A Fun $L^2$ Inequality?

Question

I'm having trouble with a step in a paper which I believe boils down to the following inequality: $$ \left\| \sum_{k\in\mathbb{Z}} f(\cdot+k) \right\|_{L^2(0,1)} \leq c \|f\|_{L^2(\mathbb{R})}. $$ I haven't come up with many ideas. Hitting the left-hand side with Minkowski, for example, produces something which can exceed $\|f\|_{L^2(\mathbb{R})}$.

I also put a bit of effort this afternoon into falsifying the above inequality (it may be that I'm misunderstanding the omitted steps in the original paper). Begin with a power series $g(x)=\sum a_kx^k$ which has a local $L^2$ singularity. It's then not too hard to use this representation to construct $f$ satisfying $$ \sum_{k\in\mathbb{Z}} f(x+k) = g(x). $$ However, the few times I attempted this did not result in an $L^2(\mathbb{R})$ function.

Any help one way or the other is appreciated.

Answer 1

The inequality is not true: counterexample:

$$f(x) = \sum_{n=1}^\infty \frac 1n \chi_{[n, n+1]}.$$

Then $f\in L^2(\mathbb R)$, but

$$g(x):= \sum_{k\in \mathbb Z} f(\cdot + k) = \sum_{n=1}^\infty \frac 1n = \infty$$

is not in $L^2(0,1)$.

Answer 2

The counterexample from Artic Char suggested that the statement is true with weight added: \begin{align} \left\| \sum_{k\in\mathbb{Z}} f(\cdot+k) \right\|_{L^2(0,1)} &\leq \sum_{k\in\mathbb{Z}} \|f\|_{L^2(k,k+1)} \\\\ &= \sum_{k\in\mathbb{Z}} \left(\int_k^{k+1} \left|\frac{1+x^2}{1+x^2} \cdot f(x)\right|^2\right)^{1/2} \\\\ &\leq \sum_{k\in\mathbb{Z}} \frac{1}{1+k^2} \|(1+x^2) \, f(x)\|_{L^2(k,k+1)} \\\\ &\leq c \, \|(1+x^2) \, f(x)\|_{L^2(\mathbb{R})}. \end{align} Or, you could choose weight $|x|^\alpha$ for any $\alpha>1$.

Answer 3

Inspiration from the Shannon sampling theorem: if you assume $\mathcal F_{\Bbb R} f$ is Schwartz with $\text{supp}\ f$ contained in $(-1/2,1/2)$ then $$ LHS^2 = \sum_k |\mathcal F_{\Bbb R}f(k)|^2 = \sum_k \int_{\Bbb R} f(y) e^{-2\pi i ky} \ \text dy = \sum_k ∫_{-1/2}^{1/2} f(y) e^{-2\pi i ky} \ \text dy = \sum_k |\mathcal F_{\Bbb T}\tilde f(k)|^2 = ‖\tilde f‖^2_{L^2(\mathbb T)}= ‖f‖^2_{L^2(-1/2,1/2)} = ‖f‖_{L^2(\Bbb R)}^2 $$ The first equality is Poisson summation, $\sum_k f(x+k) = \sum_k \mathcal F_\Bbb R f(k) e^{-2\pi i k x}$. Also, $\tilde f ∈ C^∞ (\Bbb T)$ is the 1-periodic extension of $f∈ C_c^∞(-1/2,1/2) $.

Unfortunately the assumption on the support of $f$ means that there is only one summand in LHS. I'll just leave this here anyway.