Define $u_n: \mathbb{R}^+ \to \mathbb{R}^+$ as $u_n(x) = n1_{(0,1/n)}$. I need to show that this has no dominant. A dominant is defined as an integrable function $w: X \to \mathbb{R}^+$ such that $\forall n, \forall x, |u_n(x)| \leq w(x)$.

My attempt so far: suppose it has a dominant $w$. Then since $w$ is integrable we have $\int|w|d\lambda < \infty$. Fix $x$. Then we have $$ |n1_{(0,1/n)}(x)| \leq w(x) \ \ \forall n. $$ This means that $w(x) \geq n$ for all $x \in (0,1/n)$. Now pick the sequence $x_n \in (0,\frac{1}{n})$. Then we have $w(x_n) \geq n$ for all $n$.

I don't know how to proceed here. Normally I would use continuity and consider $\lim_{n \to \infty} x_n$ but we don't have continuity, only integrability of $w$.


If $w(x) \ge u_n(x)$ for all $n$ and $x$, then in particular, $w(x) \ge n$ on the interval $[\frac{1}{n+1},\frac{1}{n})$. So $$\int_{\mathbb{R}^+} w\ d\lambda \ge \sum_{n=1}^\infty \int_{[\tfrac{1}{n+1},\tfrac{1}{n})} n\ d\lambda = \sum_{n=1}^\infty n\left( \frac{1}{n} - \frac{1}{n+1} \right) = \sum_{n=1}^\infty \frac{1}{n+1} = \infty.$$

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  • $\begingroup$ Nice solution thanks! This was indeed something I was looking for. $\endgroup$ – Sigurd Nov 11 '18 at 15:22

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