Prove that $ \lim_{t\to\infty} \frac{\int_{-\infty}^\infty\rho_t(x)\,e^{xt}\cos(\omega xt)\, dx}{\int_{-\infty}^\infty\rho_t(x)\,e^{xt}\, dx}=0.$

Question

I have a continuous family of functions $\rho_t(x)$ that converge pointwise to a Gaussian $$ \lim_{t\to\infty}\rho_t(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}. $$ I would like to prove that $$ \lim_{t\to\infty} \frac{\int_{-\infty}^\infty\rho_t(x)\,e^{xt}\cos(\omega xt)\, dx}{\int_{-\infty}^\infty\rho_t(x)\,e^{xt}\, dx}=0. $$ Usually, one would argue that the numerator has a rapidly oscillating function at large $t$ multiplied by a function that is becoming smooth and should decay to zero, but the $e^{xt}$ term makes that argument difficult.

You can assume $\rho_t(x)$ is well-behaved (whatever smoothness condition is necessary).

Answer 1

I'll start with the answer to another question: Given a sequence $(f_t)_t$ of probability density functions (pdfs), what property ensures that $\lim_{t\rightarrow\infty}\int_{-\infty}^\infty f_t(x)\cos(tx)\mathrm{d}x=0$? The answer is that $f_t$ should not fluctuate too much locally, compared to the increasing frequency $t$. Let $\ell=2\pi/t$ be the period, fix the start point $i_t\in\mathbb Z$, interval sizes $I_t\in\mathbb Z_{>0}$ and the interval count $L_t\in\mathbb Z_{>0}$. For $k=1,...,L_t$ let $\mathcal I_{t,k}=[i_t\ell+(k-1)I_t\ell,i_t\ell+kI_t\ell)$ and let $\mathcal I_{t,0}=\mathbb R\setminus\bigcup_k\mathcal I_{t,k}$ (We stick to the grid $\ell\mathbb Z$ and partition an interval into smaller parts somewhere). The discretization $d_t:\mathbb R\rightarrow\mathbb R_{\ge 0}$ of $f_t$ is given by $d_t(x)=0$ for $x\in\mathcal I_{t,0}$ and $d_t(x)=\frac{1}{I_t\ell}\int_{\mathcal I_{t,k}}f(y)\mathrm dy$ for $x\in\mathcal I_{t,k}$ and $k=1,\dots,L_t$, a simple step function. Finally, we want the approximation to be good, so assume that $\lim_{t\rightarrow\infty}\|f_t-d_t\|_1=0$ (the $\mathcal L_1$-norm $\|f_t-d_t\|_1=\int|f_t-d_t|\mathrm dt$, or total variation, is just the integral over $\mathbb R$). This cumbersome description only says that we reasonably approximate $f_t$ with rectangles (cf. CLT & integral definitions). Finally, we present the first result. $$ \lim_{t\rightarrow\infty}\left|\int f_t(x)\cos(tx)\mathrm dx\right|\le \lim_{t\rightarrow\infty}\left(\left|\int d_t(x)\cos(tx)\mathrm dx\right|+\|f_t-d_t\|_1\right)=0, $$ where we used linearity of the integral, the triangle inequality, monotonicity of the integral for the latter part (and our assumption that the discretization error vanishes), and for the former part we simply use that the standard Lebesgue integral of $\cos$ over a multiple of the period is $0$, i.e. $\int_{\mathcal I_{t,k}}c\cos(tx)\mathrm dx=0$, for any $c\in\mathbb R$. This very simple and intuitive result is at the core of the problem. The grid $\ell\mathbb Z$ is intuitive but not necessary, we could place the points anywhere, as long as the distances are not too small, then the "rounding errors" have little effect.

What are sufficient conditions for $f_t$ to have such a property? I will discuss two options, since both were already mentioned.

1. The functions $f_t$ are well-behaved. I will only discuss the simplest case, to make a point. Assume that there exists $C\in\mathbb R_{>0}$ such that all $f_t$ are $C$-Lipschitz and an interval $\mathcal I^*_t$ such that $\lim_{t\rightarrow\infty}t^{-2}\int_{\mathcal T_t}f(t)\mathrm{d}t=0$ on the tails $\mathcal T_t=\mathbb R\setminus\mathcal I^*_t$. Set $i_t=\inf\mathcal I^*_t$ to be the beginning of the interval. By the mean value theorem for each $t$ and $k$ we find $x\in\mathcal I_{t,k}$ such that $f_t(x)=d_t(x)$ is the mean of the interval. Since $f_t$ is $C$-Lipschitz and the maximum distance of points in the interval is $I_t\ell$, we have $|f(x')-d(x')|=|f(x')-d(x)|=|f(x')-f(x)|\le C|x'-x|\le CI_t\ell$. This further yields that $\int_{\mathcal I_{t,k}}| f_t(x)-d_t(x)|\mathrm dx\le C(I_t\ell)^2$ and thereby $\|f_t-d_t\|_1\le C(I_t\ell)^2L_t+\int_{\mathcal I_{t,0}}f_t(x)\mathrm dx$ Since the total number of segments is $L_tI_t$ in any case, we minimize the error by setting $I_t=1$. Now, we can cover almost $t^2$ segments, and by the assumption all of $\mathcal I^*_t$, while still achieving $\lim_{t\rightarrow\infty}\|f_t-d_t\|_1=0$. Yes, this can be further improved, significantly. A pathological case is the uniform distribution with huge bounded support and sharp ends. This yields a large uniform Lipschitz constant and a huge area to cover. Clearly, this can be dealt with by considering local Lipschitz constants (or other properties that ensure the function to be well-behaved).
2. There exists a $C$-Lipschitz limit distribution with pdf $f^*$, i.e. $\lim_{t\rightarrow\infty}\|f_t-f^*_t\|_1=0$, where $f^*_t(x)=f^*(x-s_t)$ is $f^*$ shifted by some value $s_t$. Let $d_t$ be the discretization for $f^*_t$, then for any sequence of intervals with $\ell i_t\rightarrow-\infty$, $I_t=1$, $L_tt^{-2}\rightarrow 0$ slowly, $\ell i_t+L_t\ell\rightarrow\infty$ centered at $s_t$ (notice that the total length is almost $t$, so this is easily feasible) we have $\|f^*_t-d_t\|_1\le C\ell^2L_t+\int_{\mathcal I_{t,0}}f^*_t(x)\mathrm dx\rightarrow 0$, where the second contribution vanishes because our interval of length almost $t$ eventually occupies all of $\mathbb R$. But now, since $f_t$ "converges" to $f^*_t$, the triangle inequality yields $\|f_t-d_t\|_1\rightarrow 0$. Notice that $\|f_t-f^*_t\|_1\rightarrow 0$ is weaker than the uniform convergence $\|\frac{f_t}{f^*_t}-1\|_\infty\rightarrow 0$ of the Radon-Nikodym derivative.

The second case is what we are trying to accomplish here, the first case is what is possible to achieve with continuity/differentiability/boundedness assumptions on the sequence, without even requiring a limit in any sense. We finally turn to the problem. Let $\phi(x)=\frac{1}{\sqrt{2\pi}}$ be the pdf of the normal distribution, $w_t(x)=\rho_t(x)e^{tx}$, $Z_t=\int w_t(x)\mathrm dx$, and $f_t(x)=w_t(x)/Z_t$. With $r_t(x)=\rho_t(x)/\phi(x)$, $w^*_t(x)=\rho_t(x)e^{tx-\frac{1}{2}t^2}$ and $Z^*_t=\int w^*_t(x)\mathrm d x$ the problem is given by $$ \lim_{t\rightarrow\infty}\int f_t(x)\cos(t\omega x)\mathrm dx,\, f_t(x)=\frac{w_t(x)}{Z_t}=\frac{w^*_t(x)}{Z^*_t},\\ w^*_t(x)=r_t(x)\phi(x)e^{tx-\frac{1}{2}t^2}=\frac{r_t(x)}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-t)^2}=r_t(x)\phi(x-t). $$ Clearly, this reflects the second case above, where $f_t$ converges to a shifted limit distribution. Thus, we are almost done, we only need a reasonable assumption that yields the convergence $f_t(x)\rightarrow\phi(x-t)$ in total variation. So, using $\phi_t(x)=\phi(x-t)$ we assume that $\|\phi_t-r_t\phi_t\|_1\rightarrow 0$, then we have $$ |Z^*_t-1|=\left|\int r_t(x)\phi_t(x)\mathrm dx-\int\phi_t(x)\mathrm dx\right|\le\|\phi_t-r_t\phi_t\|_1\rightarrow 0, $$ which gives $\|f_t-\phi_t\|_1\le\|f_t-Z_t^*f_t\|_1+\|r_t\phi_t-\phi_t\|_1\rightarrow 0$, since all $f_t$ are probability distributions and hence $\|f_t-Z^*_tf_t\|_1=\int f_t(x)|1-Z^*_t|\mathrm dx=|1-Z^*_t|\rightarrow 0$. Finally, notice that $\|r_t-1\|_\infty\rightarrow 0$ implies $\|\phi_t-r_t\phi_t\|_1=\int\phi_t(x)|r_t(x)-1|\mathrm dx\le\int\phi_t(x)\|r_t-1\|_\infty\mathrm dx=\|r_t-1\|_\infty\rightarrow 0$ (although, honestly, the fact that most of the mass is at an arbitrarily large and variable distance $t$ from the expectation, essentially requires that the RN derivative is uniformly bounded, at least on one side).

As a very last remark: On the very long way here I passed this post, which covers a very nice integration using the differential equation approach, in case you're interested.

Answer 2

We could do the following. Denoting $\displaystyle \rho_0(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ $$I_1=\int_{-\infty}^\infty\rho_0(x)\,e^{xt}\cos(\omega xt)\, dx=\frac{1}{\sqrt{2\pi}}\Re\int_{-\infty}^\infty e^{-x^2/2+xt+i\omega t x}\, dx$$ Making the substitution $x=ts$ $$=\frac{t}{\sqrt{2\pi}}\Re\int_{-\infty}^\infty e^{-s^2t^2/2+t^2s+t^2i\omega s}\, ds=\Re\frac{t\,e^{\frac{t^2(1+i\omega)^2}{2}}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{t^2(s-1-i\omega)^2}{2}}\, ds$$ Making the substitution $s\to s-1$ $$I_1=\Re\frac{t\,e^{\frac{t^2(1+i\omega)^2}{2}}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{t^2(s-i\omega)^2}{2}}\, ds$$ Now we can change the integration path in the complex plane, switching from $s$ to $s-i\omega$.

Formally it comes from the consideration of the closed rectangular contour, clockwise (at $R\to\infty$): $-R\to R\to R-i\omega\to-R-i\omega\to-R$. There are no singularities inside the contour; therefore, $\oint=0$.

We also note that the side integrals tend to zero as $R\to \infty$; for example, $$\Big|\int_R^{R-i\omega}e^{-\frac{t^2(z-i\omega)^2}{2}}\, dz\Big|=\Big|\int_{-\omega}^0e^{-\frac{t^2(R^2+2iyR-y^2)}{2}}\, dy\Big|<\omega\,e^{-\frac{t^2R^2}{2}+\frac{t^2\omega^2}{2}}\to 0\,\,\text{at}\,\,R\to\infty$$ Therefore, $$I_1=\Re\frac{t\,e^{\frac{t^2(1+i\omega)^2}{2}}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{t^2s^2}{2}}\, ds=e^{\frac{t^2(1-\omega^2)}{2}}\Re e^{i\omega t^2}=e^{\frac{t^2(1-\omega^2)}{2}}\cos(\omega t^2)\tag{1}$$ Now, we can present $\rho_t(x)$ in the form $$\rho_t(x)=\rho_0(x)g(x,t);\,\,g(x,t)\to 1 \,\,\text{at}\,\, t\to\infty$$ We will use the condition of uniform convergence to proceed (it would also be interesting to see a rigorous prove for a weaker condition - of pointwise convergence alone). So, we suppose that at $t>T_0\,$ there is such $\epsilon(T_0)\,$ that $\,\,\displaystyle 1-\epsilon (T_0)<g(x,t)<1+\epsilon(T_0)\,\,$ for all $x\in(-\infty;\infty)$.

Evaluating also $$I_2=\int_{-\infty}^\infty\rho_0(x)\,e^{xt}\, dx=e^{\frac{t^2}{2}}\tag{2}$$ we can make the following evaluation of the integrals ratio for all $t>T_0$ $$0<\frac{\big|\int_{-\infty}^\infty\rho_t(x)\,e^{xt}\cos(\omega xt)\, dx\big|}{\int_{-\infty}^\infty\rho_t(x)\,e^{xt}\, dx}<\frac{1+\epsilon}{1-\epsilon}\bigg|\frac{I_1}{I_2}\bigg|=\frac{1+\epsilon}{1-\epsilon}\,\bigg|\frac{e^{\frac{t^2(1-\omega^2)}{2}}\cos(\omega t^2)}{e^{\frac{t^2}{2}}}\bigg|$$ $$<\frac{1+\epsilon}{1-\epsilon}e^{-\frac{t^2\omega^2}{2}}\to0\,\,\text{at}\,\,t\to\infty\,\, \text{and}\,\,\omega\neq 0$$