Suppose there is a sequence $\{f_n(x)\}$ such that $\lim_{n\rightarrow\infty}f_n(x)=f(x)$.
I've previously used the dominated convergence theorem for interchanging the limit and the integral in $\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty} f_n(x)dx$ when $f_n(x)$ was a real-valued function $f_n:\mathbb{R}\rightarrow\mathbb{R}$. Per the usual steps, I would find an integrable function $g:\mathbb{R}\rightarrow\mathbb{R}$ such that $|f_n(x)|<g(x)$ for all $n$ in the index set and $x\in\mathbb{R}$. I would thus justify $\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty} f_n(x)dx=\int_{-\infty}^{\infty} \lim_{n\rightarrow\infty}f_n(x)dx=\int_{-\infty}^{\infty}f(x)dx$.
I am wondering what happens when $f_n(x)$ as well as the limiting function $f(x)$ are complex-valued, i.e., $f_n:\mathbb{R}\rightarrow\mathbb{C}$, but their integral is real-valued. How do I safely interchange the limit and the integral?
Specific example
The semi-classical theory of optical homodyne detection (see section on homodyne detection, for example, here) involves subtracting two independent Poisson random variables. The resulting random variable, when appropriately normalized, converges to a Gaussian random variable in distribution. I am wondering if a stronger result holds, where the density function converges pointwise to the Gaussina density as well.
Consider Poisson random variables $N_-$ and $N_+$ with respective means $a^2_-=\frac{1}{2}(a_S-a_L)^2$ and $a^2_+=\frac{1}{2}(a_S+a_L)^2$. Here, $a_S$ is the amplitude of the signal field, and $a_L$ is the amplitude of the much-stronger local oscillator field. $N_-$ and $N_+$ are the photon counts at the two arms of the homodyne detector. To recover the signal at the output, we subtract the two counts (and normalize), which effectively cancels the local oscillator field.
Thus, consider the random variable $A=\frac{N_+-N_-}{2a_L}$. Its characteristic function is just the product of the characteristic functions of the Poisson random variables $\frac{N_+}{2a_L}$ and $-\frac{N_-}{2a_L}$:
$$\phi_A(t)=\exp\left[a_+^2(e^{it/2a_L}-1)+a_-^2(e^{-it/2a_L}-1)\right],$$
where $i=\sqrt{-1}$. Now, taking the limit as $a_L\rightarrow\infty$ yields:
$$\begin{align}\lim_{a_L\rightarrow\infty}\phi_A(t)&=\lim_{a_L\rightarrow\infty}\exp\left[a_+^2(e^{it/2a_L}-1)+a_-^2(e^{-it/2a_L}-1)\right]\\
&=\lim_{a_L\rightarrow\infty}\exp\left[a_+^2\left(\frac{it}{2a_L}-\frac{t^2}{8a_L^2}+\mathcal{O}(a_L^{-3})\right)+a_-^2\left(-\frac{it}{2a_L}-\frac{t^2}{8a_L^2}+\mathcal{O}(a_L^{-3})\right)\right]\\
&\begin{aligned}=\lim_{a_L\rightarrow\infty}\exp\left[\frac{1}{2}(a_S^2+2a_Sa_L+a_L^2)\left(\frac{it}{2a_L}-\frac{t^2}{8a_L^2}+\mathcal{O}(a_L^{-3})\right)\\\qquad\qquad\qquad\qquad\qquad+\frac{1}{2}(a_S^2-2a_Sa_L+a_L^2)\left(-\frac{it}{2a_L}-\frac{t^2}{8a_L^2}+\mathcal{O}(a_L^{-3})\right)\right]\end{aligned}\\
&=\lim_{a_L\rightarrow\infty}\exp\left[ita_S-\frac{t^2}{8}-\frac{t^2a_S^2}{8a_L^2}+\mathcal{O}(a_L^{-1})\right]\\
&=\exp\left[ita_S-\frac{t^2}{8}\right],
\end{align}$$
which is the characteristic function of Gaussian random variable with mean $a_S$ and variance $\frac{1}{4}$, thus proving convergence of $A$ to Gaussian in distribution.
Now, applying the inverse Fourier transform to $\phi_A(t)$ yields the density function of $A$. I am wondering if it's Gaussian in the limit of large $a_L$. It would be if:
$$\begin{align}\lim_{a_L\rightarrow\infty}\int_{-\infty}^{\infty}\exp\left[-it(x-a_S)-\frac{t^2}{8}-\frac{t^2a_S^2}{8a_L^2}+\mathcal{O}(a_L^{-1})\right]dt=\\
\qquad\int_{-\infty}^{\infty}\lim_{a_L\rightarrow\infty}\exp\left[-it(x-a_S)-\frac{t^2}{8}-\frac{t^2a_S^2}{8a_L^2}+\mathcal{O}(a_L^{-1})\right]dt.\end{align}$$
Can someone help?
Best Answer
Use bounded convergence theorem for each coordinate. Since $|f_n|\leq g$ implies $|\Re f_n| \leq g$ and $|\Im f_n| \leq g$ ($\Re z$ is the real part of $z $ for $z \in \Bbb{C}$, and $\Im z$ is the immaginary part of $z$) $$\Re\bigg(\int f_n (x)\, dx\bigg) = \int \Re f_n(x)\, dx \\ \Im\bigg(\int f_n (x)\, dx\bigg) = \int \Im f_n(x)\, dx$$
Then the bounded convergence theorem yields $$\lim_n\Re\bigg(\int f_n (x)\, dx\bigg) = \int \lim_n\Re f_n(x)\, dx \\ \lim_n\Im\bigg(\int f_n (x)\, dx\bigg) = \int \lim_n\Im f_n(x)\, dx$$
And conclude noting that
$$\lim_n\int f_n (x)\, dx = \lim_n\bigg(\Re\bigg(\int f_n (x)\, dx\bigg) + i \Im\bigg(\int f_n (x)\, dx\bigg)\bigg) \\= \int \lim_n\Re f_n (x)\, dx +i \int \lim_n\Im f_n(x)\, dx $$