Analysis Problem of the Day 45

Today’s problem comes all the way from the UCLA Fall 2001 Analysis Qual:

Problem 4: a) State how the Fourier transform is defined for a function $f \in L^2(\mathbb{R}).$

b) A deep fact proved by Carleson and Hunt is that the maximal function

$(Mf)(\xi) := \sup_{r>0} \left|\int_{-r}^r f(x) e^{-2\pi i x \xi} dx\right|$

satisfies

$\|Mf\|_2 \leq C\|f\|_2$

for some $C>0.$ Show that this implies the Carleson-Hunt theorem, namely, that

$\widehat{f}(\xi) = \lim_{r \to \infty} \int_{-r}^r f(x) e^{-2\pi i x \xi} dx$

for almost every $\xi.$

Solution: a) The Fourier transform is certainly defined for $L^1$ functions, and therefore for the class $\mathcal{S}$ of Schwartz functions (that is, smooth functions decaying faster than any polynomial at infinity). It is also a standard fact that $\mathcal{S}$ is dense in $L^2.$ Finally, one has Plancherel’s theorem, which says that $\|f\|_2 = \|\widehat{f}\|_2$ for any $f \in \mathcal{S}.$ Thus, for any $f \in L^2, f_n \to f$ in $L^2, f_n \in \mathcal{S},$ since $f_n$ is Cauchy in $L^2,$ by Plancherel, so is $\widehat{f_n},$ and since $L^2$ is a Banach space, one may define

$\widehat{f}:= \lim_{n \to \infty}^{L^2} \widehat{f_n}.$

Notice that this definition is unique since for any other sequence $g_n \to f$ in $L^2, \|\widehat{g_n}-\widehat{f_n}\|_2 = \|g_n-f_n\|_2 \to 0.$ In fact, this also proves Plancherel’s theorem on all of $L^2.$

b) Note that this problem is asking about a.e. convergence, which is generally very difficult to show. In fact, the only standard result regarding a.e. convergence is concerned with the Hardy-Littlewood maximal function. Motivated by this, we try to follows a similar approach. Notice that $Mf(\xi) = \sup_{r>0}|\widehat{f \chi_{[-r,r]}}(\xi)|.$ In particular, for $f$ compactly supported and large $r,$ one has $\widehat{f \chi_{[-r,r]}} = \widehat{f}.$ Since $C_c^\infty$ is dense in $L^2,$ for $g \in C_c^\infty, \|g-f\|_2 <\epsilon,$ we may thus approximate

$|f(\xi)-\widehat{f \chi_{[-r,r]}}| \leq |f(\xi)-g(\xi)| + |\widehat{(f-g)} \chi_{[-r,r]}|,$

where the third term in the triangle inequality vanishes for large enough $r$ by the remark above. In particular,

$\mu(|f(\xi)-\widehat{f \chi_{[-r,r]}}|>\epsilon) \leq \mu(|f(\xi)-g(\xi)| > \frac{\epsilon}{2})+\mu(|\widehat{(f-g) \chi_{[-r,r]}}|>\frac{\epsilon}{2}).$

Now, notice that as $\epsilon \to 0,$ the first term on the right becomes small by Chebyshev’s inequality since $\|f-g\|_2 \to 0,$ and the second term is bounded uniformly in $r$ by $\|f-g\|_2$ by Chebyshev as well. Thus, $\mu(|f(\xi)-\widehat{f \chi_{[-r,r]}}|>\epsilon) \leq C\epsilon$ for large enough $r$ which implies the desired result.

Remark: This result is equivalent to a.e. convergence of Fourier series on $L^2(\mathbb{T}).$

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