$A lower bound$

Question

Let $\gamma >0$ and $r=(r_n)_{n \in \mathbb{Z} } \in \ell^\infty(\mathbb{Z})$ (a bounded and real-valued sequence). Consider a function $f$ such that $f(x)=\sum\limits_{n \in \mathbb{Z}} r_n \exp \left (-2 \gamma \left (x- \frac{n}{2} \right)^2\right)$ for $x \in \mathbb{R}$.

Prove that there exists a constant $C>0$ such that $\sup\limits_{x \in \mathbb{R} }|f(x)| \geq C \cdot ||r||_\infty$ for all $x\in \mathbb{R}$.

Real Analysis

Ichbinanonym

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Answer 1

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Let $m\in \mathbb{Z}$. Then we have

$$ \sup_{x\in \mathbb{R}} f(x) \geq f(\frac{m}{2})= \sum\limits_{n \in \mathbb{Z}} r_n \exp \left (-2 \gamma \left (\frac{m}{2}- \frac{n}{2} \right)^2\right)$$
\[=\sum\limits_{n \in \mathbb{Z}, n\neq m} r_n \exp \left (-2 \gamma \left (\frac{m}{2}- \frac{n}{2} \right)^2\right)+ r_m \exp \left (-2 \gamma \left (\frac{m}{2}- \frac{m}{2} \right)^2\right)\]
\[=\sum\limits_{n \in \mathbb{Z}, n\neq m} r_n \exp \left (-2 \gamma \left (\frac{m}{2}- \frac{n}{2} \right)^2\right)+r_m\geq r_m. \]
Here we have used the fact that the exponential function is always positive and $r_n \geq 0$. Note that without assuming that $r_n \geq 0$, the statement your want to prove is false, for instance take $r_n=-1$.
Hence
\[(1) \sup_{x\in \mathbb{R}} f(x) \geq r_m, \forall m \in \mathbb{Z}.\]
It follows from (1) that
\[\sup_{x\in \mathbb{R}} f(x) \geq \sup _{m \in \mathbb{Z}}r_m=|r|_{\infty}.\]
Indeed one can take $C=1$.

Nirenberg

Nirenberg

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Ichbinanonym

sorry for the late reply. I made a mistake, the sum is supposed to be equal to the absolute value of f. I posted another question with the corrected version.

$A lower bound$

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