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Let $f_\alpha$ be a family of continuous positive functions $\mathbb R\to \mathbb R$ where the index $\alpha$ runs in a compact metric space and the map $\alpha\to f_\alpha$ is continuous with respect to compact-open topology on the target. Suppose there is a uniform upper bound on the integrals of $f_\alpha$'s over $\mathbb R$.

Question. Is $\underset{\alpha}{\sup} f_\alpha$ necessarily an integrable function?

Apology. This sure sounds like a homework level question, but after looking at for a while I am not even sure what the answer is.

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    $\begingroup$ Note that in your hypotheses the family $\{f_\alpha\}$ is equicontinuous on bounded intervals, so the supremum is a continuous function (possibly non integrable). $\endgroup$ – Pietro Majer May 16 '11 at 16:48
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    $\begingroup$ Also, every nonnegative continuous function $f\colon\mathbb{R}\to\mathbb{R}$ can occur as $\sup\_\alpha f_\alpha$. Consider the case where the metric space is the extended reals $\mathbb{\bar R}$, $f_\alpha=0$ for $\alpha=\pm\infty$ and $f_\alpha(x)=f(x)(1-g(\alpha)\vert\alpha-x\vert)\_+$ otherwise, with $g\colon\mathbb{R}\to\mathbb{R}$ a continuous function tending to inifinity as $\vert\alpha\vert\to\infty$ fast enough that $\int f_\alpha(x)\,dx$ is bounded. $\endgroup$ – George Lowther May 16 '11 at 17:21
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No. Let the compact metric space be $[0,1]$ with the standard topology and define $$ f_\alpha(x) =\alpha\max(1-\alpha \vert x\vert,0) $$ for $\alpha\in[0,1]$. This satisfies the properties asked for, with the upper bound $\int f_\alpha(x)\\,dx\le1$ (and, equality for $\alpha\not=0$). But, $$ \sup\_\alpha f_\alpha(x)=1_{\{\vert x\vert\ge1/2\}}\frac{1}{4\vert x\vert}+1_{\{\vert x\vert < 1/2\}}(1-\vert x\vert) $$ is not integrable.

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As you stated, no: take e.g. the unit closed interval $[0,1]$ as a compact metric space and define for all $\alpha\in [0,1]$ the function $f_\alpha(x)= \alpha (1-\alpha|x|) _ +$, that depends continuously on $\alpha$ even in the uniform norm on $\mathbb{R}$. Then for $|x|\ge 1$ we have $ \sup_ {\alpha\in[0,1]} f_ \alpha (x)=1/4x $, which is not integrable.

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    $\begingroup$ Thanks so much! I accepted George Lowther's answer because he was there first. :) $\endgroup$ – Igor Belegradek May 16 '11 at 16:48
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Just one more example: let $\phi \in C_c(\mathbb{R})$ with, say, $\sup_{\mathbb{R}} \phi = 1$. For $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, set $$f_\alpha(x) = \begin{cases} \phi(x - \tan \alpha), & \alpha \in (-\frac{\pi}{2}, \frac{\pi}{2}) \\\\ 0, &\alpha = \pm \frac{\pi}{2}. \end{cases}$$ Then $\int f_\alpha = \int \phi < \infty$ for each $\alpha \ne \pm \frac{\pi}{2}$, and $\int f_\alpha = 0$ otherwise, but $\sup_\alpha f_\alpha = 1$.

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