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4 completed proof

The answer to your modified question is yes, ! $\tilde C_n$ tends to infinity as n goes to infinity. Edit: I'm not so sure (Phew! It took me a couple of revisions to prove this, but hopefully the calculations below are now correct).

In fact, there was a mistake in my original calculation$\tilde C_n\ge c\sqrt{n}$ for some positive constants c. What I think that you can say is that there is a positive constant $a$ such also show that the result holds as long as you restrict to $\vert x\vert^2\lt\vert y\vert^2-a$. Whether or \tilde C_n\le C\sqrt{n}$for some other constant C but I'm not we can take$a=1$(which would imply the full resultcompletely sure yet, as$\vert x\vert^2-\vert y\vert^2\in\mathbb{Z}$) requires although it should follow from a more careful consideration closer examination of the my expression below for$p_n(x)$. You can derive an asymptotic expansion for$p_n(x)$in 1/n. Evaluating this to second order terms in my calculationis enough to answer your question. In any case, After n steps the rate will be distribution of the order of$\sqrt{n}$. That is,$\tilde C_n\ge b\sqrt{n}$for some constant$b$. This can particle will be proven by evaluating approximately normal with variance 2n/5 in both dimensions, so we expect to get$p_n(x)$p_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}$ to leading orderin 1/n.(assuming I haven't made any errors below).

where $e_1=(1,0)$, $e_2=(0,1)$. In finite dimensional spaces, you would solve this by decomposing $p_0$ into a sum of eigenvectors and for large n, the dominant term of $L^np_0$ will be that corresponding to the largest eigenvalue. In this case, we can diagonalize the infinite dimensional operator L has a continuous spectrum, and is diagonalized by applying a Fourier transform.

The integral can be computed Using a Taylor expansion to leading second order. After n steps,the standard deviation of the particle's distance from the origin grows of the order of $\sqrt{n}$, so most of its distribution is spread over an area of size the order of n$\left(\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)\right)^n=e^{-\frac45\pi^2n\vert u\vert^2}\left(1+\frac{8\pi^4n}{75}(7\vert u\vert^4-20u_1^2u_2^2)+O(n\vert u\vert^6)\right).This means we expect expansion is valid over any domain on which$p_n(x)$to go to zero at rate 1/nn\vert u\vert^6$ is bounded. Let's discard all terms which vanish faster than thisSay, $\vert u\vert\le n^{-1/6}$.

The range Outside of integration can be replaced this domain, the integrand above is bounded by $[-\epsilon,\epsilon]$ e^{-cn(n^{-1/6})^2}=e^{-cn^{2/3}}$for any$0<\epsilon<1/2$, creating an error only of order$e^{-cn}$(some positive c). On such a rangeconstant c, $$\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)=\exp\left(-\frac{4\pi^2}{5}u^2+O(u^4)\rightwhich is much smaller than O(1/n^3)$$Substituting into the integral and changing variablescan be neglected. Then,p_n(x)&=\int_{[-\epsilon,\epsilon]^2}\exp\left(-\frac{4n\pi^2}{5}\vert u\vert^2+O(n\vert u\vert^4)+2\pi p_n(x)=\int_{\mathbb{R}^2}\left(1+\frac{8\pi^4n}{75}(7\vert u\vert^4-20u_1^2u_2^2)+O(n\vert u\vert^6)\right)e^{-\frac45\pi^2n\vert u\vert^2+2\pi ix\cdot u\right)\,du+O(e^{-cn})\\&=\frac{5}{8n\pi^2}\int_{[-\epsilon'\sqrt{n},\epsilon'\sqrt{n}]^2}\exp\left(-\frac12\vert v\vert^2+O(\vert v\vert^4/n)+i\sqrt{\frac{5}{2n}}x\cdot v\right)\,dv+O(e^{-cn})Where$\epsilon'=\epsilon\sqrt{8\pi^2/5}$. Rearranging the v^4/n term Here I not only substituted in the exponentialsecond order approximation to the integrand, it is possible but also extended the range of integration out to show that infinity. This is fine, because it only contributes can be shown that the value of this integral over$\vert u\vert\ge n^{-1/6}$has size of the order of 1/n to no more than$e^{-cn^{2/3}}$, so vanishes much faster than$O(1/n^3)$. Substituting in$v=\sqrt{\frac{8n}{5}}\pi u$also shows that the integral. Integrals like$\int_{\epsilon\sqrt{n}}^\infty\exp(-x^2/2)\,dx$vanish O(nu^6)$ term in the integrand vanishes at rate $\exp(-\epsilon^2n/2)/n$, 1/n^3$, giving the following.p_n(x)&=\frac{5}{8n\pi^2}\int_{\mathbb{R}^2}\exp\left(-\frac12\vert p_n(x)=\frac{5}{8\pi^2n}\int_{\mathbb{R}^2}\left(1+\frac{1}{24n}(7\vert v\vert^4-20v_1^2v_2^2)\right)e^{-\frac12\vert v\vert^2+i\sqrt{\frac{5}{2n}}x\cdot v\right)\,dv+O(1/n^2)\\&=\frac{5}{4n\pi}\exp\left(-\frac{5\vert x\vert^2}{4n}\right)+O(1/n^2)SoThis integral can be computed,if$$p_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+\frac{1}{24n}\left(36-\frac{90}{n}\vert x\vert^2+\frac{175}{4n^2}\vert x\vert^4-\frac{125}{n^2}x_1^2x_2^2\right)\right)+O(n^{-3}).This is a bit messy, but the exact coefficients are not too important. What matters is the general form of the expression.The leading order term also agrees with the guess above based on it being approximately normal.Also, for any fixed$\vert x\vert\lt\vert x\vert \lt\vert y\vert$then , the leading order term in$p_n(x)\gt p_n(y)$p_n(x)-p_n(y)$ will dominate for large n, giving $p_n(x)\gt p_n(y)$. You can check that if So, $\tilde C_n\to\infty$.

Consider $\vert x\vert\lt\vert y\vert-a$ x\vert\le c\sqrt{n}$for some constant a$c\le1$. Then,p_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+\frac{3}{2n}\right)+O(c^2n^{-2}).If$\vert x\vert\lt\vert y\vert\le c\sqrt{n}$then$\vert y\vert^2-\vert x\vert^2\ge 1$(as it is a nonzero integer)p_n(x)-p_n(y)&=\frac{5}{4n\pi}e^{-\frac{5\vert x\vert^2}{4n}}(1-e^{\frac{5(\vert x\vert^2-\vert y\vert^2)}{4n}})+O(1/n^2)\&=\frac{5}{4\pi n}\left(1+\frac{3}{2n}\right)e^{-\frac{5}{4n}\vert x\vert^2}\left(1-e^{-\frac{5}{4n}(\vert y\vert^2-\vert x\vert^2)}\right)+O(c^2n^{-2})\\&\ge\frac{5}{4n\pi}e^{-\frac{5\vert x\vert^2}{4n}}(1-e^{-\frac{5a}{4n}})+O(1/n^2)\&\ge\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}(1-e^{-\frac{5}{4n}})+O(c^2n^{-2})\\&\ge\frac{25}{4n^2\pi}\left(ae^{-\frac{5\vert x\vert^2}{4n}}+O(1)\right)&=\frac{25}{16\pi n^2}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+O(c^2)\right).To be certain that this As long as c is positive, we would need to evaluate chosen small enough that the constant$O(c^2)$term in the final is always greater than -1, this expression and compare to will be positive. So$a$.p_n(x)\gt p_n(y)$ for all $\vert x\vert\lt\vert y\vert\le c\sqrt{n}$, giving $\tilde C_n\ge c\sqrt{n}$.

3 fixed mistake

The answer to your modified question is yes, $\tilde C_n$ tends to infinity as n goes to infinity. AlsoEdit: I'm not so sure now, the rate there was a mistake in my original calculation. What you can say is at least that there is a positive constant $\sqrt{\frac45n\log n}$a$such that the result holds as long as you restrict to$\vert x\vert^2\lt\vert y\vert^2-a$. That is,Whether or not we can take$a=1$(which would imply the full result, as$\liminf_{n\to\infty}\frac{\tilde C_n}{\sqrt{\frac45n\log n}}\ge1\vert x\vert^2-\vert y\vert^2\in\mathbb{Z}$) requires a more careful consideration of the second order terms in my calculation. In any case, the rate will be of the order of$\sqrt{n}$. That is,$\tilde C_n\ge b\sqrt{n}$for some constant$b$. So, if$\vert x\vert\lt\vert y\vert$then$p_n(x)\gt p_n(y)$for large n. You can check from this expression that if$\vert x\vert\lt c\sqrt{\frac45n\log n}$(any x\vert\lt\vert y\vert-a$ for some constant a,$c\lt1$) $p_n(x)-p_n(y)&=\frac{5}{4n\pi}e^{-\frac{5\vert x\vert^2}{4n}}(1-e^{\frac{5(\vert x\vert^2-\vert y\vert^2)}{4n}})+O(1/n^2)\\&\ge\frac{5}{4n\pi}e^{-\frac{5\vert x\vert^2}{4n}}(1-e^{-\frac{5a}{4n}})+O(1/n^2)\\&\ge\frac{25}{4n^2\pi}\left(ae^{-\frac{5\vert x\vert^2}{4n}}+O(1)\right)To be certain that this is enough for positive, we would need to evaluate the first constant term to dominate in the inequality.final expression and compare to$a$. 2 added proof of extended question The answer to your modified question is yes,$\tilde C_n$tends to infinity as n goes to infinity. Also, the rate is at least$\sqrt{\frac45n\log n}$. That is,\liminf_{n\to\infty}\frac{\tilde C_n}{\sqrt{\frac45n\log n}}\ge1.This can be proven by evaluating$p_n(x)$to leading order in 1/n. (assuming I haven't made any errors below). The idea is to note that you are repeatedly applying a linear operator,p_{n+1}=Lp_n,\ Lp(x) \equiv (p(x)+p(x-e_1)+p(x+e_1)+p(x-e_2)+p(x+e_2))/5where$e_1=(1,0)$,$e_2=(0,1)$. In finite dimensional spaces, you would solve this by decomposing$p_0$into a sum of eigenvectors and for large n, the dominant term of$L^np_0$will be that corresponding to the largest eigenvalue. In this case, we can diagonalize by applying a Fourier transformp_0(x)=1_{\lbrace x=0\rbrace}=\int_{-[\frac12,\frac12]^2}e^{2\pi ix\cdot u}\,du.Noting that$e^{2\pi ix\cdot u}$(as a function of x) is an eigenvector of L,Le^{2\pi ix\cdot u}=\left(\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)\right)e^{2\pi ix\cdot u}gives the following for$p_n$,p_n(x)=L^np_0(x)=\int_{[-\frac12,\frac12]^2}\left(\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)\right)^ne^{2\pi ix\cdot u}\,du.The term inside the parentheses is less than 1 in absolute value everywhere away from the origin, so looks like a Dirac delta when raised to a high power n. The integral can be computed to leading order. After n steps, the standard deviation of the particle's distance from the origin grows of the order of$\sqrt{n}$, so most of its distribution is spread over an area of size the order of n. This means we expect$p_n(x)$to go to zero at rate 1/n. Let's discard all terms which vanish faster than this. The range of integration can be replaced by$[-\epsilon,\epsilon]$for any$0<\epsilon<1/2$, creating an error only of order$e^{-cn}$(some positive c). On such a range,\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)=\exp\left(-\frac{4\pi^2}{5}u^2+O(u^4)\right)Substituting into the integral and changing variables,p_n(x)&=\int_{[-\epsilon,\epsilon]^2}\exp\left(-\frac{4n\pi^2}{5}\vert u\vert^2+O(n\vert u\vert^4)+2\pi ix\cdot u\right)\,du+O(e^{-cn})\\&=\frac{5}{8n\pi^2}\int_{[-\epsilon'\sqrt{n},\epsilon'\sqrt{n}]^2}\exp\left(-\frac12\vert v\vert^2+O(\vert v\vert^4/n)+i\sqrt{\frac{5}{2n}}x\cdot v\right)\,dv+O(e^{-cn})Where$\epsilon'=\epsilon\sqrt{8\pi^2/5}$. Rearranging the v^4/n term in the exponential, it is possible to show that it only contributes of the order of 1/n to the integral. Integrals like$\int_{\epsilon\sqrt{n}}^\infty\exp(-x^2/2)\,dx$vanish at rate$\exp(-\epsilon^2n/2)/n$, givingp_n(x)&=\frac{5}{8n\pi^2}\int_{\mathbb{R}^2}\exp\left(-\frac12\vert v\vert^2+i\sqrt{\frac{5}{2n}}x\cdot v\right)\,dv+O(1/n^2)\\&=\frac{5}{4n\pi}\exp\left(-\frac{5\vert x\vert^2}{4n}\right)+O(1/n^2)So, if$\vert x\vert\lt\vert y\vert$then$p_n(x)\gt p_n(y)$for large n. You can check from this expression that$\vert x\vert\lt c\sqrt{\frac45n\log n}$(any$c\lt1\$) is enough for the first term to dominate in the inequality.

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