Skip to main content
12 events
when toggle format what by license comment
Nov 26, 2022 at 22:55 comment added Iosif Pinelis Do you have a response to the answers below?
Jul 8, 2018 at 16:32 comment added Deane Yang Quick commentary on how to find this example (using natural log instead of log base 2): Note that $\|f\|_p$ is an increasing function of $p$ and its limit as $p\rightarrow 0$ is $\exp -H(f)$. So $f$ must be in $L^1$ but not $L^p$ for any $p > 1$. $f = x^{-\alpha}$, $\alpha > 0$, does not work, so the next thing to try is $x^{-1}(1-\log x)^{-\beta}$, where we add 1 to $-\log x$ to prevent the function from blowing up at $x = 1$. This blows up at $0$ slower than $x^{-1}$ but faster than $x^{-\alpha}$ for any $\alpha < 1$. Testing this, it works if $1 < \beta \le 2$.
Jul 8, 2018 at 15:41 comment added Tejas Bhojraj @Deane. Yes, your function does work, it has a negative infinity entropy!
Jul 8, 2018 at 15:25 answer added R W timeline score: 1
Jul 8, 2018 at 14:06 review Close votes
Jul 9, 2018 at 21:06
Jul 8, 2018 at 12:34 answer added Iosif Pinelis timeline score: 7
Jul 8, 2018 at 5:30 comment added Tejas Bhojraj Yes, you are right; with $1-log(x)$, atleast $f \in L_{1}([0,1])$. $\int_{0}^{1} f = (1-log(x))^{-1}|_{0}^{1}= 1$. I still need to work out $H(f)$. And my previous comment was wrong, sorry!
Jul 8, 2018 at 5:14 comment added Deane Yang Sorry, it has to be $1-\log x$ and not $1 + \log x$, if it works at all.
Jul 8, 2018 at 5:11 comment added Tejas Bhojraj Thanks. However, $f \notin L_{1}([0,1])$. $\int_{0}^{1} f = -(1+log(x))^{-1}|_{0}^{1}= \infty$, right?
Jul 8, 2018 at 4:29 comment added Deane Yang Try $f(x) = (x(1+\log x)^2)^{-1}$
Jul 8, 2018 at 3:18 review First posts
Jul 8, 2018 at 3:38
Jul 8, 2018 at 3:14 history asked Tejas Bhojraj CC BY-SA 4.0