follow your nose.

Example {-\Delta \frac{1}{2\pi} \log \frac{1}{|z|} = \delta} in {{\mathbb R}^2 = {\mathbb C}} distributionally, i.e.

\displaystyle  		\int_{{\mathbb C}} \frac{1}{2\pi} \left(\log \frac{1}{|z|} \right) (-\Delta \varphi) \; dz = \varphi(0) \text{ for all } \varphi \in \mathcal{S}({\mathbb R}^d).

Lemma (Transpose of the Fourier Transform) For {f,g \in \mathcal{S}({\mathbb R}^d)},

\displaystyle  	\int \hat f g = \int f \hat g,

i.e., the Fourier transform is its own transpose.

Proof: Fubini’s theorem gives

\displaystyle  \begin{array}{rcl}  	LHS &=& \displaystyle \int_{{\mathbb R}^d} \int_{{\mathbb R}^d} f(x) e^{-2\pi i x \cdot \xi} g(\xi) \; dx \; d\xi \\ \\ 		&=& \displaystyle \int_{{\mathbb R}^d} \int_{{\mathbb R}^d} f(x) e^{-2\pi i x \cdot \xi} g(\xi) \; d\xi \; dx \\ \\ 		&=& RHS 	\end{array}


Corollary If {f,g \in \mathcal{S}({\mathbb R}^d)} then

\displaystyle  	\int \hat f \; \overline{\hat g} = \int f \bar g

Theorem (Plancherel) The Fourier transform extends continuous from {\mathcal{S}({\mathbb R}^d)} to a unitary map on {L^2({\mathbb R}^d)}.

Proof: Suppose {\{f_n\}} is a sequence in {\mathcal{S}({\mathbb R}^d)} converges to {f \in L^2} in the {L^2}-norm. By preceding Lemma,

\displaystyle  	|| \hat f_n - \hat f_m||_{L^2} = || \widehat{f_n - f_m} ||_{L^2} = ||f_n - f_m||_{L^2},

which tends to {0} as {n,m} tends to {\infty}. Namely, {\hat f_n} is a Cauchy sequence in {L^2}, which is complete, and so converges. Let us call the limit {\hat f}. By intertwining sequences, for example, we see that the limit does not depend on the particular sequence {\{f_n\}}.

Note that {f \mapsto \hat f} is an isometry because

\displaystyle  	\begin{matrix} 	||f_n||_{L^2} & = & ||\hat f_n||_{L^2} & \text{ (Lemma)} \\ 	\downarrow & & \downarrow \\ 	||f||_{L^2} & & ||\hat f ||_{L^2} 	\end{matrix}

Now, recall that unitary means an onto isometry. As the Fourier transform is an isometry, its range is closed and thus we just need to check that the range is dense. Well, {\mathcal{S}({\mathbb R}^d)} is in the range, which is dense. \Box

Lemma (Riemann–Lebesgue) If {f \in L^1}, then

  1. {|| \hat f ||_{L^\infty} \leq ||f||_{L^1}}.
  2. {\hat f} is uniformly continuous.
  3. {\hat f(\xi) \rightarrow 0} as {|\xi| \rightarrow \infty}.

Proof: All the assertions are true for {f \in \mathcal{S}({\mathbb R}^d)}. By the (obvious) inequality {||\hat f ||_{L^\infty} \leq ||f||_{L^1}}, these assertions are stable under taking limits. \Box


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