Showing $||T||_{op}$ is bounded Let $X$ be a Banach space and $Y$ a normed vector space over the field $\mathbb{K}$.
Let $F \subset L(X,Y)$ (where $L(X,Y) = \{{A ∈ Hom(X, Y) | A  \ \text{continuous}} \} =
\{A ∈ Hom(X, Y) \ |  \ \ ||A||_{op} < ∞ \}$ )
such that ,
$\forall x \in X, f \in Y' (= L(Y,\mathbb{K}))$ and $T \in F$: 
$\{|f(Tx)| \}$ is bounded
Show that $\{||T||_{op}  \ | \  T \in F \}$ is bounded.
Is this fairly trivial by the uniform boundedness theorem? and also by the fact that,
If you have a subset $M \subset Y$ of a normed vector space then,
$M  \ \ \text{bounded} \iff \forall f \in Y' : \{f(m) | m \in M\} $ is bounded
 A: First I will prove $$M \ \ \text{bounded} \iff \forall f \in Y' : \{f(m) | m \in M\} \text{ is bounded$$}$$ 
Forward implication  is trivial.
For the converse let $f(M) $ is bounded which means there exists a $M_f>0$ such that $$|f(m)|<M_f   \forall m$$
fix any $m\in M$. Concider $m^*\in X^{**}$ given by
$$m^*(f)=f(m)$$. 
$$|m^*(f)|=|f(m)|\leq ||f||||m||$$
So $$||m^*||\leq ||m||$$
Also concider the linear functional on the subspace spanned by $\{m\}$ given by
$$f_m(cm)=c||m||$$ 
Then it is easy to see that $||f_m||=1$ and $f_m(m)=||m||$
ByHahn-banach theorem it has a norm preserving extension say $g_m$. Then 
$$||g_m||=1,  m^*(g_m)=g_m(m)=f_m(m)=||m||$$
Which proves that $||m^*||=||m||$
Now apply Uniform boundedness principle on $\{m^*(f):m\in M\}=\{f(m):m\in M\}$. For each $f$ the above set is bounded. Hence by uniform boundedness principle $$\text{sup}_{M}||m^*||<\infty\implies \text{sup}_M ||m||<\infty$$
Hence $M$ is bounded.
Now your question $\{f(T(x))\}$ bounded means $\{T(x)\}$ bounded by above result. Now again applying unifform bounded principle we can show $$\text{sup}_F||T||<\infty $$
