Prove: $T \in L(Y,\ell^\infty(\mathbb K))$ is bounded $\implies$ $\forall n \in \mathbb N:g_n \in L(Y,\mathbb K)$ is bounded Assume $Y$ is a Banach space over $\mathbb{K}$ and $T : Y → \ell^{\infty}$ is a linear operator, such that $Ty = (g_1(y), g_2(y), g_3(y), . . .)$, where  $g_n \in L(Y, \mathbb{K})$.
I am trying to prove/disprove the following statements:

*

*($T$ is bounded) $\rightarrow$ ($\forall n \in \mathbb{N}:g_n$ is bounded )


*($\forall n \in \mathbb{N}:g_n$ is bounded ) $\rightarrow$  ($T$ is bounded)
If it is bounded then $\exists$M $\geq 0$ such that $\|Ty\|_{\infty} \leq$M$\|y\|_{\infty}$ $\forall$y$\in Y$.
Then. $\|Ty\| = \|g_n(y)\|$$ = \sup_{n\in\mathbb{N}}|g_n(y)| \forall n \geq1$, so we can easily take $M=1$
Hence, $\|Ty\|_{\infty} \leq$$\|y\|_{\infty}$ and so $\|T\| \leq 1$.
I am not convinced that I am on the right track. Any help would be grateful for both "arrows" in the question. Thanks in advance.
 A: *

*($T$ is bounded) $\rightarrow$ ($\forall n \in \mathbb{N}:g_n$ is bounded )

Assume $T$ is bounded, i.e.
$$
\exists M \geq 0:\forall y\in Y: \|Ty\|_{\infty} \leq M\|y\|_{\infty}.
$$
Since $\|Ty\|_\infty  = \sup_{n\in\mathbb{N}}|g_n(y)| \leq M$, it follows $\exists M \geq 0:\forall y\in Y: \sup_{n\in\mathbb{N}}|g_n(y)| \leq M\|y\|_{\infty}$. But $\forall n \in \mathbb N: |g_n(y)| \le \sup_{n\in\mathbb{N}}|g_n(y)|$, that yields
\begin{align}
\exists M \geq 0:\forall n \in \mathbb N:\forall y\in Y: |g_n(y)| \leq M\|y\|_{\infty}
\end{align}
which is the same as saying each $g_n$ is bounded

Added:


*($\forall n \in \mathbb{N}:g_n$ is bounded ) $\rightarrow$  ($T$ is bounded)

Since for every $y$ in $Y$ we have $Ty \in \ell^\infty$, then $\forall y \in Y: \|Ty\|_\infty = \sup_{n\in \mathbb N} |g_n(y)| < \infty$. Also since $Y$ is Banach and $(g_n)$ are bounded we have by Banach-Steinhaus theorem (unifrom boundedness principle) that $\sup_{n \in \mathbb N}\|g_n\| < \infty$. So
$$
\sup_{y:\|y\| =1} \|Ty\|_\infty = \sup_{y:\|y\| =1} (\sup_{n \in \mathbb N}|g_n(y)|) = \sup_{n \in \mathbb N} (\sup_{y:\|y\| =1}|g_n(y)|) = \sup_{n \in \mathbb N} \|g_n\| < \infty
$$
(since both suprema exist,) so it follows $T$ is bounded.
A: (1) is true. Let $y\in Y$. Let $n\in\mathbb{N}$ be arbitrary, then
$|g_{n}(y)|\leq||Ty||_{\infty}\leq||T||\cdot||y||$. This shows that
$g_{n}$ is bounded and $||g_{n}||\leq||T||$.
(2) is true. For each $n\in\mathbb{N}$, define $T_{n}:Y\rightarrow l^{\infty}$
by $T_{n}y=(g_{1}(y),g_{2}(y),\ldots g_{n}(y),0,0,\ldots)$. Clearly
$T_{n}$ is linear. Let $y\in Y$, then
\begin{eqnarray*}
||T_{n}y|| & = & \max_{1\leq k\leq n}|g_{k}(y)|\\
 & \leq & \max_{1\leq k\leq n}||g_{k}||\cdot||y||\\
 & = & M||y||
\end{eqnarray*}
where $M=\max_{1\leq k\leq n}||g_{k}||<\infty$. Therefore, $T_{n}$
is a bounded linear map. Consider the family of bounded linear maps
$\{T_{n}\mid n\in\mathbb{N}\}$. We assert that for each $y\in Y$,
$\{T_{n}y\mid n\in\mathbb{N}\}$ is a bounded subset of $l^{\infty}$.
Let $y\in Y$. Since $T:Y\rightarrow l^{\infty}$ is a well-defined
linear map, $Ty\in l^{\infty}$. That is, $M_{1}:=\sup_{k\in\mathbb{N}}|g_{k}(y)|<\infty$.
For each $n$, we have that $||T_{n}y||_{\infty}=\sup_{k\leq n}|g_{k}(y)|\leq M_{1}$.
This shows that $\{T_{n}y\mid n\in\mathbb{N}\}$ is a bounded subset
of $l^{\infty}$. By Uniform Boundedness Principle, $M_{2}:=\sup_{n}||T_{n}||<\infty$.
Let $y\in Y$ be arbitrary. For each $n\in\mathbb{N}$, we have that
$\sup_{k\leq n}|g_{k}(y)|=||T_{n}y||_{\infty}\leq M_{2}||y||$. It
follows that $||Ty||_{\infty}=\sup_{k}|g_{k}(y)|\leq M_{2}||y||$.
Hence, $T$ is bounded.
