In this post, we shall outline a brief introduction to Fourier analysis.

${L^p(\mathbb R)}$ (henceforth denoted by ${L^p}$) is the set of all square-integrable functions, i.e.,

$\displaystyle L^p = \{ f : \mathbb R \rightarrow \mathbb R \vert \int_{\mathbb R} |f(x)|^p \;\text{d} x<\infty \}.$

Note that ${L^p}$ functions are defined upto a measure zero set, so are not functions but equivalence classes of functions that are equal except on a measure-zero set.

For every ${L^1}$ (i.e., integrable) function, we define the Fourier transform operator by

$\displaystyle \mathcal F : L^1 \rightarrow \mathcal C_0(\mathbb R),$

$\displaystyle f(x) \longmapsto \widehat f(\xi) := \int_{\mathbb R} f(x) \exp{(2 \pi i x \xi)} \;\text{d} x .$
It is clear the integral is well-defined a.e. ${x\in\mathbb R}$, in fact,

$\displaystyle \|\mathcal F(f)\|_\infty \leq \|f\|_1.$
However, that ${\mathcal F(f)}$ eventually decays to zero as ${|x|\rightarrow\infty}$ is not so obvious and the result and goes by the name of “Riemann-Lebesgue Lemma”.

Our objective is to define ${\mathcal F}$ on ${L^2}$ functions, because ${L^2}$ is special and we can hope that ${\mathcal F}$ is an operator on ${L^2}$. How nice it would be, if ${\mathcal F}$ turned out to be a bijection, or better an isometry on ${L^2}$!

Just like the Fourier transform ${f \mapsto \widehat f}$, we can define the Inverse Fourier transform, ${f \mapsto \mathcal G(f) = \check f}$ in the obvious way:

$\displaystyle \mathcal G (f) = \check f(x) = \int_{\mathbb R} f(\xi) \exp{2 \pi i \xi x} \;\text{d} x.$

Unfortunately, its not true that ${\mathcal G(f) \in L^1}$ whenever ${f}$ is. However, if ${f}$ and ${\widehat f}$ are both ${L^1}$ functions, then

$\displaystyle \mathcal G\circ \mathcal F (f) = -f \qquad \text{a.e. } \mathbb R.$
Moreover, ${f}$ is continuous except possibly at a measure-zero set.

The proof that ${\mathcal G \circ \mathcal F (f) = -f}$ does not follow directly from Fubini, since

$\displaystyle \mathcal G \circ \mathcal F (f) = \int_{\mathbb R}\int_{\mathbb R} f(y) \exp{-2 \pi i \xi y} \exp{2 \pi i \xi x} \;\text{d}y \;\text{d}\xi$
need not be integrable. The trick is to bring in some good ${L^1}$ function ${g}$ whose FT and Inverse-FT exist and are equal and use the “change-of-hat” trick:

$\displaystyle \int \hat f g = \int f \hat g, \qquad \forall f, g \in L^1.$
If ${g = \widehat g}$ and ${f}$ are all in ${L^1}$, then it follows that ${\widehat f}$ is in ${L^1}$. The function ${g}$ has done its work and we can throw it away! But notice that the integrable function ${g}$ whose FT is itself is, upto a scalar, the Gaussian ${e^{-\pi x^2}}$!

We now have a big tool with us, namely the Fourier Inversion theorem:

Theorem 1 If ${f, \widehat f\in L^1}$ then ${\check{\hat{f}} = -f }$ a.e., i.e. ${\mathcal G\circ \mathcal F(f) = -f}$ a.e. and ${f}$ is ${\mathcal C^0}$ a.e. ${\mathbb R}$.

We now use the fact that ${\mathcal C^\infty_0}$ functions (smooth functions decreasing to zero) are dense in ${L^p}$ for ${p <\infty}$ to define the FT on ${L^2}$.

Theorem 2 If ${f}$ is smooth enough, then its FT ${\widehat f}$ is also smooth of that order.

In proving this theorem, we use a fundamental property of FTs viz, it converts derivatives into polynomials:

$\displaystyle f \in \mathcal C^k \cap L^1\;\; \text{and} \;\; f^{(\alpha)} \in \mathcal C_0 \Rightarrow \widehat{f^{(\alpha)}}(\xi) = (2 \pi i \xi)^\alpha \widehat f(\xi)\quad \text{for } \alpha \leq k-1.$

Now given an ${f}$ in ${L^2}$, approximate it by functions ${f_n}$ in ${L^1 \cap L^2 \cap \mathcal C^\infty_0}$. Since ${\widehat{f_n} \in L^1 \cap \mathcal C^\infty_0}$, by completeness of ${L^2}$, they will converge to some function. Call it ${\widehat f}$. Well-definedness of ${\widehat f}$ is left to standard texts. But now, here comes the result we had secretly hoped for:

Theorem 3 (Plancherel) If ${f \in L^1 \cap L^2}$ then ${\widehat f\in L^2}$ and the Fourier transform ${\mathcal F}$ is an isometry (bijective, continuous, norm-preserving, homeomorphism ${\cdots}$) onto its image and moreover,

${ \mathcal F|_{L^1 \cap L^2} }$ extends uniquely to ${L^2}$.

This is fantastic, because the uniqueness assures us that if two square-integrable functions share the same FT, then they must coincide almost everywhere! Only the little theory developed so far can help us to solve some PDEs using Fourier Transforms. I discuss solving the transport equation using FTs in this post. Although I have to yet understand precisely how, I have learnt that the techniques of Fourier analysis help in solving problems in number theory. The ${p}$-adic fields ${\mathbb Q_p}$ and the real field and their finite extensions are locally compact so we have a unique Haar measure that allows us to define Fourier transforms on them. I look forward to reading Ramakrishnan & Valenza’s Fourier Analysis on Number Fields soon.

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Taipiece: Prof. Athavale had taught this in a course at IIT Bombay. But I didn’t quite follow it then, since my understanding of real analysis was not so developed then. I once wrote an amusing post on him.