I have trouble while dealing with the following problem which I found as a proof step in a theoretical paper without further comments:
Suppose that for $\alpha \ge 0$, $(D_n(\alpha))_{n\in\mathbb{N}}$ is a sequence with the property $D_n(\alpha) \to 0$. (*)
Furthermore, for all $n\in\mathbb{N}$, $D_n(\alpha)$ is increasing in $\alpha$, that is, for $\alpha_1 \ge \alpha_2$, $D_n(\alpha_1) \ge D_n(\alpha_2)$.
Show that there exists a sequence $(\alpha_n)_{n\in\mathbb{N}}$ with $\alpha_n \to \infty$ ($n\to\infty$) such that $D_n(\alpha_n) \to 0$.
In fact, $D_n(\alpha)$ was of the form $\sup_{|x| \le \alpha}|f_n(x)|$ and it was shown (*) for it. I do not think the author used the specific form of $D_n$ for his argument, but he thought its somehow a folklore result.
I think the proof should use a diagonal argument, but i struggle to formalize it. I first thought the following way: For each $\alpha\in\mathbb{N}$ there exists $N(\alpha) \in \mathbb{N}, N(\alpha) \ge N(\alpha-1)+1$ such that
$\forall n\ge N(\alpha): \quad D_n(\alpha) \le \frac{1}{\alpha}$.
Then I would arrange the sequence as follows:
$D_1(0),...,D_{N(0)-1}(0),D_{N(1)}(1),...,D_{N(2)-1}(1),D_{N(2)}(2),...,D_{N(3)-1}(2),D_{N(3)}(3),...$
In other words, I define $\alpha_n := \inf\{\alpha\in\mathbb{N}_0:n < N(\alpha)\}$.
For each $\varepsilon > 0$, choose $\alpha \in \mathbb{N}$ with $\frac{1}{\alpha} < \varepsilon$. Then by definition, for all $n \ge N(\alpha)$ it holds that
$D_n(\alpha_n) \le \frac{1}{\alpha} < \varepsilon$.
Is that correct?
Thank you very much for a reply.