If $Y$ is complete show that $T_nx\to T(x) ;T\in B(X,Y)$ 
*

*Let $X$ be a Banach Space ,$Y$ a normed space and $T_n\in B(X,Y)$ such that $(T_nx)$ is Cauchy in $Y$ for every  $x\in X$ .Show that $(||T_n||)$ is bounded.

*If in addition $Y$ is complete show that $T_nx\to T(x) ;T\in B(X,Y)$ where $B(X,Y)$ denotes the set of bounded linear operators from $X$ to $Y$.


Attempt:


*

*$(T_nx)$ is Cauchy $\implies $ 
$(T_nx)$ is bounded for each $x$ and $X$ is Banach Space hence by Uniform Boundedness Principle we have $T_nx$ is uniformly bounded and hence so $(||T_n||)$ is bounded.

*Unable to do it.
Since $Y$ is complete  and  $T_nx$ is Cauchy ,so $T_n(x)\to c_x$ for each $x$.
But how to show that $c_x=T(x)$.


Please help me in this case and moreover is the 1st one correct?
 A: First we define a map $T \colon X \to Y$ by
$$T(x) := \lim_{n\to\infty} T_n(x).$$
The limit exists, since by assumption $\bigl(T_n(x)\bigr)$ is a Cauchy sequence in $Y$, and $Y$ is complete. By the uniqueness of limits in Hausdorff spaces, we have thus defined $T$ as a map.
As the next step, we show that $T$ is a linear map. That follows immediately from the linearity of the $T_n$ and general limit algebra,
$$T(x+y) = \lim_{n\to\infty} T_n(x+y) = \lim_{n\to\infty}\:\bigl(T_n(x) + T_n(y)\bigr) = \lim_{n\to\infty} T_n(x) + \lim_{n\to\infty} T_n(y) = T(x) + T(y)$$
for $x,y\in X$, and
$$T(\lambda x) = \lim_{n\to\infty} T_n(\lambda x) = \lim_{n\to\infty} \:\bigl(\lambda T_n(x)\bigr) = \lambda \lim_{n\to\infty} T_n(x) = \lambda T(x)$$
for $x\in X$ and $\lambda \in \mathbb{C}$ (or $\lambda \in \mathbb{R}$ if we're dealing with real scalars).
Finally, we show that $T\in B(X,Y)$, using the first part. By the first part, there is an $M$ such that $\lVert T_n\rVert \leqslant M$ for all $n$, and that gives
$$\lVert T(x)\rVert = \lim_{n\to\infty} \lVert T_n(x)\rVert \leqslant \limsup_{n\to\infty} \:\bigl( \lVert T_n\rVert\cdot \lVert x\rVert\bigr) \leqslant M\lVert x\rVert,$$
whence $\lVert T\rVert \leqslant M$ follows.
