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# Bounded or Not? (TIFR 2013 problem 16)

Question:

True/False?

Suppose $$\left \{a_i\right\}$$ is a sequence in $$\mathbb{R}$$ such that $$\sum|a_i||x_i|<\infty$$ whenever $$\sum|x_i|<\infty$$. Then $$\left \{a_i\right\}$$ is a bounded sequence.

Hint:

For any $$r\in(0,1)$$, $$\sum r^n <\infty$$.

Also, if the radius of convergence of a power series is R, then R is  given by $$limsup|a_n|^{1/n}=\frac{1}{R}$$

Discussion:

Of course, $$\sum|a_n||r|^{n}<\infty$$ for any $$r\in(-1,1)$$.

Recall that, $$\sum|a_n|x^{n}<\infty$$ for $$|x|<m$$ means that radius of convergence of the power series is atleast m.

If the radius of convergence of  $$\sum|a_n|x^{n}$$ is R then $$R\ge1$$.

i.e, $$limsup|a_n|^{1/n} = \frac{1}{R} \le 1$$

Hence, there exists $$N\in \mathbb{N}$$ such that $$sup\{|a_n|^{1/n} : n \ge k\} \le 1$$ for all $$k \ge N$$ (This is from the definition of limsup of a sequence).

Hence,  $$sup\{|a_n|^{1/n} : n \ge N\} \le 1$$. Therefore, for each $$n \ge N$$ we have $$|a_n|^{1/n} \le 1$$ for all $$n \ge N$$. (Because sup is supremum which is least upper bound).

A real number which is in between 0 and 1 when raised to any power stays in between 0 and 1.

This allows us to state that $$|a_n| \le 1$$ for all $$n \ge N$$.

There are only finitely many terms left in the sequence which may not bounded by 1. But taking the maximum of their absolute value and 1 together we get a bound for the whole sequence.

For any $$n\in \mathbb{N}$$,

$$|a_n| \le max\{1,|a_1|,|a_2|,…,|a_{N-1}| \}$$