A friend of mine gave me the following problem recently:

Let $(a_n)$ be a sequence of real numbers. Suppose that $\sum_{n=1}^\infty a_nb_n$ converges for any sequence $(b_n)$ such that $\lim\limits_{n\to\infty} b_n=0$. Show that $\sum |a_n|<\infty$.

Hint he gave me was to use Banach-Steinhaus theorem (a.k.a. uniform boundedness principle).

  • I would like to know whether there are different solutions to this, in particular, whether there is an elementary solution. (Using just calculus, without resorting to more advanced results from functional analysis or other areas.)

The argument using Banach-Steinhaus theorem goes as follows:

Let $a=(a_n)$ be a given (fixed) sequence such that $\sum a_nb_n$ converges for each $b=(b_n)\in c_0$.

For every $n$ we can define $f_n\colon c_0\to{\mathbb R}$ as $\sum_{k=1}^n a_nb_n$. We consider $c_0$ with the usual sup-norm, which makes $c_0$ into a Banach space.

Each function $f_n$ is obviously linear.

For any $b\in c_0$ such that $\|b\|=\sup\limits_{n\in\mathbb N} |b_n| \le 1$ we have $$|f_n(b)| \le \sum_{k=0}^n |a_k|.$$ This shows that all functions $f_n$ are bounded linear functionals.

Now if we fix some $b=(b_k)\in c_0$ we have $$\left|\sum_{k=0}^n a_kb_k\right| \le \sum_{k=0}^\infty |a_k|\cdot |b_k|.$$ The sum on the RHS is finite, since it can be expressed as $\sum a_kc_k$, where $c_k=\pm b_k$. (The choice of sign here depends on the signs of $b_k$ and $a_k$.) So we have shown that at each point $b\in c_0$ the values $f_n(b)$ are bounded by the same constant (depending only on $b$).

By Banach-Steinhaus theorem we have that there exists a constant $M$ such that $\|f_n\|<M$ for each $n\in\mathbb N$.

Now it only remains to show that $\|f_n\|=\sum_{k=0}^n |a_k|$. We already know that $\|f_n\|\le\sum_{k=0}^n |a_k|$. The other inequality can be obtained by choosing $b$ in unit ball, such that $b_k$ is either $1$ or $-1$, depending on the sign of $a_k$ for $k=1,2,\dots,n$; and $b_k=0$ for $k>n$.

Now we have that $\sum_{k=0}^n |a_k|<M$ for each $n$, which implies $\sum_{k=0}^n |a_k|\le M$.


We prove it by contradiction. We may as well suppose that $a_n \ge 0$ for all $n$, to avoid cluttering up the proof.

Suppose that $\Sigma a_n = \infty$. Then we can break up $(a_n)$ into finite sub-sequences $(a_n)_{0 \le n \le s_1},(a_n)_{s_1 < n \le s_2},(a_n)_{s_2 < n \le s_3}$ etc. such that each sub-sequence has a sum $\ge 1$.

Now just set $b_n = 1/r$ for $s_r < n \le s_{r+1}$. Then for all $r \ge 1$, the sub-sequence $(a_nb_n)_{s_r < n \le s_{r+1}}$ has $\Sigma_{}a_nb_n \ge 1/r$, so $\Sigma a_nb_n$ diverges although $b_n \to 0$.


Let $s_n=\sum\limits_{k=1}^n|a_k|$. If $s_n\to\infty$, then one knows that the series $\sum\limits_n|a_n|/s_n$ diverges hence, defining $b_n=\mathrm{sgn}(a_n)/s_n$ if $a_n\ne0$ and $b_n=0$ if $a_n=0$, one sees that $b_n\to0$ and that the series $\sum\limits_nb_na_n$ diverges.

Contraposition yields the result you ask about.


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