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Recently, I've been asked to prove that this function is continuous on $\mathbb{R}$

$$f(x)= \sum _ {n=0} ^{\infty } \left ( \frac{x^n}{n!} \right )^2$$

I do recognize its very similar to the numerical series that defines $e$, yet I am working here with sequences and series of functions, so I don't know wheter this is helpful or not. Also, I think it could be helpful to try to use the uniform limit theorem and Weierstrass' M test.

I'm blank, and I would appreciate some little advices about how I can start working on the proof and what I should aim to in order to get the proof done and what thorems should I give a look. Thanks for your attention!!!

Sebastian Giraldo Gomez
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    A power series is not only continuous, but also infinitely differentiable inside the interval of convergence. – Sangchul Lee Jul 01 '20 at 04:29
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    Use the ratio test to show that the series converges uniformly on any compact interval. – copper.hat Jul 01 '20 at 04:46
  • Ok. So using the ratio test as suggested shows me that for any chosen $x \in \mathbb{R}$ the limit of the test is $L=0<1$. So the power series converges for any $x$ around $0$,i.e. any $x \in \mathbb{R}$. As this proves pointwise convergence for any arbitrary point, do we have uniform convergence on $\mathbb{R}$? – Sebastian Giraldo Gomez Jul 01 '20 at 05:16
  • You don't need uniform convergence on $\mathbb{R}$ to show continuity. As Sangchul Lee pointed out, it's infinitely differentiable at every point. – saulspatz Jul 01 '20 at 05:56
  • This is $I_0(2 x)$. – user64494 Jul 01 '20 at 07:05
  • Ok, so the main idea would be
    1. Prove its convergence radius is infinite
    2. Every power series is differentiable inside its convergence raidus
    3. Differeniable implies continuity

    Then $f$ is continuous since each $f_{n}$ for every $n \in \mathbb{N}$ and every $x \in \mathbb{R}$

     – Sebastian Giraldo Gomez Jul 01 '20 at 15:32

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For $A\in \Bbb N$ let $f_A=\sum_{n=0}^A(x^n/n!)^2.$ For $M>0$ let $\|f-f_A\|_M=\sup \{|f(x)-f_A(x)|: |x|<M\}.$

Then for any $M>0$ we have $\lim_{A\to \infty}\|f-f_A\|_M=0.$

Because for each $|x|<M$ we have $$|f(x)-f_A(x)|=|\sum_{n>A}(x^n/n!)^2|\le \sum_{n>A}(|x|^n/n!)^2<$$ $$<\sum_{n>A}(M^n/n!)^2=^{def} g_A(M)$$ and for a fixed $M$ we have $\lim_{A\to \infty}g_A(M)=0.$

In other words $f_A\to f$ uniformly (as $A\to \infty$) on any $(-M,M).$

Theorem: If a sequence $(f_A)_{A\in \Bbb N}$ of continuous real functions converges uniformly to $f$ on every bounded subset of $\Bbb R$ then $f$ is continuous.

Proof: Given $x\in \Bbb R$ and $\epsilon >0,$ we can find $\delta>0$ such that $|y-x|<\delta\implies |f(y)-f(x)|<\epsilon$ as follows:

Take $M>|x|.$ Take some big enough $A\in \Bbb N$ such that $\sup \{|f(y)-f_A(y)|: |y|<M\}<\epsilon/3.$ Now $f_A$ is continuous at $x$ so take some $\delta>0$ small enough that $(-\delta+x,\delta+x)\subset (-M,M)$ and such that $|y-x|<\delta\implies |f_A(y)-f_A(x)|<\epsilon/3.$

So if $|y-x|<\delta$ then $$|f(y)-f(x)|\le |f(y)-f_A(y)|+|f_A(y)-f_A(x)|+|f_A(x)-f(x)|<$$ $$<\epsilon/3+\epsilon/3+\epsilon/3.$$

This theorem applies verbatim to $\Bbb C$ if we replace $(-M,M)$ with $\{z\in \Bbb C:|z|<M\}$ and replace $(\delta+x,\delta+x)$ with $\{z: |z-x|<\delta\}.$

You should also study the proof of the Cauchy-Hadamard Radius Formula for more insight about power series.

DanielWainfleet
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