Let $X,Y,Z$ be normed-spaces above same field. Assume that $Y$ is a banach space. Let $F:X×Y\to Z$ a function that satisfy:
$\forall x\in X$ $F(x,-)$ linear and continuous above $Y$. $\forall y\in Y$ $F(-,y)$ continuous. Assume that $(x_n)_n \subset X$ converges to 0 and $(y_n)_n\subset Y$ converges to 0. $\forall n\in N$ define: $T_n:Y\to Z$ , $T_n=F(x_n,-)$.
A.Show that $\forall n\in N$ , $T_n\in L(X,Y)$ and that there is $0 \leq M<\infty$ such that $||T_n||\leq M$ $\forall n\in N$.
B.Prove that: $lim_{n\to \infty} F(x_n,y_n)=0$.
My work:
Part A. $T_n$ is linear: Let $y_1,y_2\in Y$ and $a,b\in Z$ then: $T_n(ay_1+by_2)=F(x_n,ay_1+by_2)=${ F(-,y) is linear} = $aF(x_n,y_1)+bF(x_n,y_2)=aT_n(y_1)+bT_n(y_2)$ But
$T_n$ is continuos: This is concluded from the fact that $F(x,-)$ is continuous for all $x\in X$.So $T_n\in L(X,Y)$. $T_n$ is continuous thus bounded so by definition we get that there's a constant $C$ such that $||T_n(y)||\leq C*||y||$ then $||T_n||\leq C$. But, how to use the uniform bounded theory to finish it ?
Part B. $|F(x_n,y_n)|=||T_n(y_n)||\leq ||T_n||*||y_n||\leq C*||y_n||$. Here we used part A. Now, since $(y_n)_n$ converges to 0, then $(y_n)$ is bounded. So $|y_n)|\leq \epsilon/C$ . We get that $||F(x_n,y_n)||\leq \epsilon$, thus $F(x_n,y_n)\to 0$. Or we can say directly that, $||y_n||\ to 0$ since $(y_n)$ converges to zero and the norm is continuous. So $F(x_n,y_n)\to 0$ as $n\to \infty$.
Thanks in advance for any help.
Hints: The linear map $T_n$ is continuous because $F(x_n, .)$ is continuous. For fixed $y$ use the facts that $(x_n)$ is a bounbded sequence and $F(.,y)$ is continuous to conclude that $\sup_n \|T_ny\| <\infty$. By UBP we get $\sup \|T_n|| <\infty$. Let $M= \sup_n \|T_n\| $. Conclude that $|F(x,y) | \leq M\|x\| \|y||$ to prove part B.