I have this non linear operator $$H(p) = -\sum_{n=0}^ {\infty} p_n ln(p_n)$$ where $p_n$ are given by a function $p(n)$ when $n$ is a whole number.
I want to find what set of $p(n)$ makes $H(p)$ diverge, but that also have the following properties:
- Decreasing monotonic
- Positive in x>0
- No singularities
Some examples of functions in this set are:
- $\frac{1}{x+1}$
- $\frac{1}{ln(x)}$
- $\frac{1}{x^a+1}$ where $0<a<1$
Working on this I have developed the following hypothesis:
- Any function that decreases faster than $\frac{1}{n+1}$ makes H converge. For example all functions of the form $1/n^a$ with a>1 converge to $0$ faster than $1/n$ and yet make $H$ converge. $H(1/n^3)\approx-0.59$
- Any function that decreases slower than $\frac{1}{n+1}$ makes H diverge.
- $\frac{1}{n+1}$ makes H diverge the slowest
I haven't been able to disprove these hypothesis, but I haven't been able to prove them either, I just keep finding more examples that support them.
One idea I had was to transform $H(p)$ into an integral, solve it, and then I would just have to see what $p(n)$ makes $H(p)$ diverge.
$$H(p)=-\int_a^b p(n) ln(p(n))$$
But when I try to use integration by parts I just get the same thing I started with.
I've also tried to turn this problem around:
$$J(p)=H(1/p)=-\int_a^b \frac{1}{p_n} ln(\frac{1}{p_n})$$
and try to look for functions that make $J(p)$ converge, but I have been equally unsuccessful.
Finally I've tried to find a function that tells me how fast $H(p)$ grows based on how fast $p(n)$ grows, given that I have reasons to believe this function grows as the growth of p(n) decreases until it reaches a maximum at $p(n)=\frac{1}{n}$ where it begins to decrease. But I haven't been able to find that function despite gleaming its behavior.
How can I prove or disprove my hypothesis?
Edit:
Since $\int_0^\infty log(x)dx$ diverges I think all I need is to multiply it by a function that also diverges in that range. So my set would be the functions that:
- Decreasing monotonic
- Positive in x>0
- No singularities
- Diverges in $\int_0^\infty f(x)dx$
'Anything slower that $(n+1)^{-1}$' is essentially $(n+1)^{-1+\epsilon}\quad(\epsilon>0)$, if I'm not missing something.
So plugging it in, and switch to integration:
$$ \begin{align} -\sum_{n=0}^ {\infty} p_n \ln(p_n)&= -\sum_{n=0}^ {\infty} (n+1)^{-1+\epsilon} \ln(n+1) (-1+\epsilon)\\ &\ge -\int_1^\infty (m)^{-1+\epsilon} \ln m (-1+\epsilon) \mathrm dm\\ &= \frac{1+\epsilon}{\epsilon^2} \end{align} $$
Which tends to infinity as $\epsilon\to0^+$.