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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?2011-01-02
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    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.2011-01-02
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    Oh right, now I see what you meant. Fixed.2011-01-02
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    Is( n!)1/n is convergentor not?2011-03-19

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