Linear functional on $B(X,Y)$ with Compact-Open topology 
Theorem(Grothendieck): Let $X$ and $Y$ be two Banach space. Let $B(X,Y)$ denotes the set of Bounded liner operator from $X$ to $Y$. Equipped $B(X,Y)$ with the standard Compact-Open topology, denote by $\tau$. Then every continuous linear functional on $(B(X,Y),\tau )$ has the following form: $$f: B(X,Y)\rightarrow \mathbb{C} : A \mapsto \sum_{n=1}^{\infty} <y_n^*, Ax_n>$$ where $\{x_n \} \subseteq X$, $\{ y_n^*\}\subseteq Y^*$ such that $\sum_{n=1}^{\infty}||x_n|| \cdot ||y_n^*||< \infty$.


It is easy to check if $f$ has above from, then $f$ continuous with $\tau$. This theorem claims there is no others. Below is my thought. It may be long, and it is not necessary to read them, if you don’t want to. Any help, hint, discussion or reference are helpful. Thanks!

My thought: Suppose $f$ continuous, then $\{A\in B(X,Y): |f(A)|<1  \}$ is a open neighborhood of zero(with $\tau$). Use the form of local base, and note that union of finite compact set is compact. There is a compact set $K\subseteq X$ and $\epsilon > 0$, such that $\{A\in B(X,Y): ||Ax||<\epsilon, \forall x\in K \} \subseteq \{A\in B(X,Y): |f(A)|< 1 \}$. By Grothendieck compactness principal, without lose of generality, we can assume $K$ is a closed convex hull of a null sequence $\{ x_n\}, i.e. $$x_n \rightarrow 0$ in X. That is: $K=\{\sum_{n=1}^{\infty} a_n x_n: \sum_{n=1}^{\infty}|a_n| \leq 1\}$. By convexity, it is clear that $Ax_n=0$ implies $A|_{K}=0$. Also note that, $L=\{A\in B(X,Y): A|_{K}=0 \}$ is a subspace of $B(X,Y)$, and f acts on $L$ is bounded, so $f(L)=0$.
Now we consider the map $\phi$ from $B(X,Y)$ to $\Pi_x Y:=\{(y_n)_{n=1}^{\infty}\in \Pi_{n=1}^{\infty}Y:||(y_n)||:=\textrm{sup}_n \frac{||y_n||}{||x_n||}< \infty  \}$, via $$\phi: A \mapsto (Ax_1,Ax_2,...,Ax_n,...).$$ It’s routine to check $\Pi_x Y$, with this “weighted-$\infty$-norm”, is Banach and $\phi$ is continuous linear map. The first paragraph shows $\textrm{ker}\phi \subseteq \textrm{ker}f$. There is a will defined linear map $g$ from $\textrm{Range}\phi$ to $\mathbb{C}$ by setting $g=f(\phi^{-1}((y_n)))$.
I tend to claim the following:

*

*Every continuous functional on $\Pi_x Y$ has form $$(y_n)\mapsto \sum_{n=1}^{\infty}<y_n^*, y_n>, \hspace{0.2cm}\textrm{with its norm} \hspace{0.1cm} \sum_{n=1}^{\infty}||x_n||\cdot||y_n^*||<\infty.$$

*$g$ is a continuous linear functional on subspace $\textrm{Range} \phi$, so use Hahn-Banach theorem, $g$ can be extension to $\Pi_x Y$.
Together with 1 and 2, the theorem established, since $f=g\circ \phi$.


I’m confident with the first claim. Because $||x_n||\rightarrow 0$, so does $||y_n||$, thus $\Pi_x Y$ is just like the “wighted-$c_0$” by replace each component $\mathbb{C}$ with $Y$(maybe just like tensor?). And the dual of $c_0$ is exactly $l^1$, so they are agree in form. However, for the second claim, I’m not so confident. Because it seems no reason for $\phi$ must be open or closed. I tried to use Arzela-Ascoli theorem on K, to prove $\textrm{Range}\phi$ is closed, then use open-mapping theorem, or something else. But all fail. I think I may miss something important.
 A: $\newcommand{\bf}[1]{\mathbb #1}$Picking up your argument from the point you say

"Now we consider the map $\phi$ from $B(X, Y)$ to $\Pi_xY$ ...",

let me instead consider
$$
  c_0(Y) = \Big\{(y_n)_{n\in {\bf N}} \in  Y^{\bf N}: \lim_{n\to \infty } y_n = 0\Big\},
  $$
equipped with the sup norm, and put
$$
  \phi:T\in B(X,Y)\mapsto \big (T(x_n)\big )_{n\in {\bf N}} \in  c_0(Y).
  $$
As you observed,  since
$$
  \phi(T)=0 \Rightarrow f(T)=0,
  $$
there exists a linear map $g$ from $\text{Ran}(\phi)$ to ${\bf C}$, such that $g\circ \phi=f$.
I claim that $g$ is continuous with norm no bigger than $2/\varepsilon $.  To see this,  let $y = (y_n)_{n\in {\bf N}} \in  \text{Ran}(\phi)$,
with $\|y\|\leq 1$.  Choose $T\in B(X,Y)$ such that $\phi(T)=y$, and notice that
$$
  \|T(x_n)\|=\|y_n\|\leq 1, \quad\forall n\in {\bf N}.
  $$
By convexity we then have that
$$
  \|T(x)\|\leq 1, \quad\forall x\in K.
  $$
Setting $A=\varepsilon T/2$, it follows that
$$
  \|A(x)\|<\varepsilon , \quad\forall x\in K,
  $$
so
$$
  1>|f(A)|=\frac{\varepsilon |f(T)|}2 = \frac{\varepsilon |g(\phi(T))|}2 = \frac{\varepsilon |g(y)|}2,
  $$
whence
$$
  |g(y)|<\frac2\varepsilon,
  $$
proving the claim.
Using Hahn-Banach we may extend $g$ to the whole of $c_0(Y)$ and, imitating the proof that $c_0^*=\ell ^1$, we may prove
that the extended $g$ has the form
$$
  g(y) = \sum_{n\in {\bf N}}\langle y_n^*, y_n\rangle ,\quad\forall y = (y_n)_{n\in {\bf N}} \in  c_0(Y),
  $$
where $y^*=(y^*_n)_{n\in {\bf N}}\in \ell ^1(Y)$.
The grand conclusion is then that
$$
  f(T) = g(\phi(T)) = \sum_{n\in {\bf N}}\langle y_n^*, T(x_n\rangle \rangle ,\quad\forall T\in B(X,Y),
  $$
while
$$
  \sum_{n\in {\bf N}}\|x_n\|\|y_n^*\| \leq      \Big (\sup_{n\in {\bf N}}\|x_n\|\Big )\Big (\sum_{n\in {\bf N}}\|y_n^*\|\Big )<\infty .
  $$
