For $c\in (0,1)$, consider the probability density $$f_c := \frac{1-c^2}{1+c^2-2c\cos(2\pi x)}.$$

Goal. Evaluate the limit as $c\rightarrow 1$ of $$E(f_c) := \int_{-\frac12}^{\frac12}\log(f_c)f_c dx - \pi\|f_c\|_{\dot{H}^{-\frac12}}^2,$$ where $\dot{H}^{-\frac12}$ denotes the usual homogeneous Sobolev norm of index $-\frac12$. Alternatively, one can write $$\pi\|f_c\|_{\dot{H}^{-\frac12}}^2 = -\int_{-\frac12}^{\frac12}\int_{-\frac12}^{\frac12}\log|\sin(\pi x)|f_c(x)f_c(y)dxdy + \int_{-\frac12}^{\frac12}\log|\sin(\pi y)|dy.$$ For variational reasons, I know that $E(f_c)>0$.

I know that for $k\ge 1$, the $k$-th Fourier coefficient $\hat{f}_c(k) = c$. This implies that $$\pi\|f_c\|_{\dot{H}^{-\frac12}}^2 = 2\pi\sum_{k=1}^\infty \frac{c^{2k}}{2\pi k} = -\log(1-c^2), \tag{1}$$ where I've used the power series for $\log(1+z)$ to obtain the final equality.

Now inserting the definition of $f_c$ and using that $\int_{-\frac12}^{\frac12}f_c=1$, I can compute $$\int_{-\frac12}^{\frac12}\log(f_c)f_cdx = \log(1-c^2) - (1-c^2)\int_{-\frac12}^{\frac12}\frac{\log\Big(1+c^2-2c\cos(2\pi x)\Big)}{1+c^2-2c\cos(2\pi x)}dx. \tag{2}$$ Now inserting (1) and (2) into the RHS for $E(f_c)$ yields $$E(f_c) = 2\log(1-c^2) - (1-c^2)\int_{-\frac12}^{\frac12}\frac{\log\Big(1+c^2-2c\cos(2\pi x)\Big)}{1+c^2-2c\cos(2\pi x)}dx. $$ I am stuck with what to do with the second term. Naively, I might try to write $$1+c^2-2c\cos(2\pi x) = (1-c)^2 + 2c[1-\cos(2\pi x)]$$ and pull out a factor of $(1-c)^2$ to write \begin{multline} (1-c^2)\int_{-\frac12}^{\frac12}\frac{\log\Big(1+c^2-2c\cos(2\pi x)\Big)}{1+c^2-2c\cos(2\pi x)}dx = 2\log(1-c) \\ + (1-c^2)\int_{-\frac12}^{\frac12}\frac{\log\Big(1+\frac{2c[1-\cos(2\pi x)]}{(1-c)^2}\Big)}{1+c^2-2c\cos(2\pi x)}dx. \end{multline} The first term is OK, because I can combine it with $2\log(1-c^2)$. But the second term, I am still stuck. My guess is to try some sort of Taylor expansion of $1-\cos(2\pi x)$ when $|x|\lesssim 1-c$, but not sure how to make this work.

  • $\begingroup$ $$\int\frac{\log\Big(1+c^2-2c\cos(2\pi x)\Big)}{1+c^2-2c\cos(2\pi x)}\,dx$$ doest exist (even if it is a monster). The key problem if the evaluation at the bounds. $\endgroup$ Nov 21 at 10:30
  • $\begingroup$ OK, but I am not interested in an exact formula for the antiderivative, only for the limit as $c\rightarrow 1$ after you normalize by $(1-c^2)$. $\endgroup$
    – Nik Quine
    Nov 21 at 14:18

1 Answer 1


If we naively assume that we can pass the limit inside the integral, the result would be $$\lim_{c\to 0} E(f_c) = 0.$$ Notice that for $c$ small enough, $$c^2<|1-2c|=\inf_{x\in(-1/2,1/2)} |1-2c\cos(2\pi x)|<|1-2c\cos(2\pi x)|.$$ Hence, we have for $c<1/4,$ $$|c^2 + 1-2c\cos(2\pi x)|\ge ||1-2c\cos(2\pi x)|- c^2|=|1-2c\cos(2\pi x)|- c^2\ge 1-2c-c^2>7/16.$$ Additionally, again for $c<1/4$, $$|c^2 + 1-2c\cos(2\pi x)|\le (1+c)^2\le 25/16.$$ Let me denote $f(x)=c^2 + 1-2c\cos(2\pi x)$. Thus, we have $7/16\le f(x)\le 25/16$ and \begin{aligned} \frac{|\log\Big(1+c^2-2c\cos(2\pi x)\Big)|}{|1+c^2-2c\cos(2\pi x)|}dx &\le \frac{16}{7}\left( \log f(x)1_{f(x)\ge 1}(x) +\log(1/f(x))1_{f(x)<1}(x)\right) \\ &\le \frac{16}{7}(\log(25/6)1_{f(x)\ge 1}(x) +\log(16/7) 1_{f(x)<1}(x)). \end{aligned} The latter quantity is in $L^1(0,1/2)$ and thus by Lebesgue's Dominated Convergence Theorem, we may pass the limit inside to find that $E(f_c)\to 0$ as $c\to 0$.

  • $\begingroup$ Grr...I am sorry. As $c\rightarrow 0$ is a misprint. It should be as $c\rightarrow 1$. The other limit is not interesting for my purposes. $\endgroup$
    – Nik Quine
    Nov 20 at 19:36
  • 1
    $\begingroup$ You should probably make another question then for the limit $c\to 1$ since it is a significant edit. $\endgroup$
    – Dispersion
    Nov 20 at 19:37
  • $\begingroup$ I agree. It's obnoxious when the OP significantly changes the question after it was answered. $\endgroup$
    – Gonçalo
    Nov 21 at 20:09
  • $\begingroup$ @Gonçalo People make mistakes. I apologized to Dispersion, and I fail to see how you were harmed. $\endgroup$
    – Nik Quine
    Nov 22 at 15:10

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