A lower bound for an exponential series
Let $\gamma >0$ and $r=(r_n)_{n \in \mathbb{Z} } \in \ell^\infty(\mathbb{Z})$ (a bounded and real-valued sequence), where $r_n \geq 0 \ \forall \ n \in \mathbb{Z}$. Consider the function $f(x)=\sum\limits_{n \in \mathbb{Z}} r_n \exp \left (-2 \gamma \left (x- \frac{n}{2} \right)^2\right)$ for $x \in \mathbb{R}$.
Prove that there exists a constant $C>0$ such that $\sup\limits_{x \in \mathbb{R} }|f(x)| \geq C \cdot ||r||_\infty$.
Ichbinanonym
116
Answer
Answers can only be viewed under the following conditions:
- The questioner was satisfied with and accepted the answer, or
- The answer was evaluated as being 100% correct by the judge.
Mathe
3.5K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 1004 views
- $17.00
Related Questions
- [Real Analysis] Let $a>1$ and $K>0$. Show that there exists $n_0∈N$ such that $a^{n_0}>K$.
- real analysis
- Infinite Geometric Series
- Determine the Closed Form of a Recurrance Relation
- Mathematical modeling
- Convergence of $\sum\limits_{n=1}^{\infty}(-1)^n\frac{n+2}{n^2+n+1}$
- Show that $\sum_{n=1}^{\infty} \frac{\sin n}{n}$ is convergent
- Math and graph representing a competitive struggle between competitors with a fixed number of supporters.
The statement doesn't make much sense. You are taking the supremum is the left had side over all x \in R and they say for all x \in R. Did you mean having just f(x) without supremum in the left hand side of what you wish to prove?
yes my bad, I corrected it. thanks
I still think "sup" should be replaced by "inf" for the problem to make sense.
Proving the statement with "sup" is trivially easy.
For inf it is equally trivial.
would it be equally trivial if I didn't require each r_n to be non-negative?