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Consider the following function: $ f(x) = \sum_{i,j>=0} a_{ij} x^i (1-x)^j$ , where the coefficients satisfy $a_{ij} \in [0,1]$. Observe that $f(x)$ converges for $x\in [0,1]$. Is $f$ analytic over $[0,1]$? If not, how about over $[c,d]$ for $c,d \in (0,1)$?

I think I can show that the answer is yes for the $[c,d]$ case by showing uniform convergence (EDIT: I was wrong... Not sure I can anymore). Is there a simpler way? More importantly, are there nice general conditions that guarantee that the infinite sum of polynomials is analytic over the interval (or a proper sub-interval) of convergence? This motivates the following questions:

Question 1: Let $f(x) = \sum_{i,j} a_{ij} x^i (1-x)^j$, where there are no guarantees on $a_{ij}$. Assume $f$ converges on interval $[a,b]$. Is $f(x)$ analytic over $[a,b]$? What about over $[c,d]$ where $c,d \in (a,b)$?

Question 2: Let $f(x) = \sum_k p_k(x)$, where $p_k$ is a polynomial. Assume $f$ converges on interval $[a,b]$, and that $p_k(x) \geq 0$ for $x \in [a,b]$. Is $f(x)$ analytic over $[a,b]$? What about over $[c,d]$ where $c,d \in (a,b)$?

Question 3: Same as question 2, but there are no guarantees on the sign of $p_k(x)$ anymore. (EDIT: As Jonas Meyer pointed out, this is false by the Weierstrass approximation theorem)

I guess what I'm trying to get at is understanding which natural conditions, short of manually showing uniform convergence, guarantee that a convergent sum of polynomials is analytic over its region of convergence (or a sub-interval of that). Question 1 starts with a very specific sum of polynomials that I somewhat understand, and Questions 2 and 3 are grasping at a generalization. These questions are my best guesses as to what generalizations may look like, though please feel free to suggest others if I'm off-base here.

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Questions 2 and 3 are the same. You might as well take all $a_k=1$, because $a_kp_k(x)$ is also a polynomial. Perhaps that isn't what you meant to ask? – Jonas Meyer Jan 2 '11 at 7:22
Yes, you are correct. I meant to say that $a_k$ and $p_k(x)$ have the same sign over the region of convergence. Will edit accordingly. – srd Jan 2 '11 at 7:49
Oh right, now I see what you meant. Fixed. – srd Jan 2 '11 at 7:55
Is( n!)1/n is convergentor not? – user8455 Mar 19 '11 at 7:57

This answer doesn't answer all of your specific questions.

A uniform limit of polynomial functions on $[0,1]$ need not be analytic. Every continuous function on $[0,1]$ is a uniform limit of polynomial functions, by the Weierstrass approximation theorem. Of course, if you only have pointwise convergence, things could be worse. This means that the answer to $2$ and $3$ is no. [Edit after question edit: In the edited version, $2$ is a different question, but the answer is still no, and it can still be answered using Weierstrass's theorem.]

Things work out better if you work over the complex numbers instead of the real numbers. If a sequence of analytic functions defined on an open subset of $\mathbb{C}$ converges uniformly on compact subsets, then the limit function is analytic. There are useful criteria for determining whether a sequence converges uniformly on compact sets, such as Montel's theorem. However, there are examples of non-analytic pointwise limits of complex analytic functions. See for example Davidson's "Pointwise limits of analytic functions."

If you are comfortable going to the complex case but start out with functions defined on an interval $[c,d]$ in $\mathbb{R}$, you might want to try extending all of your functions to an open subset of $\mathbb{C}$ that contains $[c,d]$. If the sequence converges uniformly on some such open set, then the limit function will be complex analytic, and therefore its restriction to $[c,d]$ will be real analytic.

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Ah I see. I naively thought that showing my original function uniformly converges over the reals was sufficient -- I knew it seemed too easy to be true! – srd Jan 2 '11 at 7:23
However, for power series, convergence in an open interval automatically implies analyticity. One perspective for seeing why this is true is that the radius of convergence actually tells you the radius of an open disk in the complex plane on which the series converges uniformly on compact subsets, so you get a lot more than meets the eye by going beyond the real line. I suspect that things still work out for your first example due to its special form, but I'm not sure and don't plan to think about it soon. – Jonas Meyer Jan 2 '11 at 7:30
Right, I think it will work out as well due to the special form. Namely, observe that that function is simply a power series in two variables $x$ and $y$, restricted to the line $y=1-x$. That has to give something... I'll think about it / dig it up... – srd Jan 2 '11 at 7:37

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