It is a famous result of Kolmogorov that there exists a (Lebesgue) integrable function on the torus such that the partial sums of Fourier series of $f$ diverge almost everywhere (a.e.). More specifically, he exhibited an $f\in L^{1}(\mathbb{T})$ such that

\begin{align*} \sup_{N\geq 1}\left|S_{N}f(x)\right|=\sup_{N\geq 1}\left|(f\ast D_{N})(x)\right|=\infty, \qquad\forall \text{ a.e. } x\in\mathbb{T}, \end{align*} where $S_{N}$ is the $N^{th}$ partial sum and $D_{N}$ is the $N^{th}$ Dirichlet kernel given below. \begin{align*} S_{N}f(x):=\sum_{\left|n\right|\leq N}\widehat{f}(n)e^{2\pi inx}, \quad D_{N}(x):=\dfrac{\sin 2\pi(N+\frac{1}{2})x}{\sin \pi x} \end{align*} and we identify $\mathbb{T}$ with the unit interval $[0,1]$.

I have read, for example pg. 118 in [Pinsky], that Kolmogorov's counterexample can be replicated in the context of the Fourier transform on the real line $\mathbb{R}$, showing that $L^{1}$ pointwise Fourier inversion can fail quite horribly. If my understanding is correct, then the following claim is true:

Claim. There exists a function $f\in L^{1}(\mathbb{R})$ such that \begin{align*} S_{R}f(x):=\int_{\left|\xi\right|\leq R}\widehat{f}(\xi)e^{2\pi i\xi x}\mathrm{d}\xi=\int_{\mathbb{R}}f(y)\dfrac{\sin 2\pi R(x-y)}{\pi (x-y)}\mathrm{d}y\not\rightarrow f(x), \qquad\forall\text{ a.e. }x\in\mathbb{R} \end{align*} as $R\rightarrow\infty$.

I have not seen a proof of this claim in the literature. Is there a way to directly take Kolmogorov's counterexample for the Fourier series and turn it into a counterexample for the Fourier transform? Or does one have to go back to the original proof and make the appropriate changes for the different setting? I haven't tried the latter yet, but my efforts at quicker, "details under the rug" approach have so far been unsuccessful.

I believe that it follows from the weak type (1,1) bound for the Hilbert transform that the partial sums of the Fourier integrals converge in measure to $f\in L^{1}(\mathbb{R})$ as $R\rightarrow\infty$. Whence, there exists a subsequence $\left\{R_{k}\right\}$ tending to $\infty$ such that $S_{R_{k}}f(x)\rightarrow f(x)$ a.e., as $R\rightarrow\infty$. So it cannot be true that

\begin{align*} \lim_{R\rightarrow\infty}\left|S_{R}f(x)\right|=\infty, \qquad\forall \text{ a.e. }x\in\mathbb{R} \end{align*}

But I don't see how this observation helps me in my original task.

[Kolmogorov] A.N. Kolmogorov, "Une série de Fourier-Lebesgue divergente presque partout," Fundamenta mathematicae 4.1 (1923), 324-328.

[Pinsky] M.A. Pinsky, An Introduction to Fourier analysis and Wavelets, Pacific Grove: Brooks/Cole, 2002.

  • $\begingroup$ Did K actually show that $|S_N|\to\infty$ almost everywhere? I thought it was just that $|S_N|$ was unbounded (actually tending to infinity seems implausible). No doubt the reason you've never seen a proof for the line is that if all else fails we could simply repeat the proof for the circle with simple modifications, and nobody's going to get irl reputation points for that. Does the result for the line follow from the result for the circle (as opposed to repeating the proof)? I believe so, but right now the details escape me, as they sometimes do. $\endgroup$ – David C. Ullrich Jun 25 '15 at 17:58
  • $\begingroup$ @DavidC.Ullrich My apologies for belatedly responding. Skimming the intro of K's paper, it appears that he just shows $\sup_{N\geq 1}\left|S_{N}(x)\right|=\infty$ for a.e. $x$. This would be consistent with the exposition (L. Grafakos, Classical Fourier Analysis), from which I originally learned K's result. Actually, I don't think it can be true that $\left|S_{N}(x)\right|\rightarrow\infty$ for a.e. $x$. That would imply that the partial sums of the Fourier integrals of your example tended to $\infty$, which can't be true by the weak type (1,1) bound of the Hilbert transform. $\endgroup$ – Matt Rosenzweig Jun 26 '15 at 20:01
  • $\begingroup$ Thanks. I thought for a minute that that weak-type boundedness for the Hilbert transform made $|S_n|\to\infty$ impossible, but I don't see exactly how it goes. ??? (An explanation on the circle or the line would be fine...) I see how we get a bound on the measure of the set where $|S_N|>\lambda$, but not for the set where $M>\lambda$, $M$ being the sup of the partial sums/integrals... $\endgroup$ – David C. Ullrich Jun 26 '15 at 20:08
  • $\begingroup$ I was being dense. Yes, the weak-type bound on the Hilbert transform implies convergence in measure. (Details on request...) $\endgroup$ – David C. Ullrich Jun 27 '15 at 17:02

Ok. Note that $2\pi=1$ here...

Say $f$ is Kolomogorov's example. Regard $f$ as a periodic function defined on the entire line. Let $c_n$ be the $n$-th Fourier coefficient of $f$ (we want to reserve \hat for the Fourier transform).

Say $\psi\ne 0$ is a smooth function with support in $(-1/2,1/2)$. Say $\psi$ is even just so we don't have to worry about the FT versus the inverse FT. There is a Schwarz function $\phi$ with $\psi=\hat\phi$.

Now define $g$ by $g(n+t)=c_n\psi(t)$ for $n\in\mathbb Z$ and $|t|<1/2$, $g(\xi)=0$ for other $\xi$.

Note that $f\phi\in L^1(\mathbb R)$; this is going to be the bad $L^1$ function. You can show without too much trouble that in fact $$g=\widehat{f\phi}.$$

Now consider $\int_{-(N+1/2)}^{N+1/2}g(\xi)e^{ix\xi}\,d\xi$. If we show that this is unbounded (as $N\to\infty$) for almost every $x$ we're done. But $$\int_{-(N+1/2)}^{N+1/2}g(\xi)e^{ix\xi}\,d\xi=\sum_{-N}^Nc_n\int_{-1/2}^{1/2}\psi(t)e^{ix(n+t)}\,dt=C(x)\sum_{-N}^Nc_ne^{ixn}=C(x)S_N(x),$$which is unbounded at any point where $S_N$ is unbounded and $C(x)\ne0$. Here $C(x)=\int_{-1/2}^{1/2}\psi(t)e^{ixt}=\phi(x)$; hence $C(x)=0$ on at most a countable set, since $\phi$ is the restriction to $\mathbb R$ of an entire function. QED

Detail: Why is $\widehat{f\phi}=g$? Seems clear, took me a minute to give an actual proof.

Let $\epsilon>0$. Choose a trigonometric polynomial $p$ with $\int_0^1|f-p|<\epsilon$. Then $\|p\phi-f\phi\|_1<c\epsilon$, so $$\|\widehat{f\phi}-\widehat{p\phi}\|_\infty<c\epsilon.$$

Now say $p(x)=\sum_{-\infty}^\infty a_ne^{int}$, where all but finitely many $a_n$ vanish. Then $|a_n-c_n|<\epsilon$ for all $n$. This shows that $$\|g-\sum_na_n\tau_n\hat\phi\|_\infty<c\epsilon,$$where $\tau_n$ denotes translation by $n$. Since $p$ is really a finite sum, $$\widehat{p\phi}=\sum_na_n\tau_n\hat\phi.$$So $\|g-\widehat{f\phi}\|_\infty<c\epsilon$.

  • $\begingroup$ It is a result that any periodic distribution $T$ of period $h$ is tempered, and (underyour definition of the Fourier transform) $\widehat{T}$ is supported on the integer multiples of $2\pi/h$. If $T$ is a measure $\mu$ of period $2\pi$, then it follows from an application of the Poisson summation formula that $\widehat{\mu}=\sum_{n\in\mathbb{Z}}c_{n}\delta_{2\pi n}$, where $c_{n}$ are the Fourier coefficients of $\mu$. So in our case, $\mu=fdx$, where $f\in L^{1}(\mathbb{T})$. The rest of the computing $\widehat{f\phi}$ is just convolving $\psi=\widehat{\phi}$ with dirac masses. $\endgroup$ – Matt Rosenzweig Jun 26 '15 at 19:14
  • $\begingroup$ Well yes - that's why I said it seemed clear. Not knowing where you were at regarding all this I decided to add a more elementary justification. $\endgroup$ – David C. Ullrich Jun 26 '15 at 19:21
  • $\begingroup$ Btw you might answer the question I asked yesterday: Did K actually get $|S_N|\to\infty$ or just $|S_N|$ unbounded? (If the former then (i) I'm surprised, (ii) I should probably add a little bit to my answer, showing why the "partial integral" tends to infinity - what I posted just gives a subsequence tending to infinity.) $\endgroup$ – David C. Ullrich Jun 26 '15 at 19:23

This answer is intended to complement David's very nice solution by showing that the partial sums of the Fourier integrals are at most be pointwise unbounded almost everywhere (a.e.), when $f\in L^{1}(\mathbb{R})$; there exists a subsequence which converges to $t$ pointwise a.e. Combining this with David's solution, we see that the partial sums of the Fourier series are at most be pointwise unbounded a.e. This resolves an error I made in the original question by incorrectly claiming that Kolmogorov showed there exists an $f\in L^{1}(\mathbb{R})$ such that $\left|S_{N}f\right|\rightarrow\infty$ a.e.

The proof is a consequence of the weak type (1,1) boundedness of the Hilbert transform.

Lemma 1. The operators $S_{R}$ defined above are bounded $L^{1}(\mathbb{R})\rightarrow L^{1,\infty}(\mathbb{R})$, uniformly in $R>0$.

Proof. One can verify that for $f\in\mathcal{S}(\mathbb{R})$, \begin{align*} S_{R}f=(\chi_{[-R,R]}\widehat{f})^{\vee}=\dfrac{i}{2}\left(M_{-R}HM_{R}-M_{R}HM_{-R}\right)f, \end{align*} where $M_{a}$ is the operator defined by pointwise multiplication with $e^{2\pi i ax}$, for $a\in\mathbb{R}$. It it is evident that $M_{a}$ is a bounded operator $L^{p}\rightarrow L^{p}$, for $1\leq p\leq\infty$, with constant independent of $a$. For $f\in L^{1,\infty}(\mathbb{R})$ and $\lambda>0$ fixed, \begin{align*} \left|\left\{x\in\mathbb{R} : \left|M_{a}f(x)\right|>\lambda\right\}\right|=\left|\left\{x\in\mathbb{R} : \left|f(x)\right|>\lambda\right\}\right|\leq\dfrac{\left\|f\right\|_{L^{1,\infty}}}{\lambda}, \end{align*} whence $M_{a}$ is a bounded operator $L^{1,\infty}\rightarrow L^{1,\infty}$ with constant independent of $a$.

I claim that $M_{a}HM_{-a}$ is uniformly weak type (1,1) bounded. For $\lambda>0$, \begin{align*} \left|\left\{\left|M_{a}HM_{-a}f\right|>\lambda\right\}\right|&=\left|\left\{\left|HM_{-a}f\right|>\lambda\right\}\right|\\ &\lesssim\dfrac{\left\|M_{-a}f\right\|_{L^{1}}}{\lambda}\\ &=\dfrac{\left\|f\right\|_{L^{1}}}{\lambda}, \end{align*} where the implied constant is independent of $a$. Since by the triangle inequality, \begin{align*} \left\{\left|S_{R}f\right|>\lambda\right\}\subset\left\{\left|M_{-R}HM_{R}f\right|>\lambda\right\}\cup\left\{\left|M_{R}HM_{-R}f\right|>\lambda\right\}, \end{align*} we conclude that $S_{R}$ is uniformly weak type $(1,1)$ bounded.

By the density of $\mathcal{S}(\mathbb{R})$ in $L^{1}(\mathbb{R})$, $S_{R}$ extends to a bounded operator $\widetilde{S_{R}}:L^{1}(\mathbb{R})\rightarrow L^{1,\infty}(\mathbb{R})$. To see that $\widetilde{S_{R}}f=S_{R}f$ a.e., when $f\in L^{1}(\mathbb{R})$, let $f_{n}\in\mathcal{S}(\mathbb{R})$ and $f_{n}\rightarrow f$ in $L^{1}(\mathbb{R})$, and observe that \begin{align*} \left|\left\{\left|S_{R}f-\widetilde{S_{R}}f\right|>\lambda\right\}\right|&\leq\limsup_{n\rightarrow\infty}\left|\left\{\left|S_{R}(f-f_{n})\right|>\dfrac{\lambda}{2}\right\}\right|+\left|\left\{\left|\widetilde{S_{R}}(f_{n}-f)\right|>\dfrac{\lambda}{2}\right\}\right|\\ &\lesssim\limsup_{n\rightarrow\infty}\dfrac{2\left\|f-f_{n}\right\|_{L^{1}}}{\lambda}\\ &=0, \end{align*} where we use that $\left\|S_{R}(f-f_{n})\right\|_{L^{\infty}}\leq R\left\|f-f_{n}\right\|_{L^{1}}\rightarrow 0, n\rightarrow\infty$. Since $\lambda>0$ was arbitrary, we conclude that $S_{R}f=\widetilde{S_{R}}f$ a.e. $\Box$

Theorem. For $f\in L^{1}(\mathbb{R})$, $S_{R}f\rightarrow f$ in measure, and there exists a subsequence $R_{k}\rightarrow\infty$, which depends on $f$, such that $S_{R_{k}}f\rightarrow f$ a.e.

Proof. First, observe that if $f\in \mathcal{S}(\mathbb{R})$, then it follows from Fourier inversion that \begin{align*} \left|S_{R}f(x)-f(x)\right|=\left|\int_{\left|\xi\right|\leq R}\widehat{f}(\xi)e^{2\pi i \xi x}\mathrm{d}\xi-\int_{\mathbb{R}}\widehat{f}(\xi)e^{2\pi i \xi x}\mathrm{d}\xi\right|\leq\int_{\left|\xi\right|>R}\left|\widehat{f}(\xi)\right|\mathrm{d}\xi,\quad\forall x\in\mathbb{R} \end{align*} Since $\widehat{f}\in L^{1}(\mathbb{R})$, the RHS tends to zero as $R\rightarrow\infty$. So $S_{R}f\rightarrow f$ uniformly.

Now assume $f\in L^{1}(\mathbb{R})$, fix $\delta>0$, and let $\epsilon>0$ be given. By the triangle inequality, \begin{align*} \left\{\left|S_{R}f-f\right|\geq\delta\right\}\subset\underbrace{\left\{\left|S_{R}(f-g)\right|\geq\dfrac{\delta}{3}\right\}}_{E_{1,R}}\cup\underbrace{\left\{\left|S_{R}g-g\right|\geq\dfrac{\delta}{3}\right\}}_{E_{2,R}}\cup\underbrace{\left\{\left|g-f\right|\geq\dfrac{\delta}{3}\right\}}_{E_{3}}, \end{align*} where $g\in\mathcal{S}(\mathbb{R})$. Clearly, $\left|E_{3}\right|\leq 3\left\|f-g\right\|_{L^{1}}/\delta$. And by the uniform $L^{1}\rightarrow L^{1,\infty}$ boundedness of the operators $S_{R}$, we have that \begin{align*} \left|E_{1,R}\right|\lesssim\dfrac{3\left\|f-g\right\|_{L^{1}}}{\delta},\quad\forall R>0 \end{align*} By density, we can choose $g$ so that $\left\|f-g\right\|_{L^{1}}<\delta\epsilon/9$. Now choose $R_{0}>0$ sufficiently large so that $E_{2,R}=\emptyset$ for all $R\geq R_{0}$. Putting these results together, we conclude that

\begin{align*} \left|\left\{\left|S_{R}f-f\right|\geq\delta\right\}\right|\lesssim \dfrac{3(\delta\epsilon/9)}{\delta}+0+\dfrac{3(\delta\epsilon/9)}{\delta}<\epsilon,\quad\forall R\geq R_{0} \end{align*}

Since $\mathbb{R}$ is a $\sigma$-finite measure space, convergence in measure implies converge a.e. of a subsequence. We conclude that there exists a subsequence $\left\{R_{k}\right\}$, which may depend of $f$, such that $S_{R_{k}}f\rightarrow f$ a.e. $\Box$


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