I wish to prove that $g(x,p)=|f(x)|^p\ln|f(x)|$ is a bounded function of $(x,p)$, where $0<|f(x)|\leq M$ for all $x$, and $p\in[p_1,p_2]$, where $0<p_1<p_2<\infty$.

$f$ is measurable but not necessary continuous.

My attempt: Let $(x,p)$ be an arbitrary point. We wish to show $|g(x,p)|\leq K$ where $K$ is independent of $x, p$.

Case 1) Suppose $|f(x)|\geq 1$. Then $0<|f(x)|^p\ln|f(x)|\leq M^{p_2}\ln M$. The bounds are independent of $(x,p)$ so we are done for this case.

Case 2) Suppose $|f(x)|<1$. Then $0<|f(x)|^p\leq M^p\leq M^{p_2}$.

The $\ln|f(x)|$ part is kind of tricky as it is unbounded. However since $t^p\ln t\to 0$ as $t\to 0^+$, $|f(x)|^p\ln|f(x)|$ ought to be bounded. I don't know how to write it rigorously though.

Thanks for any help.


If $t<1$, then $$|t^p \ln t| \le |t|^{p_1} |\ln t|.$$ Since $|t|^{p_1} |\ln t| \to 0$ as $t\to 0$, $|t|^{p_1} |\ln t|$ can be extended to a continuous function defined on $[0,1]$ and thus is bounded on $[0,1]$. Thus $t^p \ln t$ is also bounded on $[0,1] \times [p_1\times p_2]$. Together with case one you have that $t^p \ln t$ is bounded on $[0,M] \times [p_1,p_2]$.

  • $\begingroup$ I see... Here we implicit use the fact that $|t|^{p_1}|\ln t|$ is a continuous function of $t$ right? $\endgroup$ – yoyostein Sep 17 '16 at 6:17
  • $\begingroup$ Can you briefly explain "Since $|t|^{p_1} |\ln t| \to 0$ as $t\to 0$, $|t|^{p_1} |\ln t|$ is bounded on $[0,1]$". This is the part where I am a bit confused. The rest are ok. Is it something like for $t<\delta$, $||t|^{p_1} |\ln t||<\epsilon$, for $\delta \leq t\leq 1$ it is bounded by extreme value theorem since $[\delta,1]$ is compact. $\endgroup$ – yoyostein Sep 17 '16 at 6:23
  • $\begingroup$ Please see the edit, is that ok? @yoyostein $\endgroup$ – user99914 Sep 17 '16 at 6:27
  • 1
    $\begingroup$ Ok I think I get it $\endgroup$ – yoyostein Sep 17 '16 at 6:31

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