Replicating Kolmogorov's Counterexample for Fourier Series in Context of Fourier Transforms

Question

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.

Answer 1

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

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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 $\|g-\widehat{f\phi}\|_\infty<c\epsilon$.

Answer 2

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. $\Box$