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I want to prove the following (I believe it is true but I am not sure)

If $f\in \mathrm{AP}(\mathbb{R})$, where the space of almost periodic functions $\mathrm{AP}(\mathbb{R})$ is defined in the sense of Bohr https://en.wikipedia.org/wiki/Almost_periodic_function with the mean value $$ M(f) = \lim_{T\longrightarrow \infty} \frac{1}{T}\int_{0}^T f(x)\;dx = 0 $$ then its anti-derivative \begin{equation*} F(x) = \int_0^x f(t)\;dt \end{equation*} is bounded.

The analog of this in the case $f$ is periodic is clear, but I don't know how to do with this case. One idea is using approximation by trigonometric polynomials but there is still some difficulties.

Sean
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2 Answers2

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This is actually false.

$AP$ is a Banach space with the uniform norm, being precisely the uniformly closed span of a set of functions. It turns out that $AP_0$, the space of functions in $AP$ with mean zero, is a closed subspace. (It's not hard to show this directly (details below). Or if you want to be cool, you get into the not-quite-trivial theory: In fact $f\in AP$ has mean zero if and only if $f$ is in the closed span of the functions $e_a(t)=e^{iat}$ for $a\ne0$.)

Say $Tf(x)=\int_0^x f(t)\,dt$.

If $Tf$ were bounded for every $f\in AP_0$ the Closed Graph Theorem would show that $||Tf||_\infty\le c||f||_\infty$ for $f\in AP_0$, which is clearly not so.

Details: Define $Af(x)=\frac1x\int_0^x f(t)\,dt$. Say $f_n\in AP_0$ and $f_n\to0$ uniformly. Then $Af_n\in C_0([0,\infty))$ and $Af_n\to Af$ uniformly, so $Af\in C_0$, hence $Mf=0$.

Note Usually the mean value is defined as $\lim\frac1{2T}\int_{-T}^T$. Wasn't clear to me immediately, but that's actually equal to the mean value defined above. Proof: The two are the same if $f(t)=e^{iat}$.

David C. Ullrich
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  • I think AP is a Hilbert space but not with the uniform norm, the norm induced by the mean value? – Sean Mar 24 '18 at 18:04
  • If you use the mean value to define an inner product then $AP$ is an incomplete inner product space; its completion is a Hilbert space. – David C. Ullrich Mar 24 '18 at 18:12
  • @DavidC.Ullrich About the last note: for (weakly) almost periodic functions it is actually true that $\frac{1}{vol(A_n)} \int_{A_n} f(x) dx$ converges to the same Mean for all Folner sequences. The convergence is actually stronger: $\frac{1}{vol(A_n)} \int_{A_n} f(x+y) dx$ converges to the mean uniformly in $y$, and it is really this uniform convergence which makes the limit independent of the choice of the Folner sequence (this is well hidden behind the details in the proofs). – N. S. Mar 24 '18 at 18:26
  • @Sean It's a very cool story that you should look up somewhere: $AP$ is a Banach algebra (with pointwise multiplication). Say its maximal ideal space is $B$. Then $B$ is the Bohr compactification of $\Bbb R$; turns out $B$ is a compact abelian group, $AP$ "is" $C(B)$, and that Hilbert space is $L^2(B)$; the functions $e^{iat}$ give a complete orthonormal set. – David C. Ullrich Mar 24 '18 at 18:26
  • Can you suggest some good source to learn them? Thank you very much! – Sean Mar 24 '18 at 18:29
  • @Sean I like A.S. Besicovitch, "Almost periodic functions" and C. Corduneanu, "Almost periodic functions" – N. S. Mar 24 '18 at 18:31
  • @Sean Not sure. I suspect you'd find links to most of what I said in the Wikipedia article. I also suspect it's all in Rudin Fourier Analysis on Groups. – David C. Ullrich Mar 24 '18 at 18:31
  • @N.S. Cool. What an age we live in, that I can find the definition without having to stand up. What you said seems pretty clear, again just because it holds for characters. – David C. Ullrich Mar 24 '18 at 18:33
  • Thank you to @ N.S and @ David very much! – Sean Mar 24 '18 at 18:35
  • Is it correct if I replace $f$ to be quasi-periodic? I.e. $f(x) = F(\xi_1x,\xi_2x,\ldots,\xi_mx)$ where $F\in C(\mathbb{T}^m$)? – Sean Apr 04 '18 at 00:40
  • @Sean In that case $F$ is bounded, by compactness, so $f$ is bounded - nothing to do with mean zero. – David C. Ullrich Apr 06 '18 at 15:04
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Hint

$$f(x)=\sum_{n} \frac{1}{n^2} e^{i\frac{x}{n^2}}$$ is almost periodic as the uniform limit of trig polynomials.

Assume now by contradiction that $F(x)=\int_0^x f(t) dt$ is almost periodic. Then the Fourier Bohr coefficient at $e^{in^2x}$ is given by $$b_{\frac{1}{n^2}}=\langle F(x), e^{-i\frac{x}{n^2}} \rangle =\lim_{T\longrightarrow \infty} \frac{1}{T}\int_{0}^T F(x) e^{-i\frac{ x}{n^2}}\;dx$$

Use integration by parts to conclude that $b_{\frac{1}{n^2}}=\frac{1}{i}$. But this contradicts the Bessel inequality $$\sum_{n} |b_{\frac{1}{n^2}}|^2 \leq \langle F(x), F(x) \rangle $$

P.S. If you need any details for any step let me know.

Added: $$\frac{1}{T}\int_{0}^T F(x) e^{-i\frac{ x}{n^2}}=\frac{1}{T} \left( F(x) \frac{n^2}{-i}e^{-i\frac{ x}{n^2}}|_0^T-\frac{n^2}{-i}\int_0^T f(x) e^{-i\frac{ x}{n^2}} \right)$$

Now, since we assumed that $F(x)$ is bounded, we have $$\lim_{T \to \infty} \frac{1}{T} \left( F(x) \frac{n^2}{-i}e^{-i\frac{ x}{n^2}}|_0^T \right)=0$$

Therefore $$\langle F(x), e^{-i\frac{x}{n^2}} \rangle= \lim_T \frac{n^2}{i}\int_0^T f(x) e^{-i\frac{ x}{n^2}} =\frac{1}{i}$$

N. S.
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  • This shows the antiderivative cannot be AP; the question was whether the antiderivative is bounded. (Yes, it's pretty clear that the antiiderivative is unbounded, but I don't see how you can use those cool little arguments to that effect, since it's not AP, or $L^2$.) – David C. Ullrich Mar 24 '18 at 18:45
  • I think N.S is correct, since by Bohr theorem, if the anti-derivative is bounded then it belongs to AP. – Sean Mar 24 '18 at 18:48
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    @Sean Yeah, I just saw that theorem in the link he posted above. Of course then we need to prove it. The antiderivative is $F(t)=\sum(e^{it/n^2}-1)$. Calculus exercises: (i) Show directly that that series actually converges (ii) show directly the sum is unbounded. – David C. Ullrich Mar 24 '18 at 18:52
  • @ David C. Ullrich: That is interesting! – Sean Mar 24 '18 at 19:07
  • @N.S I think you have a mistake when computing the Fourier coefficient of $F$. – Sean Mar 25 '18 at 19:28
  • @N.S can you carry out the integration by parts step? – Sean Mar 25 '18 at 19:56
  • @Sean Yes I did, and you are right there is an $i$ missing in the denominator. $F(t)$ is $F(t)=\frac{1}{i}\sum(e^{it/n^2}-1)$. – N. S. Mar 25 '18 at 21:34
  • Is it correct if I replace $f$ to be quasi-periodic? I.e. $f(x) = F(\xi_1x,\xi_2x,\ldots,\xi_mx)$ where $F\in C(\mathbb{T}^m$)? – Sean Apr 04 '18 at 00:40
  • @Sean Unless I make a stupid mistake, if $t_n \in \mathbb R$, then by compactness of $\mathbb T$ you can find a subsequence such that $(\xi_1t_{k_n},.., \xi_m t_{k_n}) \to (s_1,..., s_m))$ on $\mathbb T^m$. Then, the set of translates $T_{t_{k_n}} f$ converges in sup norm to $g(x)= F(\xi_1(x+s_1),\xi_2(x+s_2),\ldots,\xi_m(x+s_n)$. It follows from here that the closed hull of translates of $f$ is compact with respect to the sup norm, and hence $f$ is Bochner almost periodic. But Bochner almost periodic is equivalent to Bohr almost periodic, so any such $f$ belongs to $AP(\mathbb R)$. – N. S. Apr 04 '18 at 02:02
  • To make the argument a bit more clear: since $\mathbb T^n$ is a compact group, any continuous function is Bochner almost periodic on $\mathbb T^n$. It follows that the hull $$T_t F = \overline{ { T_t F | t \in \mathbb T^n } }$$ is compact in $(C(\mathbb T^n), | , |\infty)$. It follows that the set $$\overline{ { T{(\xi_1 t,\xi_2t,\xi_3t,.., \xi_mt)} F | t \in \mathbb R} }$$ is compact, and from here it follows that the closed hull of $f$ is compact. This implies that $f$ is (Bochner hence Bohr) almost periodic. – N. S. Apr 04 '18 at 02:08
  • I don't get this, how is it related to the the question, if mean value of $F\in C(\mathbb{T}^m) = 0$ then the function $f(x) = F(\xi_1 x,\xi_2x,\ldots,\xi_mx)$ has bounded anti-derivative? I know of course $f$ defined this way is almost periodic for sure (thanks to your comments). – Sean Apr 04 '18 at 02:17
  • @Sean Sorry, misunderstood the question.I still think that the answer is no, but the problem is more subtle. The idea is the following: Pick some $b_n$ of the form $b_n=k_n+l_n\sqrt{2}$ with $k_n, l_n \in \mathbb Z$ which converges fast enough to $0$. Then, as in my answer $$f(x)=\sum_{n} b_n e^{ib_n x}$$ is almost periodic with unbounded antiderivative. I show next that $f(x)$ is quasiperiodic. Indeed, define $$F : \mathbb T^2 \to \mathbb T^2\ F(s,t)= \sum_{n} b_n e^{2\pi i(k_n s+l_nt)}$$ If I didn't make any mistake $$f(x)=F(\xi_1x, \xi_2 x)$$ – N. S. Apr 04 '18 at 02:46
  • with $\xi_1=\frac{1}{2\pi}, \xi_2=\frac{\sqrt{2}}{2\pi}$$.... The idea is that to make $f$ quaiperiodic, I made the Fourier Bohr module finitely generated. – N. S. Apr 04 '18 at 02:47