# Computing limit of an energy of a probability measure

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.

• $$\int\frac{\log\Big(1+c^2-2c\cos(2\pi x)\Big)}{1+c^2-2c\cos(2\pi x)}\,dx$$ doest exist (even if it is a monster). The key problem if the evaluation at the bounds. Nov 21 at 10:30
• 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 $(1-c^2)$. Nov 21 at 14:18

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 $$c\to 0$$.
• 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. Nov 20 at 19:36
• You should probably make another question then for the limit $c\to 1$ since it is a significant edit. Nov 20 at 19:37