Showing the compactness of a limit operator. I was trying to solve this exercise from Kreyszig's book, section 8.1 exercise number 10. My attempt was try to show that the operators in the sequence are bounded, but I don't find it. If this fact were possible, then we can apply Theorem 8.1.4 (Finite dimensional domain or range) to the $T_n$ showing that these operators are compact and after use the theorem 8.1.5. 
Any help would be greatly appreciated. Thanks!

Let $Y$ be a Banach space and let $T_n:X\rightarrow Y$, $n=1,2,\ldots$, be operators of finite rank. If $(T_n)$ is uniformly operator convergent, show
  that the limit operator is compact.

 A: This doesn't necessarily use the theorems you refer to (I don't own that book), but this is the one proof I used to know. Hope it helps you anyway. Remember that in a complete metric space, totally bounded (having finite $\varepsilon$-net for any $\varepsilon$) and relatively compact are the same.
I also assumed that your $T_n$ are continuous (i.e. bounded in this case), because, as mentioned in the comments, if they're not the result is false. My apologies if this is not the answer you needed.


To show that $T$ is a compact operator we will show that $T$ is continuous and that $T(K)$ is relatively compact for $K \subset X$ bounded. 
Since $T$ is a uniform limit of continuous operators, it is also continuous. In fact, for any $x \in X$, so we have
\begin{align}
\| T(x) - T(y) \| &= \| T(x) -T_n(x) + T_n(x) - T_n(y) + T_n(y) - T(y)\| \leq \\
& \leq \| T(x) - T_n(x)\| + \| T_n(x) - T_n(y)\| + \| T_n(y) - T(y)\|  \to 0
\end{align}
if $\| x-y\| \to 0$, since $T_n$ is continuous (middle term) and $T_n \to T$ uniformly (first and third term).
Let $K \subset X$ be a bounded subset.
Since $T_n(K)$ is relatively compact in $Y$ Banach, it has a finite $\frac{1}{n}$-net $y_i$, so we have, for $x \in K$,
$$
\| T(x) - y_i \| \leq  \| T_n(x) - T(x)\| + \| T_n(x) - y_i\| \leq \tfrac{1}{n} + \tfrac{1}{n} = \tfrac{2}{n};
$$
that is $T(K)$ has a finite $\frac{2}{n}$-net, and hence is relatively compact, because $n$ is arbitrary. Thus $T$ is compact.
