Return to Answer

Suppose that $\mu_n(1)=1-t_n$, $\mu_n(2)=\cdots=\mu_n(n+1)=t_n/n$, and $\mu_n(n+2)=\mu_n(n+3)=\cdots=0$, where $t_n:=1/\ln n$ and $n\ge3$. Then $\mu_n(1)\to1$ and $H_1(\mu_n)\to1$. So, 1' and 2' hold; one may say 2' holds with an infinitely slow rate. It is easy to modify this example to have 2' hold with an arbitrarily slow rate.

After the editing of the question, the answer becomes no. Indeed, suppose that $\mu_n(1)=1-t_n$, where $t_n\downarrow0$, and suppose that $H_1(\mu_n)\to0$ so slowly that $$H_1(\mu_n)\ge1\Big/\sqrt{\ln\frac1{t_n}}$$ eventually (i.e., for all large enough $n$).

Then $\mu_n(j)\le t_n$ for $j\ge2$ and therefore $$\ln^2\frac1{\mu_n(j)}\ge\ln\frac1{t_n}\;\ln\frac1{\mu_n(j)},$$ whence eventually $$H_2(\mu_n)\ge\sum_{j\ge2}\mu_n(j)\ln^2\frac1{\mu_n(j)} \ge\sum_{j\ge2}\mu_n(j)\ln\frac1{\mu_n(j)}\; \ln\frac1{t_n} =\Big(H_1(\mu_n)-(1-t_n)\ln\frac1{1-t_n}\Big)\ln\frac1{t_n} \ge \Big(1\Big/\sqrt{\ln\frac1{t_n}}(1-o(1))\Big)\ln\frac1{t_n}\to\infty.$$

Thus, whenever 1' and 3' hold, 2' cannot hold.

Suppose that $\mu_n(1)=1-t_n$, $\mu_n(2)=\cdots=\mu_n(n+1)=t_n/n$, and $\mu_n(n+2)=\mu_n(n+3)=\cdots=0$, where $t_n:=1/\ln n$ and $n\ge3$. Then $\mu_n(1)\to1$ and $H_1(\mu_n)\to1$. So, 1' and 2' hold; one may say 2' holds with an infinitely slow rate. It is easy to modify this example to have 2' hold with an arbitrarily slow rate.

After the editing of the question, the answer becomes no. Indeed, suppose that $\mu_n(1)=1-t_n$, where $t_n\downarrow0$, and suppose that $H_1(\mu_n)\to0$ so slowly that $$H_1(\mu_n)\ge1\Big/\sqrt{\ln\frac1{t_n}}$$ eventually (i.e., for all large enough $n$).

Then $\mu_n(j)\le t_n$ for $j\ge2$ and therefore $$\ln^2\frac1{\mu_n(j)}\ge\ln\frac1{t_n}\;\ln\frac1{\mu_n(j)},$$ whence eventually $$\begin{aligned} H_2(\mu_n)&\ge\sum_{j\ge2}\mu_n(j)\ln^2\frac1{\mu_n(j)} \\ &\ge\sum_{j\ge2}\mu_n(j)\ln\frac1{\mu_n(j)}\; \ln\frac1{t_n} \\ &=\Big(H_1(\mu_n)-(1-t_n)\ln\frac1{1-t_n}\Big)\ln\frac1{t_n} \\ &\ge \Big(1\Big/\sqrt{\ln\frac1{t_n}}(1-o(1))\Big)\ln\frac1{t_n}\to\infty. \end{aligned} $$

Thus, whenever 1' and 3' hold, 2' cannot hold.

Suppose that $\mu_n(1)=1-t_n$, $\mu_n(2)=\cdots=\mu_n(n+1)=t_n/n$, and $\mu_n(n+2)=\mu_n(n+3)=\cdots=0$, where $t_n:=1/\ln n$ and $n\ge3$. Then $\mu_n(1)\to1$ and $H_1(\mu_n)\to1$. So, 1' and 2' hold; one may say 2' holds with an infinitely slow rate. It is easy to modify this example to have 2' hold with an arbitrarily slow rate.

After the editing of the question, the answer becomes no. Indeed, suppose that $\mu_n(1)=1-t_n$, where $t_n\downarrow0$, and suppose that $H_1(\mu_n)\to0$ so slowly that $$H_1(\mu_n)\ge1\Big/\sqrt{\ln\frac1{t_n}}$$ eventually (i.e., for all large enough $n$).

Then $\mu_n(j)\le t_n$ for $j\ge2$ and hence $$\ln^2\frac1{\mu_n(j)}\ge\ln\frac1{t_n}\;\ln\frac1{\mu_n(j)},$$ whence $$H_2(\mu_n)\ge\sum_{j\ge2}\mu_n(j)\ln^2\frac1{\mu_n(j)} \ge\sum_{j\ge2}\mu_n(j)\ln\frac1{\mu_n(j)} \ln\frac1{t_n} =\Big(H_1(\mu_n)-(1-t_n)\ln\frac1{1-t_n}\Big)\ln\frac1{t_n} \ge \Big(1\Big/\sqrt{\ln\frac1{t_n}}(1-o(1))\Big)\ln\frac1{t_n}\to\infty.$$

Suppose that $\mu_n(1)=1-t_n$, $\mu_n(2)=\cdots=\mu_n(n+1)=t_n/n$, and $\mu_n(n+2)=\mu_n(n+3)=\cdots=0$, where $t_n:=1/\ln n$ and $n\ge3$. Then $\mu_n(1)\to1$ and $H_1(\mu_n)\to1$. So, 1' and 2' hold; one may say 2' holds with an infinitely slow rate. It is easy to modify this example to have 2' hold with an arbitrarily slow rate.

After the editing of the question, the answer becomes no. Indeed, suppose that $\mu_n(1)=1-t_n$, where $t_n\downarrow0$, and suppose that $H_1(\mu_n)\to0$ so slowly that $$H_1(\mu_n)\ge1\Big/\sqrt{\ln\frac1{t_n}}$$ eventually (i.e., for all large enough $n$).

Then $\mu_n(j)\le t_n$ for $j\ge2$ and therefore $$\ln^2\frac1{\mu_n(j)}\ge\ln\frac1{t_n}\;\ln\frac1{\mu_n(j)},$$ whence eventually $$H_2(\mu_n)\ge\sum_{j\ge2}\mu_n(j)\ln^2\frac1{\mu_n(j)} \ge\sum_{j\ge2}\mu_n(j)\ln\frac1{\mu_n(j)}\; \ln\frac1{t_n} =\Big(H_1(\mu_n)-(1-t_n)\ln\frac1{1-t_n}\Big)\ln\frac1{t_n} \ge \Big(1\Big/\sqrt{\ln\frac1{t_n}}(1-o(1))\Big)\ln\frac1{t_n}\to\infty.$$

Thus, whenever 1' and 3' hold, 2' cannot hold.

Suppose that $\mu_n(1)=1-t_n$, $\mu_n(2)=\cdots=\mu_n(n+1)=t_n/n$, and $\mu_n(n+2)=\mu_n(n+3)=\cdots=0$, where $t_n:=1/\ln n$ and $n\ge3$. Then $\mu_n(1)\to1$ and $H_1(\mu_n)\to1$. So, 1' and 2' hold; one may say 2' holds with an infinitely slow rate. It is easy to modify this example to have 2' hold with an arbitrarily slow rate.

Suppose that $\mu_n(1)=1-t_n$, $\mu_n(2)=\cdots=\mu_n(n+1)=t_n/n$, and $\mu_n(n+2)=\mu_n(n+3)=\cdots=0$, where $t_n:=1/\ln n$ and $n\ge3$. Then $\mu_n(1)\to1$ and $H_1(\mu_n)\to1$. So, 1' and 2' hold; one may say 2' holds with an infinitely slow rate. It is easy to modify this example to have 2' hold with an arbitrarily slow rate.

After the editing of the question, the answer becomes no. Indeed, suppose that $\mu_n(1)=1-t_n$, where $t_n\downarrow0$, and suppose that $H_1(\mu_n)\to0$ so slowly that $$H_1(\mu_n)\ge1\Big/\sqrt{\ln\frac1{t_n}}$$ eventually (i.e., for all large enough $n$).

Then $\mu_n(j)\le t_n$ for $j\ge2$ and hence $$\ln^2\frac1{\mu_n(j)}\ge\ln\frac1{t_n}\;\ln\frac1{\mu_n(j)},$$ whence $$H_2(\mu_n)\ge\sum_{j\ge2}\mu_n(j)\ln^2\frac1{\mu_n(j)} \ge\sum_{j\ge2}\mu_n(j)\ln\frac1{\mu_n(j)} \ln\frac1{t_n} =\Big(H_1(\mu_n)-(1-t_n)\ln\frac1{1-t_n}\Big)\ln\frac1{t_n} \ge \Big(1\Big/\sqrt{\ln\frac1{t_n}}(1-o(1))\Big)\ln\frac1{t_n}\to\infty.$$