0
$\begingroup$

If we consider the application $T: \mathcal{D}(\mathbb{R}^{\star}) \to \mathbb{C}$ defined by $<T,\varphi>= \sum_{n \geq 1} \varphi(\dfrac{\ln n}{n})$.

My question is: $<T,\varphi>$ converge for all $\varphi \in \mathcal{D}(\mathbb{R}^\star)$?

I'm lost beacause $Supp \varphi \subset \mathbb{R}^\star$ and in the same times, $n$ can take $1$.

Thank you for the help.

$\endgroup$
  • $\begingroup$ So basically you are asking, whether $T$ is well-defined ? $\endgroup$ – Vobo Mar 4 '17 at 15:52
0
$\begingroup$

Supposing $\mathbb{R}^\star:=(0,\infty)$ your sum converges. Since $\lim_{n\to\infty}\frac{\ln{n}}{n}=0$ and $\mathrm{supp}{\varphi}\underset{\text{compact}}\subset(0,\infty)$, $\frac{\ln{n}}{n}$ doesn't belong to $\mathrm{supp}{\varphi}$ for large $n$. Thus the sum is a finite one.

EDIT: a rigorous explanation

Put $\varepsilon:=\min\mathrm{supp}{\varphi}$. Then clearly $\varepsilon>0$. Since $\lim_{n\to\infty}\frac{\ln{n}}{n}=0$, there exists $n_0\in\mathbb{N}$ such that $n\geq n_0\Rightarrow \frac{\ln{n}}{n}<\varepsilon$. Thus, $<T,\varphi>=\varphi\left(\frac{\ln{1}}{1}\right)+\dots+\varphi\left(\frac{\ln{n_0}}{n_0}\right)<\infty$.

$\endgroup$
  • $\begingroup$ Please, how we can prouve rigourously the convergence of this sum? Why there exist $n_0$ such that $\forall n \geq n_0: <T,\varphi> < +\infty$? And who is this $n_0$? Please. $\endgroup$ – user415040 Mar 4 '17 at 15:54
  • $\begingroup$ Then please accept it by clicking the green "$\surd$" mark. $\endgroup$ – stb2084 Mar 6 '17 at 15:27
  • $\begingroup$ Thank you so much. I accept it. $\endgroup$ – user415040 Mar 7 '17 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy