Suppose that $T$ is a linear operator from a normed space $X$ into a normed space $Y$ such that $\sum_{n=1}^{\infty} T (x_{n})$ is a convergent series in $Y$ whenever $\sum_{n=1}^{\infty} x_{n}$ is an absolutely convergent series in $X .$ Prove that $T$ is bounded.
Theorem. Let T be a linear operator from a normed space X into a normed space Y. Then the following statements are equivalent.
(i) $T$ is continuous on $X$
(ii) $T$ is continuous at $0 \in X$
(iii) $\sup \left\{\|T x\|: x \in B_{X}\right\}<\infty$ In this case, let $\|T\|=\sup \left\{\|T x\|: x \in B_{X}\right\}$ and $T$ is said to be bounded.
I think we have:
PROOF. Let $\sum_{n} x_{n}$ be a absolutely convergent series in $X$. We show that $\left\|T\left(\sum_{n} x_{n}\right)\right\| \leq \sum_{n}\left\|T x_{n}\right\|,$ so it may be assumed that $\sum_{n}\left\|T x_{n}\right\|$ is finite, which together with the completeness of $Y$ implies that the absolutely convergent series $\sum_{n} T x_{n}$ converges. since $\sum_{n=1}^{m} x_{n} \rightarrow \sum_{n} x_{n}$ and $T\left(\sum_{n=1}^{m} x_{n}\right)=\sum_{n=1}^{m} T x_{n} \rightarrow \sum_{n} T x_{n}$ as $m \rightarrow \infty,$ then we have
$\sum_{n} T x_{n}=T\left(\sum_{n} x_{n}\right) .$ Therefore
$
\left\|T\left(\sum_{n} x_{n}\right)\right\|=\left\|\sum_{n} T x_{n}\right\| \leq \sum_{n}\left\|T x_{n}\right\|
$.
Here's a quick proof.
If $T$ is not bounded, then for every $C>0$, there exists some $x\in B(0,1)$ such that $\|Tx\|\geq C$. Applying linearity and considering $\frac{C}{r}$ in the above, we get that for every $r>0$, there exists some $x\in B(0,r)$ such that $\|Tx\|\geq C$. Let $r_n=2^{-n}$ and let $x_1\in B(0,1/2)$ be arbitrary.
Inductively, pick $x_{n+1}\in B(0,r_{n+1})$ such that $\|Tx_{n+1}\|\geq 1$. Then, we have
$$ \sum_{n=1}^{\infty} \|x_n\|\leq \sum_{n=1}^{\infty} r_n<\infty $$ so $\sum_{n=1}^{\infty} x_n$ is absolutely convergent. However,
$$ \left\|\sum_{m=1}^{n+1} Tx_m-\sum_{m=1}^{n} Tx_m\right\|=\|T x_{n+1}\|\geq 1, $$ so that $(\sum_{m=1}^n Tx_m)_{n\in \mathbb{N}}$ is not Cauchy and hence, not convergent.