Connections between martingales and Fourier analysis

Question

I have had this strange feeling recently that somehow, the theory of martingales we study in probability, and the theory of Fourier analysis are very alike. But I am not able to formalize my thoughts.

To illustrate, let us focus on $f\in L_1(\mathbb T,\mathcal B_{\mathbb T},Leb)$. Define $S_N(f)(x)=\sum_{k=-N}^N \hat{f}(k)e^{ikx}$, the $N$-th partial sum of the Fourier series of $f$. It seems to me that $\{S_N(f)\}_N$ is essentially like a random walk, as the increment $\hat{f}(N)e^{iNx}+\hat{f}(-N)e^{-iNx}$ at the $N$-th stage has mean $0$, and it is orthogonal to $\{1,e^{\pm x},...,e^{\pm (N-1)x}\}$, which would perhaps translate to independence and thus it seems plausible that $\{S_N(f)\}_N$ is a martingale. Of course, independence, conditional expectations, etc. are probably not meaningful terms in classical analysis, and this is one thing I am unable to formalize.

Now let me come to the convergence results. It is well known that in martingale theory, for $p>1$, an $L_p$-bounded martingale converges a.s. and in $L_p$. In Fourier analysis, we have the result that $L_p$ norm convergence of $S_N(f)$ to $f$ (if $f\in L_p$) is equivalent to $\sup_N ||S_N||<\infty$ where $S_N:L_p\to L_p$ is being treated as a linear operator. So here we have the analogy with $L_p$ bounded martingales. Further, for $p>1$, we have $S_N(f)\to f$ a.e., again in analogy.

The results for both Fourier series and martingales fail when $p=1$ and you need more conditions like uniform integrability of martingales, which, I think, translates to $\sum_n |\hat{f}(n)|<\infty$. I also do not know what happens to this little jump from 1 to $p$ that makes both these two results work/fail. I can feel these two theories go parallelly but it seems quite mysterious to me.

Maybe there is a connection which I cannot see?

Answer 1

There are links, for example in the theory of singular integrals, see section 6.2 in Stroock's book on probability theory. Also, googling "BMO and martingales" will give you information in the direction you look for.

Answer 2

To add to what Yemon wrote in his comment, and to give a concrete reference: there is some known connection giving Martingales based proofs of Littlewood-Paley type inequalities. This is discussed, for example, in

PAUL-ANDRÉ MEYER
Démonstration probabiliste de certaines inégalités de Littlewood-Paley
Séminaire de probabilités (Strasbourg), tome 10 (1976)

(There are a total of five exposes, all available on Numdam. The first is http://www.numdam.org/item?id=SPS_1976__10__125_0 .)

Answer 3

There is no way to translate mere orthogonality into independence or even into a martingale condition.

Indeed, the independence of real-valued random variables (r.v.'s) $X_1,\dots,X_N$ is a very strong condition, involving a continuum equations, say $P(X_1\le x_1,\dots,X_N\le x_N)=P(X_1\le x_1)\cdots P(X_N\le x_N)$ for all $(x_1,\dots,x_N)\in\mathbb R^N$.

The condition that $X_1,\dots,X_N$ are martingale differences is less restrictive than the independence (given that the $X_i$'s are zero-mean), but it still involves a continuum equations, including (say) $E(X_i|X_{i-1}=x_{i-1})=0$ for all $i=2,\dots,N$ and all $(x_1,\dots,x_{N-1})\in\mathbb R^{N-1}$.

On the other hand, the orthogonality of $X_1,\dots,X_N$ is a much, much weaker condition, involving only finitely many (namely, $\binom N2$) equations.

On the other hand, the r.v.'s $X_k:=e^{ikU}$, where $k=1,2,\dots$ and $U$ is a r.v. uniformly distributed on $[0,2\pi)$, which are implicitly used in Fourier analysis, are not merely mutually orthogonal. In particular, here we have the much stronger condition $E\prod_{j=1}^N X_{k_j}^{p_j}=0$ whenever $\sum_{j=1}^N {p_j}{k_j}\ne0$, which is somewhat close to the stronger condition $E\prod_{j=1}^N Y_{k_j}^{p_j}=0$ whenever $\sum_{j=1}^N ({p_j}{k_j})^2\ne0$, where the $p_k$'s and $k_j$'s are integers, $Y_k:=e^{ikU_k}$, and the $U_k$'s are independent copies of the r.v. $U$.