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Here is my question:

Let $f:[0,1]\to\mathbb{C}$ be a continuous function. Show that $$g(z) = \int_0^1 f(t)e^{tz^2} dt$$ defines an entire function.

I know that one approach can be made by using Morera's Theorem.

Let $C$ be a closed contour. Applying Fubini's Theorem, we get $$\oint_C g(z)dz = \oint_C \int_0^1 f(t)e^{tz^2}dtdz = \int_0^1 f(t) \left( \oint_C e^{tz^2} dz \right)dt = \int_0^1 0dt = 0.$$ If we show that $g$ is continuous in $\mathbb{C}$, the proof is completed. However, I am not sure how to do it. For instance, I think that the continuity follows from the fact that the integrand $f(t)e^{tz^2}$ is continuous for all $(t,z)\in[0,1]\times\mathbb{C}$, but I don't know what theorem I should use to conclude it.

Mittens
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Lord Vader
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  • @Thomas Sorry, I wrote it wrong. It's $dt$. – Lord Vader Apr 19 '23 at 18:01
  • Also in the title. – Paul Frost Apr 19 '23 at 18:10
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    Dominated convergence is one way , or really just uniform convergence on compact sets, even more direct epsilon delta proofs using uniform continuity are also possible. – peek-a-boo Apr 19 '23 at 18:22
  • @peek-a-boo In order to use DCT, what integrable function should I consider to be the dominant one? I don't see how to bound the exponential function in that case. – Lord Vader Apr 19 '23 at 18:27
  • Continuity is a local property. So fix a point and a compact set around it. You can bound by a constant. Since the point was arbitrary, you’re done. btw here is a more complicated example, but the OP there also had the same confusion. Often times, exploiting locality of the properties makes things extremely easy. – peek-a-boo Apr 19 '23 at 18:27
  • Does this answer your question? Upper bound on difference quotient with Cauchy Integral Formula – Mittens Apr 19 '23 at 19:00
  • @peek-a-boo I get your idea. Can you tell me if the following proof is correct? Thanks in advance. Let $z$ be a fixed complex number, and let $(z_n)$ be a sequence in $\mathbb{C}$ that converges to $z$. Let $K$ be a compact subset of $\mathbb{C}$ that contains the whole sequence, which exists because the sequence is bounded. Since the funcion $f(t)e^{tz^2}$ is continuous in $[0,1]\times K$, there exists a constant $M$ such that $|f(t)e^{t(z_n)^2}|\leq M$ for all $n$. Therefore, we can apply the DCT to conclude $$\int_0^1 f(t)e^{t(z_n)^2}dt \to \int_0^1 f(t)e^{tz^2}dt.$$ – Lord Vader Apr 19 '23 at 19:03
  • All this can be seen as a particular example of a result (problem in Rudin's RCA) studied many times in MSE: Theorem Let $\mu$ be a complex measure on a measurable space $(X,\mathscr{B})$ and let $D\subset\mathbb{C}$ be open. Suppose $\varphi$ is a bounded complex valued function in $D\times X$ such that $\varphi(\cdot,x)$ is holomorphic in $D$ for each $x\in X$, and that $\varphi(z,\cdot)$ is $\mathscr{B}$--measurable for each $z\in D$. Define $$ f(z):=\int_X \varphi(z,x)\mu(dx),\qquad z\in D. $$ Then $f$ is holomorphic in $D$. – Mittens Apr 19 '23 at 19:04
  • yes, that’s right. Also, one more thing I should say is that although yes you can find a $K$ containing all the $z_n$’s, we don’t really need it. Since $z_n\to z$, if you fix a ball around $z$, then the tail of the sequence lies inside that ball, so you can take $K$ to be the closure of that ball. I think it will be good to realize that you can actually get away with things even if you don’t invoke as much. Anyway, I’ve written several times about holomorphy of integrals, you just need to search MSE. – peek-a-boo Apr 19 '23 at 19:06
  • @OliverDíaz I had not thought it in that way. Thanks!! – Lord Vader Apr 19 '23 at 19:09

2 Answers2

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This is a standard Complex Analysis theorem:

Theorem: Let $U$ be an open subset of $\Bbb C$, let $[a,b]$ be a closed and bounded interval of $\Bbb R$, and let $F\colon[a,b]\times U\longrightarrow\Bbb C$ be a continuous function such that, for each $t\in[a,b]$, the function$$\begin{array}{ccc}U&\longrightarrow&\Bbb C\\z&\mapsto&F(t,z)\end{array}$$is analytic. Then the function$$\begin{array}{rccc}f\colon&U&\longrightarrow&\Bbb C\\&z&\mapsto&\int_a^bF(t,z)\,\mathrm dt\end{array}$$is also analytic.

And it's not hard to prove. Take $z_0\in U$ and take $r>0$ such that $\overline{B_r(z_0)}\subset U$. For each $t\in[0,2\pi]$, let $\gamma(t)=z_0+re^{it}$. Then, for each $t\in[a,b]$ and each $z\in B_r(z_0)$,$$F(t,z)=\frac1{2\pi i}\int_\gamma\frac{F(t,u)}{u-z}\,\mathrm du$$and therefore\begin{align}f(z)&=\int_a^bF(t,z)\,\mathrm dt\\&=\int_a^b\frac1{2\pi i}\int_\gamma\frac{F(t,u)}{u-z}\,\mathrm du\,\mathrm dt\\&=\frac1{2\pi i}\int_\gamma\frac{\int_a^bF(t,u)\,\mathrm dt}{u-z}\,\mathrm du.\end{align}So, since the function $u\mapsto\int_a^bF(t,u)\,\mathrm dt$ is continuous, $f|_{B_r(z_0)}$ is analytic, because then the function$$\begin{array}{ccc}B_r(z_0)&\longrightarrow&\Bbb C\\z&\mapsto&\displaystyle\int_\gamma\frac{\int_a^bF(t,u)\,\mathrm dt}{u-z}\,\mathrm du\end{array}$$is analytic. Since the restriction of $f$ to each disk $B_r(z_0)$ is analytic, $f$ is analytic.

Mittens
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José Carlos Santos
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  • I think OP is unsure how to demonstrate that $u\mapsto\int_a^b F(t,u),\mathrm{d}t$ is continuous. Also maybe we can note that the differentiability of: $$z\mapsto\int_\gamma(u-z)^{-1}h(u),\mathrm{d}u$$When $h$ is continuous, is an easy thing to show via difference quotients and needs nothing else that is fancy. – FShrike Apr 19 '23 at 18:28
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The problem the OP is can be deduced from the following general result which is good to know, prove and remember, as it used often in analysis:

Proposition: Let $(X,\mathcal{A}$, $\mu$) be a measure space, $\Omega \subseteq \mathbb{C}$ an open set and $f: X \times \Omega \to \mathbb{C}$. Assume that

  • $\forall z \in \Omega$ the mapping $x \to \phi(x,z)$ is measurable,
  • For any $x\in X$, the mapping $z \to \phi(x,z)$ is holomorphic on $\Omega$,
  • there exists an integrable function $g(x)$ such that $|\phi(x,z)| \leq g(x)$ $\forall x \in X$ $\forall z \in \Omega$.

Then the function $$F(z) = \int_X \phi(x,z) d\mu(x)$$ is holomorphic on $\Omega$, and for $m \in \mathbb{Z}_+$ we have $$ \left(\frac{d}{dz}\right)^m F(z) = \int_X \left(\frac{\partial}{\partial z}\right)^m \phi(x,z) d\mu(x).$$

This is given as a problem for example in Rudin, W., Real and Complex Analysis, 3rd edition,McGraw-Hill, Chapter 10, problem 16, pp. 239. It has also been studied in MSE, for example here P1, P2.

In the OP's problem, $(X,\mathcal{A},\mu)=([0,1],\mathscr{B}([0,1]),\lambda)$ (here $\lambda$is Lebesgue measure on $[0,1]$), and $$\phi(x,z)=f(x) e^{xz^2}$$ For any open ball$B(0;R)$, $|\phi(x,z)|\leq |f(x)|e^{R^2}\equiv g(x)$. Consequently, $F$ is holomorphic in any ball $B(0;R)$ and so, $F$ is entire.

Mittens
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