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This post is just a collection of basic results I have compiled for referring to in desperate times Nothing too deep except Haar-von Neumann’s theorem.

A measure $\mu$ on a locally compact (Hausdorff, always) group $G$ is left-invariant if

$\boxed{ \int_G f(s^{-1}x) \text{d} \mu(x) = \int_G f(x) \text{d} \mu(x). }$

The most important theorem in this topic is the existence and uniqueness of the Haar measure (proven by Haar and von Neumann, respectively).

Theorem 1: On every locally compact group $G$ there is a unique (up to a positive constant of proportionality) left-invariant positive measure $\mu \neq 0$.

Proposition 2: $G, \mu$ as usual. For $G$ to have a finite measure, it is necessary and sufficient that $G$ be compact!

Proposition 3: There is a continuous group-homomorphism (called right modulus), $\triangle_r : G \to \mathbb R^*_+$ such that

$f(xs^{-1}) \text{d} \mu(x) = \triangle_r(s) \int f(x) \text{d} \mu(x).$

Proposition 4: Let $\mu$ (respectively, $\nu$) be a left-invariant Haar measure on locally compact groups $H$ (resp., $K$). Then the product integral on $G = H \times K$ is left-invariant and

$\boxed{\triangle^G_r(s,t) = \triangle^H_r(s) . \triangle^K_r(t).}$

Corollary 5: $G$ is unimodular (i.e., $\triangle_r =1$) precisely when $H$ and $K$ are so.

The most important example (for me) is the general linear group over number fields or $p$-adics and its subgroups. All semisimple (and more generally reductive) groups are unimodular. Compact groups are unimodular. Abelian groups are trivially so. However the measure on Borel (and parabolic) subgroups (i.e., upper triangular matrices) is not unimodular. The above proposition 4 allows one to transfer the Levi decomposition on the groups to their measures.

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.

Abhishek Parab

I? An Indian. A mathematics student. A former engineer. A rubik's cube addict. A nature photographer. A Pink Floyd fan. An ardent lover of Chess & Counter-Strike.

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