Notes on Chapter 9 (Fourier Transforms) of Walter Rudin's Real and Complex Analysis. This is the last chapter of the real analysis portion of the book! meery crimskats!

Basic Properties

In this chapter we adopt the convention that \(m\) is the Lebesgue measure divided by \(\sqrt{2\pi}\), and \(dx\) or \(dy\) will refer to the ordinary Lebesgue measure. Define:

$$\|f\|_p = \left\{\int_{-\infty}^\infty |f(x)|^p\ dm(x)\right\}^{1/p}\qquad (1\leq p < \infty)$$ $$(f*g)(x)=\int_{-\infty}^\infty f(x-y)g(y)\ dm(y)\qquad (x\in \mathbb{R})$$ $$\hat{f}(t) = \int_{-\infty}^\infty f(x)e^{-ixt}\ dm(x)\qquad (t\in\mathbb{R}).$$

We will write \(L^p\) for \(L^p(\mathbb{R})\) and \(C_0\) will denote the space of continuous functions on \(\mathbb{R}\) that vanish at infinity. See the notes on chapter 3 for more information.

We list the basic properties of the Fourier transform without proof.

Basic properties: Suppose \(f\in L^1\), and \(\alpha\) and \(\lambda\) are real numbers.

  • If \(g(x)=f(x)e^{i\alpha x}\), then \(\hat{g}(t)=\hat{f}(t-\alpha)\).
  • If \(g(x)=f(x-\alpha)\), then \(\hat{g}(t)=\hat{f}(t)e^{-i\alpha t}\).
  • If \(g\in L^1\) and \(h=f*g\), then \(\hat{h}(t)=\hat{f}(t)\hat{g}(t)\).
  • If \(g(x) = \overline{f(-x)}\), then \(\hat{g}(t)=\overline{\hat{f}(t)}\).
  • If \(g(x) = f(x/\lambda)\) and \(\lambda > 0\), then \(\hat{g}(t)=\lambda\hat{f}(\lambda t)\).
  • If \(g(x)=-ixf(x)\) and \(g\in L^1\), then \(\hat{f}\) is differentiable and \(\hat{f}'(t)=\hat{g}(t)\). □

Inversion and Uniqueness Theorems

Our next objective are the inversion and uniqueness theorems, which arises from a series of lemmas which we state without proof:

Translates: For any function \(f\) on \(\mathbb{R}\) and every \(y\in\mathbb{R}\), let \(f_y\) be the translate of \(f\) defined by \(f_y(x) = f(x-y)\). If \(1\leq p < \infty\) and if \(f\in L^p\), the mapping \(y\mapsto f_y\) is a uniformly continuous mapping of \(\mathbb{R}\) into \(L^p\). □

Bounding \(\hat{f}\): If \(f\in L^1\), then \(\hat{f}\in C_0\) and \(\|\hat{f}\|_\infty\leq \|f\|_1\). □

We introduce a couple of functions that will be involved in the proof of the inversion theorem (and are also connected to harmonic functions). Define $$H(t) = e^{-|t|}$$ and define its Fourier transform $$h_\lambda(x) = \int_{-\infty}^\infty H(\lambda t)e^{itx}\ dm(t)\qquad (\lambda > 0).$$

We observe that $$h_\lambda(x)=\sqrt{\frac{2}{\pi}}\frac{\lambda}{\lambda^2+x^2}$$ and furthermore, $$\int_{-\infty}^\infty h_\lambda(x)\ dm(x) = 1.$$

Note that also \(0<H(t)\leq 1\) and \(H(\lambda t)\rightarrow 1\) as \(\lambda \rightarrow 0\).

Properties of \(H\) and \(h\):

  1. If \(f\in L^1\), then $$(f*h_\lambda)(x)=\int_{-\infty}^\infty H(\lambda t)\hat{f}(t)e^{ixt}\ dm(t).$$
  2. If \(g\in L^\infty\) and \(g\) is continuous at a point \(x\), then $$\lim_{\lambda\rightarrow 0} (g*h_\lambda)(x) = g(x).$$
  3. If \(1 \leq p < \infty\) and \(f\in L^p\), then $$\lim_{\lambda\rightarrow 0} \|f*h_\lambda - f\|_p = 0.$$ □

We can now show the desired theorems:

Inversion theorem: If \(f\in L^1\) and \(\hat{f}\in L^1\), and if $$g(x)=\int_{-\infty}^\infty \hat{f}(t)e^{ixt}\ dm(t)\qquad (x\in\mathbb{R}),$$ then \(g\in C_0\) and \(f(x)=g(x)\) a.e. □

That \(g\) is in \(C_0\) arises from the bounding \(\hat{f}\) theorem. We apply property 1 above, and take the limit on both sides as \(\lambda\rightarrow 0\). The RHS converges to \(g(x)\) for every \(x\in\mathbb{R}\) by the dominated convergence theorem. The LHS converges to \(f\) in the \(L^p\) norm, hence there is a subsequence of \(\lambda_n\) which such that the LHS converges (in the regular sense) to \(f(x)\) a.e. Hence \(f(x)=g(x)\) a.e.

We get the uniqueness theorem immediately:

Uniqueness theorem: If \(f\in L^1\) and \(\hat{f}=0\) for all \(t\in\mathbb{R}\), then \(f(x) = 0\) a.e. □

The Plancherel Theorem

On \(\mathbb{R}\) (as opposed to \(T\), the unit circle which we studied before), \(L^2\) is not a subset of \(L^1\) as the measure of \(\mathbb{R}\) is infinite (for each \(f\in L^2\) consider \(f \geq 1\) and \(f < 1\)). Hence our definition of the Fourier transform (for \(L^1\) functions) is not immediately applicable to \(L^2\).

However, if \(f\in L^1\cap L^2\), this definition does apply, and it turns out that this is an isometry of \(L^1\cap L^2\) into \(L^2\), which extends to an isometry of \(L^2\) onto \(L^2\). This defines the Fourier transform on \(L^2\):

Fourier/Plancherel Transform: For each \(f\in L^2\) there is a \(\hat{f}\in L^2\) s.t. the following hold:

  • If \(f\in L^1\cap L^2\), then \(\hat{f}\) is the previously defined Fourier transform of \(f\).
  • For every \(f\in L^2\), \(\|\hat{f}\|_2 = \|f\|_2\).
  • The mapping \(f\mapsto \hat{f}\) is a Hilbert space isomorphism of \(L^2\) onto itself.
  • The following is true: If $$\varphi_A(t) = \int_{-A}^A f(x)e^{-ixt}\ dm(x),\qquad \psi_A(x) = \int_{-A}^A \hat{f}(t) e^{ixt}\ dm(x),$$ then \(\|\varphi_A - \hat{f}\|_2 \rightarrow 0\) and \(\|\psi_A - f\|_2 \rightarrow 0\) as \(A\rightarrow\infty\). □

As a direct corollary of the last point, we have the inversion theorem:

Inversion theorem for L^2: If \(f\in L^2\) and \(\hat{f}\in L^1\),then $$f(x)=\int_{-\infty}^\infty \hat{f}(t)e^{ixt}\ dm(t)\qquad (x\in\mathbb{R})\qquad a.e.$$ □

Note that if \(f\in L^1\), then the Fourier transform defines \(\hat{f}\) unambiguously for every \(t\). However, if \(f\in L^2\), the Plancherel theorem defines \(\hat{f}\) uniquely as an element of the Hilbert space \(L^2\), but only determined a.e. as a point function, which can cause difficulties.

The Banach Algebra \(L^1\)

A Banach space \(A\) is a Banach algebra if in addition there is a multiplication defined in \(A\) which satisfies the inequality \(\|xy\|\leq \|x\|\|y\|\), is associative, distributes over addition, and \((\alpha x)y = x(\alpha y) = \alpha(xy)\) for any scalar \(\alpha\).

In particular, \(C_0\) is a commutative Banach algebra without unit under pointwise multiplication of functions, and \(L^1\) is a Banach algebra with multiplication defined by convolution, as \(\|f*g\| \leq \|f\|_1 \|g\|_1\), associativity holds (seen via Fubini or by taking Fourier transforms), and the other requirements are simple to observe. Furthermore, convolution is commutative, so \(L^1\) is a commutative Banach algebra.

Note that the Fourier transform is an algebra isomorphism of \(L^1\) into \(C_0\) (as it takes convolutions to products), hence \(L^1\) has no unit.

Given a Banach algebra \(A\), we can consider the homomorphisms of \(A\) into the complex field, which are precisely the complex linear functionals \(\varphi\) which also preserve multiplication. Interestingly enough, no boundedness assumption is needed on \(\varphi\):

Complex homomorphisms are short: If \(\varphi\) is a complex homomorphism on a Banach algebra \(A\), then the norm \(\|\varphi\|\leq 1\). □

The proof is cute, so I will write it out in full. Suppose \(\varphi(x_0) = \lambda\) with \(|\lambda| > \|x_0\|\) for some \(x_0\in A\). Set \(x = x_0/\lambda\). Then \(\|x\| < 1\) and \(\varphi(x) = 1\).

Since \(\|x^n\| \leq \|x\|^n\) and \(\|x\| < 1\), the elements \(s_n = -x-x^2-\cdots-x^n\) form a Cauchy sequence in \(A\). Since \(A\) is complete, there exists a limit \(y\). As \(x + s_n = xs_{n-1}\), taking limits gives \(x + y = xy\).

However, this means that \(\varphi(x) + \varphi(y) = \varphi(x)\varphi(y)\), which is impossible since \(\varphi(x) = 1\).

Finally, we can prove the following characterization:

Complex homomorphisms on \(L^1\): Let \(\varphi\) be a non-zero complex homomorphism on \(L^1\). There is a unique \(t\in\mathbb{R}^1\) s.t. \(\varphi(f) = \hat{f}(t)\) for all \(f\in L^1\). □

We begin with the relation $$\varphi(f*g)=\varphi(f)\varphi(g) \qquad (f,g\in L^1).$$

By \(L^p\) duality, there is a (unique) \(\beta\in L^\infty\) s.t. $$\varphi(f) = \int_{-\infty}^\infty f(x)\beta(x)\ dm(x)\qquad (f\in L^1).$$

By using these two relations and interchanging order of integration we get $$\varphi(f)\int_{-\infty}^\infty g(y)\beta(y)\ dm(y) = \varphi(f)\varphi(g) = \varphi(f*g) = \int_{-\infty}^\infty g(y)\varphi(f_y)\ dm(y)$$ where \(f_y\) is the translate \(f_y(x) = f(x-y)\).

Fix an \(f\in L^1\) s.t. \(\varphi(f)\neq 0\). The uniqueness assertion of \(L^p\) duality implies that $$\varphi(f)\beta(y) = \varphi(f_y)\qquad a.e.$$

We have that \(y\mapsto f_y\) is a continuous mapping of \(\mathbb{R}\) into \(L^1\) (as seen earlier) and \(\varphi\) is continuous on \(L^1\), so the RHS is continuous. Hence we can assume that \(\beta\) is continuous (changing it on a set of measure 0 if necessary).

We then have $$\varphi(f)\beta(x+y) = \varphi(f_{x+y}) = \varphi((f_x)_y) = \varphi(f)\beta(x)\beta(y),$$ so that $$\beta(x+y) = \beta(x)\beta(y)\qquad (x, y\in\mathbb{R}).$$

One can show that \(\beta\) is differentiable and differentiating and solving shows that \(\beta(x) = e^{-itx}\) for some \(t\in\mathbb{R}\), hence giving us the Fourier transform as desired.

Uniqueness follows from showing that if \(t\neq s\) then there is an \(f\in L^1\) s.t. \(\hat{f}(t)\neq\hat{f}(s)\).