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?

  • 2
    $\begingroup$ There are connections between martingales and Fourier/harmonic analysis but not in the (tempting) way that you are attempting to sketch. I think most people who've studied some probablity and some fourier analysis wonder at some point if you can get the Carleson-Hunt theorem out of an L^p-martingale convergence theorem but to my knowledge no one has been able to make this work. (Your attempted martingale isn't a martingale) $\endgroup$ – Yemon Choi Sep 20 '18 at 16:30
  • $\begingroup$ How do you define a martingale out of a fourier series? Is there any reference or paper I can study, to get an idea how people have attempted it? $\endgroup$ – Landon Carter Sep 20 '18 at 16:35
  • $\begingroup$ Why the downvote? $\endgroup$ – Landon Carter Sep 20 '18 at 16:36
  • 1
    $\begingroup$ Landon, the connection is not through the trigonometric system, but usually through ideas like dyadic decomposition of an interval. Try typing "martingales harmonic analysis" into a search engine and moving from there $\endgroup$ – Yemon Choi Sep 20 '18 at 16:41
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    $\begingroup$ If you identify the unit interval with $(\mathbb{Z} / 2 \mathbb{Z})^\infty$, then the dyadic martingales coincide with the "Fourier series" of the associated over the associated dual group $\bigoplus^\infty \mathbb{Z} / 2 \mathbb{Z}$. In the other direction, results that are true for martingales are generally true for Fourier series that are "lacunary" i.e.: whose frequencies grow exponentially fast. $\endgroup$ – Adrián González-Pérez Oct 2 '18 at 11:20

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.

  • $\begingroup$ Thank you! I can see parallels between many of the arguments we use in Fourier analysis and martingales in Stroock's book. For instance, bounding the Hardy Littlewood maximal function in Lp norm by the same constant, (p/(p-1)) times the Lp norm of f... $\endgroup$ – Landon Carter Sep 23 '18 at 4:56

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

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 .)

  • $\begingroup$ You know of something in english? $\endgroup$ – user123124 Jul 21 '20 at 5:29
  • $\begingroup$ @user1: sorry, off the top of my head I don't. $\endgroup$ – Willie Wong Jul 21 '20 at 14:45

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. $U$.

  • $\begingroup$ There is no way to translate mere orthogonality into independence or even into a martingale condition. True, but on the other hand, there is a connection worth knowing: mathoverflow.net/a/16517/1044 $\endgroup$ – Mark Meckes Sep 27 '18 at 16:12
  • $\begingroup$ @MarkMeckes : Thank you for your comment. Of course, if arbitrary functions of the r.v.'s are allowed, then one can express the independence of r.v.'s $X$ and $Y$ as the orthogonality of the centered r.v.'s $f(X)$ and $g(Y)$ for all Borel-measurable $f$ and $g$ such that $f(X)$ and $g(Y)$ are in $L^2$. In fact, it is enough here to take $f$ and $g$ to be the Borel-measurable indicators. However, I don't think Fourier analysis extends even to such indicators of the functions $e^{ik\cdot}$. $\endgroup$ – Iosif Pinelis Sep 27 '18 at 20:25

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