# Showing a sequence of integrals converges to zero

Let $\varphi$ be a complex-valued function which is analytic on $\{z \in \mathbb C : |z| \leq 2\}$, let $\gamma$ be the unit circle in the complex plane, and define

$$F_n(z) = \int_\gamma \frac{1}{s-z} \int_{-2}^{2} e^{-nt^2} \frac{\varphi(t)}{t-s}\,dt\,ds,$$

where $|z| < 1/2$.

Main Question: Is it true that $F_n(z) \to 0$ uniformly for $|z| < 1/2$ as $n \to \infty$?

We should note that the double integral exists. Indeed, by properties of Cauchy-type integrals (see Gakhov, Boundary Value Problems or here), the function

$$g(s) = \int_{-2}^{2} e^{-nt^2} \frac{\varphi(t)}{t-s}\,dt$$

is analytic on $\mathbb C \setminus [-2,2]$ and has continuous extensions from the upper and lower half-planes to the interval $(-2,2)$ which satisfy

$$\lim_{\epsilon \to 0^+} g(x \pm i\epsilon) = \pm i\pi e^{-nx^2} \varphi(x) + \operatorname{P.V.} \int_{-2}^{2} e^{-nt^2} \frac{\varphi(t)}{t-x}\,dt$$

for $-2 < x < 2$. Consequently, $g(s)$ is continuous on $\gamma$ except for two jump discontinuities at $s = \pm 1$.

Idea for an approach

I have an idea for an approach which I have so far been unable to make rigorous. At the end are a couple of problems that I see with it that I would greatly appreciate some feedback on.

First I'd like to split the inner integral up into

$$\int_{-2}^{2} = \int_{|t| < 1} + \int_{1 < |t| < 2},$$

and so write

$$F_n(z) = \int_\gamma \frac{1}{s-z} \int_{|t| < 1} e^{-nt^2} \frac{\varphi(t)}{t-s}\,dt\,ds + \int_\gamma \frac{1}{s-z} \int_{1 < |t| < 2} e^{-nt^2} \frac{\varphi(t)}{t-s}\,dt\,ds. \tag{1}$$

Now switch the order of integration in both integrals. The first becomes

$$\int_\gamma \frac{1}{s-z} \int_{|t| < 1} e^{-nt^2} \frac{\varphi(t)}{t-s}\,dt\,ds = \int_{|t|<1} e^{-nt^2} \varphi(t) \int_\gamma \frac{ds}{(s-z)(t-s)}\,dt,$$

and the inner integral here is

$$\int_\gamma \frac{ds}{(s-z)(t-s)} = 2\pi i \left(\frac{1}{t-z} - \frac{1}{t-z}\right) = 0.$$

The second integral in $(1)$ becomes

\begin{align} \int_\gamma \frac{1}{s-z} \int_{1 < |t| < 2} e^{-nt^2} \frac{\varphi(t)}{t-s}\,dt\,ds &= \int_{1 < |t| < 2} e^{-nt^2} \varphi(t) \int_\gamma \frac{ds}{(s-z)(t-s)}\,dt \\ &= 2\pi i \int_{1 < |t| < 2} e^{-nt^2} \frac{\varphi(t)}{t-z}\,dt, \end{align}

so we conclude that

$$F_n(z) = 2\pi i \int_{1 < |t| < 2} e^{-nt^2} \frac{\varphi(t)}{t-z}\,dt.$$

Then

\begin{align} e^{n} |F_n(z)| &= 2\pi \left| \int_{1 < |t| < 2} e^{-n(t^2-1)} \frac{\varphi(t)}{t-z}\,dt \right| \\ &\leq 2\pi \cdot \operatorname{Length}(\{1 < |t| < 2\}) \cdot \sup_{1 < |t| < 2} \left( e^{-n(t^2-1)} \frac{|\varphi(t)|}{|t-z|} \right) \\ &\leq 2\pi \cdot 2 \cdot \sup_{1 < |t| < 2} \left(1 \cdot \frac{|\varphi(t)|}{1/2} \right) \\ &\leq C \end{align}

for some constant $C > 0$, so

$$|F_n(z)| \leq Ce^{-n} \to 0$$

uniformly for $|z| < 1/2$ as $n \to \infty$.

There are a couple of issues that I see with this:

• Can the interchange of order of integration be justified in both cases?

• Are the subsequent evaluations of the inner integrals using the residue theorem valid?

• Ignore the "wide audience" part of the bounty, I'm just looking for a detailed answer. – Antonio Vargas Feb 10 '15 at 1:59
• I wonder why dont you use residue theorem – Arashium Feb 16 '15 at 6:29
• @Arashium, The integral in $g(s)$ is not over a closed contour and the integrand of $\int_\gamma \frac{g(s)}{s-z}\,ds$ isn't meromorphic inside $\gamma$. Did you have another idea of how it could be applied? – Antonio Vargas Feb 16 '15 at 6:31

$$g_n(s) = \int_{-2}^{2} e^{-nt^2} \frac{\varphi(t)}{t-s}\,dt$$ is analytic on $\mathbb C \setminus [-2,2]$ and has continuous extensions from the upper and lower half-planes to the interval $(-2,2)$ which satisfy $$\lim_{\epsilon \to 0^+} g_n(x \pm i\epsilon) = \pm i\pi e^{-nx^2} \varphi(x) + \operatorname{P.V.} \int_{-2}^{2} e^{-nt^2} \frac{\varphi(t)}{t-x}\,dt$$ for $-2 < x < 2$ (consequently, $g_n(s)$ is continuous on $\gamma$ except for two jump discontinuities at $s = \pm 1$).

let $\epsilon >0$. For every $s$ in the unit circle such that $|s-1|\leq \epsilon$, or $|s-(-1)|<\epsilon$, it is clear that $|g_n(s)|$ is bounded by some positive real number $K$, no matter what are $n$ or $s$. [Edit: actually, this needs some justification for the P.V integral, see for example the Document here, which can be immediately adapted, or see the EDIT2 below]

Furthermore, ${1\over |s-z|}$ is bounded from above by some positive number $M$ for every $|z|\leq 1/2$ and $s$ on the unit circle. So, choosing $\epsilon$ sufficiently small, and denoting by $\gamma_\epsilon$ the small two parts of the unit circle near $\pm 1$, up to $\epsilon$, you can ensure that $$\Big|\int_{\gamma_\epsilon}{1\over s-z} g_n(s)ds\Big| \leq \int_{\gamma_\epsilon}MK ds$$ is as small as desired, independently of $z$ and $n$. Let us denote by $\gamma'$ the unit circle from which $\gamma_\epsilon$ has been removed. The problem is now to show uniform convergence on $\gamma'$.

Now, the function $\Big|{\varphi(t)\over t-s}\Big|$ is bounded by some positive number $L$ whenever $s$ is on $\gamma'$ and $t$ is between $-2$ and $2$, so there holds $$|g_n(s)|\leq \int_{-2}^2 Le^{-nt^2} dt \leq u_n,$$ where $u_n\to 0$ as $n\to 0$ (proof is easy).

Conclude that $\Big|\int_{\gamma'}{1\over s-z} g(s) dt ds\Big|\leq \int_{\gamma'} Mu_n ds < v_n$, for some sequence $v_n \to 0$ as $n\to 0$ (Q.E.D).

P.S : regarding your "solution", I think it is incorrect: you cannot break the integral representing $g(s)$ in two parts, for it is clear that each part diverges whenever $s=1$ or $s=-1$. The only sense that can be given to the integral at these points is via the Cauchy principal value.

EDIT2: I add here the justification that the $g_n(s)$ are uniformly bounded by $K$ for every $n$ and $s\in \gamma$ such that $|s-(\pm1)|<\epsilon$, with $\epsilon$ sufficiently small. Setting $s = x+iy$, and taking into account that
$$\lim_{y \to 0^+} g_n(x \pm iy) = \pm i\pi e^{-nx^2} \varphi(x) + \operatorname{P.V.} \int_{-2}^{2} e^{-nt^2} \frac{\varphi(t)}{t-x}\,dt,$$ it suffices to prove that $$\pm i\pi e^{-nx^2} \varphi(x) + \operatorname{P.V.} \int_{-2}^{2} e^{-nt^2} \frac{\varphi(t)}{t-x}\,dt$$ is bounded by some $K$, no matter what are $x\in [-1,1]$ or $n$. It is clear that $|i\pi e^{-nx^2}\varphi(x)|$ is bounded from above (since $\varphi$ is analytic in $[-2, 2]$ by hypothesis, hence continuous in this interval). So, it suffices to show that the P.V integral is bounded from above. The function $f(z)=e^{-nz^2}\varphi(z)$ is analytic in the domain $|z|\leq 2$, and plays the same role as the function $f$ in the document I have indicated. In other words, define the path $\gamma^{(\epsilon)}_1$ to be $[-2, x-\epsilon]\cup [x+\epsilon, 2]$, the path $\gamma^{(\epsilon)}_2$ to be $t\mapsto x+\epsilon e^{it}$, with $\pi\geq t\geq 0$, and $\gamma_3$ to be $t\mapsto 2e^{it}$, with $0\leq t\leq \pi$. Then exactly as indicated in the document, you have $$-P.V\int_{-2}^2 e^{-nt^2} \frac{\varphi(t)}{t-x}\,dt = \lim_{\epsilon\to 0} \int_{\gamma^{(\epsilon)}_2}{f(z)\over z-x}dz + \int_{\gamma_3}{f(z)\over z-x}dz = -i\pi f(x) + C,$$ for some constant $C$. This shows the contention.

• Thanks. I think the trickiest part from this point is justifying the uniform bound of $g_n(s)$ by $K$. This has given me some good ideas, and I think I have a proof which I'll try to post at a later date. – Antonio Vargas Feb 16 '15 at 17:03
• @Antonio Vargas I think there is no problem here (so I have not felt the need to add more explanations in my answer). I have now updated it. Anyway, thank you for the bounty. – MikeTeX Feb 16 '15 at 22:25