how to construct an absolutely convergent series which is not convergent in the space $(C[a, b], ||•||_1) $?

Solution 1:

Here is a general argument.

Since you are able to show that a normed space $X$ is not complete, I assume you reached to a point where you found a Cauchy sequence $(x_n)$ that does not converge, correct?

Since $(x_n)$ is Cauchy, find $n_1\in\mathbb{N}$ such that if $n,m\ge n_1$ then $\|x_n-x_m\|<\frac{1}{2}$. Then, find $n_2>n_1$ such that if $n,m\ge n_2$ then $\|x_n-x_m\|<\frac{1}{2^2}$. Continuing this process, we obtain indices $n_1<n_2<\dots$ such that $\|x_{n_{k+1}}-x_{n_k}\|<\frac{1}{2^k}$ for all $k\in\mathbb{N}$.

Set $y_{k}=x_{n_{k+1}}-x_{n_k}$ for all $k\ge1$. Then, $$\sum_k\|y_k\|\le\sum_k\frac{1}{2^k}<\infty$$ On the other hand, the partial sums are $s_K=\sum_{k=1}^Ky_k=x_{n_{K+1}}-x_{n_1}$, so, if the series $\sum_ky_k$ converges, then the sequence $\{x_{n_k}-x_{n_1}\}_{k=1}^\infty$ converges and thus the sequence $\{x_{n_k}\}$ converges. But a Cauchy sequence having a convergent subsequence is also convergent to the limit of the subsequence. This cannot be, since we know that $(x_n)$ does not converge.

One can come up with this by looking at the standard proof of the fact that completeness is equivalent to the implication of series convergence by absolute convergence.