$\ell^2 \to L^\infty$-inequality for almost periodic functions

Question

Suppose $u : \mathbb R \to \mathbb C$ is a smooth, (Bohr) almost periodic function. Formally, such a function admits a Fourier series expansion

$$u(x) = \sum_{\lambda \in \Lambda} \widehat u(k) e^{i \lambda x}$$

for some countable frequency set (sometimes known as the spectrum; one can in some sense think of this as the Fourier support) $\Lambda \subseteq \mathbb R$, which, for the sake of generality, we take to be a $\mathbb Z$-module, i.e. closed under addition, subtraction, and contains the zero frequency. To be concrete, let us suppose $\widehat u(\lambda) \in \ell^1(\Lambda) \cap \ell^2(\Lambda)$.

I am interested in seeing whether there is a "nice" sub-class of almost periodic functions for which one has the estimate

$$||u||_{L^\infty} \lesssim ||\widehat u||_{\ell^2}.$$

Without any restrictions, one can consider a sequence of the form

$$u_n (x) := \sum_{k = 1}^n \frac1k e^{i \lambda_k x},$$

for, say, frequencies localised to $\lambda_k \subseteq[1/2, 1]$. Then evidently $u_n(0) \to \infty$ however $||\widehat{u_n}||_{\ell^2} \lesssim 1$. More dramatically, the derivatives are also uniformly bounded $||\lambda \widehat{u_n}||_{\ell^2} \lesssim 1$, which shows that any form of Sobolev embedding fails. From this example, it seems that there are two main enemies:

1. concentration of mass near a single frequency; for periodic functions the mass is spread out discretely in frequency space so derivative control translates to decay of high frequencies,
2. "resonant" Fourier coefficients in the sense that they all have the same phase, so at $x = 0$ there is no cancellation.

I am not sure how fatal the first enemy is, while for the second I know that there are some results which state that coefficients with "randomly" distributed phases (or signs in the real-valued setting) lead to some cancellation in the periodic setting, though I am not familiar with any quantitative estimates in this direction or results for almost periodic functions.

Answer 1

Perhaps this is not what you're looking for, but: these subclasses do not constitute an infinite dimensional subspace of Bohr almost periodic functions $AP(\mathbb{R})$. The proof of this claim is verbatim similar for all LCAG, so we start with some generalities.

Let $G$ be a locally compact Abelian group and $bG$ denote its Bohr compactification. The space of Bohr almost periodic functions $AP(G)$ on $G$ is isometrically isomorphic to $C(bG)$, the space of continuous functions on $bG$. Let $\widehat{G}$ denote the Pontryagin dual of $G$. Then, $bG$ is the Pontryagin dual of $\widehat{G}_d$, $\widehat{G}$ with the discrete topology. The Fourier series of a $u\in AP(G)=C(bG)$ is $$u = \sum_{y\in\widehat{G}} \hat{u}(y) y, \hspace{8mm} \hat{u}(y) = \langle u,y\rangle = \int_{bG} u\overline{y} \ d\mu $$ where the series converge in $L^2(bG)$ norm, and $\mu$ is the normalized Haar measure on $bG$. For $G=\mathbb{R}$, every $y\in\widehat{G}=\mathbb{R}$ is of the form $y_t(x) = e^{2\pi itx}$ for some $t\in\mathbb{R}$. So, the Fourier series of a $u\in AP(\mathbb{R})$ is as you gave above. Also $$\int_{b\mathbb{R}}u\ d\mu = \lim_{N\to\infty}\frac{1}{2N} \int_{-N}^{N} u(x)\ dx \hspace{8mm} \forall u\in AP(\mathbb{R})$$

$\|u\|_{L^2(bG)}\leq \|u\|_{L^{\infty}(bG)} = \|u\|_{L^{\infty}(G)}$ since $\mu(bG)=1$; and by Plancherel theorem $\|\hat{u}\|_{\ell^2} = \|u\|_{L^2(bG)}$ for all $u\in L^2(bG)$. So $\| \hat{u}\|_{\ell^2} \leq \|u\|_{L^{\infty}(G)}$ for all $u\in AP(G)$.

Claim: There is no $K>0$ and an infinite dimensional closed subspace $W\subseteq AP(G)$ such that $\|u\|_{L^{\infty}} \leq K\|\hat{u}\|_{\ell^2}$ for all $u\in W$.

Pf. If there existed such a $W$, then the Fourier transform $F:AP(G)\to\ell^2(\widehat{G}) \simeq L^2(bG)$ restricted to $W$ would be a linear isomorphism.

Let $n>K^2$ and $\{f_1,\dots,f_n\}\subset W$ be an orthonormal set w.r.t. the inner product of $L^2(bG)$. For each $x\in G$ $$\sum_{j=1}^n |f_j(x)|^2 \leq \|\sum_{j=1}^n \overline{f_j(x)}f_j \|_{L^{\infty}} \leq K \|\sum_{j=1}^n \overline{f_j(x)}f_j \|_{L^{2}(bG)} = K\left(\sum_{j=1}^n |f_j(x)|^2 \right)^{1/2} $$ Thus, $n\leq K^2$. Contradiction.