$X$ be Banach , $T:X \to \mathcal l ^{\infty}$ be linear , $(Tx)_n$ the $n$-th term of $T(x)$;$f_n(x)=(Tx)_n$ ; if each $f_n$ is bdd then so is $T$? Let $X$ be a Banach space , $T:X \to \mathcal l ^{\infty}$ be a linear transformation , for each $x\in X$ and each $n \in \mathbb N$ , $(Tx)_n$ be the $n$-th term of $T(x)$  and for each $n \in \mathbb N$ , let $f_n:X \to \mathbb R$ be defined as $f_n(x)=(Tx)_n, \forall x \in X$ . Then is it  true that if each $f_n$ is continuous , then $T$ is continuous ? What if we replace $\mathcal l^{\infty}$ by $\mathcal l^1$ ?
I tried applying uniform-boundedness principle but with no success ; Please help . Thanks in advance 

It is true, it is a consequence of the uniform boundness principle.
https://en.wikipedia.org/wiki/Uniform_boundedness_principle
Apply the contraposition. Consider the family $T^n(x) =(Tx)_n$ and suppose that $\sup_{T^n,\|x\|=1}\|T^n(x)\|=\infty$. Then there exists $x$ such that $\sup_n\|(Tx)_n\|=\infty$. This is impossible since $T$ must be defined at $x$.
Thus $\sup_{T^n,\|x\|=1}\|T^n(x)\|<\infty$. This is equivalent to saying that $T$ is bounded.
