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 $-\frac{N_-}{2a_L}$:


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?

  • 1
    $\begingroup$ There's no problem with DCT for complex-valued functions. See any book on measure theory... (or for that matter show for yourself that DCT for complex-valued functions is immediate from DCT for real-valued functions). $\endgroup$ – David C. Ullrich Jul 16 '15 at 15:40
  • $\begingroup$ The main problem is that I don't know measure theory all that well (though I never took a formal course in it, I know probability theory pretty well at the graduate engineering student level). In this context, I am not sure how function $g$ would look like. Is $|f_n(x)|$ simply the magnitude of a complex-valued function $f_n(x)$ and $g(x)$ has to be greater than that magnitude for all $x$? $\endgroup$ – M.B.M. Jul 16 '15 at 15:59
  • 1
    $\begingroup$ You might consider learning some measure theory... Yes, DCT looks exactly the same: If $f_n\to f$ almost surely, $|f_n|\le g$ and $\int g<\infty$ then $\int f_n\to\int f$. $\endgroup$ – David C. Ullrich Jul 16 '15 at 16:05
  • $\begingroup$ @DavidC.Ullrich what if the dominating function $g$ is also complex valued ? Then $|f_n|\leq g$ doesn't make sense, perhaps better ask $|f_n|\leq |g|$ where $|.|$ is the usual complex modulus $\endgroup$ – Alonso Delfín Jul 16 '15 at 16:14
  • 2
    $\begingroup$ @AlonsoDelfín Ok, to be fair to everyone, I've just now found places online where DCT is stated only for real-valued functions. That's just stupid. Trivial proof that the real DCT implies the complex DCT: Say $f_n$ is complex-valued, $f_n\to f$ as, $|f_n|\le g$ and $\int g<\infty$. Write $f_n=u_n+iv_n$, $f=u+iv$. Then $|u_n|\le|f_n|\le g$, so the real DCT shows $\int u_n\to\int u$. Similarly for $v_n$. Hence $\int f_n\to \int f$. $\endgroup$ – David C. Ullrich Jul 16 '15 at 16:35

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

$$\lim_n\int f_n (x)\, dx = \lim_n\bigg(\Re\bigg(\int f_n (x)\, dx\bigg) + i \Im\bigg(\int f_n (x)\, dx\bigg)\bigg) \\= \int \lim_n\Re f_n (x)\, dx +i \int \lim_n\Im f_n(x)\, dx $$

  • $\begingroup$ Thanks for writing out the answer. I just want to make two notes: 1) it mirrors @DavidC.Ullrich's comment above, and; 2) the "local limit theorem"-like result in the example seems to hold using this flavor of DCT (one could use the same dominating function $g(t)=\exp[-t^2]$ for both real and imaginary parts since sine and cosine are bounded in $[-1,1]$). $\endgroup$ – M.B.M. Jul 16 '15 at 18:55
  • $\begingroup$ I don't fully understand your point 2), could you explain it in greater detail? $\endgroup$ – Conrado Costa Jul 16 '15 at 19:02
  • $\begingroup$ "Example" refers to the "specific example" I gave in my question (sorry that it's long). Your answer allows me to interchange the limit and the integral at the very end of the "specific example." This shows that the probability density function of the random variable $A$ (which you recover from the characteristic function using Fourier transform) approaches Gaussian in the limit (after dividing by $2\pi$, which I omitted for brevity). Convergence in probability described earlier in "specific example" is central-limit-theorem-like, while convergence of densities is "local". Does this explain? $\endgroup$ – M.B.M. Jul 16 '15 at 19:13

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