Proving that a space is complete There is something that bugs me about the proof I've been shown that $C(\Omega)$ (the space of continuos function on $\Omega$, a compact subset of $\mathbb R^n$) with the $\sup$ norm is complete.
Basically they say: let $\{u_m\} \in C(\Omega)$ be cauchy; $||u_n - u_m|| \to 0$.
Let $u$ be the function they converge to; with some clever inequalities we show that $u$ is continuos, hence $u \in C(\Omega)$ and we're happy.
But I don't know that the $\{u_m\}$ are converging to anything! Above they assumed that there is a "limit", $u$, to which the functions converge to, which seems weird to me because it comes out of nowhere.I don't know what kind of object $u$ is, if it exists at all!
Now I was trying to prove that the space $\mathcal L(H_1, H_2)$ (the space of all bounded linear operators from the hilbert space $H_1$ to $H_2$) is complete with the norm $$||L|| = \sup_{||x||_{H_1}=1} ||Lx||_{H_2}$$
But I don't know where to start because I don't think I can say "Let $L$ be the linear operator the cauchy sequence $\{L_m\}$ is converging to.. "
 A: You know the $u_m$ converges to a function, because $\mathbb{R}$ is complete. 
Indeed, for every $x \in \Omega$, you have
$$| u_n(x) - u_m(x) | \leq \| u_n - u_m\|_\infty \to 0$$
So $u_n(x)$ is a Cauchy sequence, and by completeness of $\mathbb{R}$, converges.
The function $u\colon\Omega\to\mathbb{R}$ is then defined by :
$$u(x) = \lim_{n\to \infty} u_n (x)$$

For your second question, you can define the operator L by :
$$L(x) = \lim_{n\to + \infty} L_n(x)$$
This limit exist, because you can show that $(L_n(x))$ is a Cauchy sequence, and by completness of $H_2$, you get the result.
It's a Cauchy sequence, because for $x\in H_1$,
$$\| L_n(x) - L_m(x) \|_{H_2} \leq \|x\|_{H_1} \| L_n-L_m\|_{\mathcal{L}(H_1,H_2)} \to 0$$
You need then to show that it's linear, but it's trivial :
$$L(x+\lambda y) = \lim_{n\to + \infty} L_n(x+\lambda y) = \lim_{n\to + \infty} L_n(x) + \lambda  L_n(y) = L(x) +  \lambda L(y)$$
Then you need to prove that it's continuous...
A: The thing about $C(\Omega)$ works because $\mathbb R$ is complete: If $(f_n)$ is a Cauchy sequence in $C(\Omega)$ then the definition of the metric shows that for every $x\in\Omega$ the sequence $(f_n(x))$ is a Cauchy sequence of reals. Since $\mathbb R$ is complete there exists a real number $y$ such that $f_n(x)\to y$; we define $f(x)=y$ and proceed.
The completeness of $\mathcal L(H_1,H_2)$ works exactly the same, using the fact that $H_2$ is complete.
Ok, not exactly the same, because the norm is not the sup over all of $H_1$. But it's easy to show that if $(T_n)$ is Cauchy in $\mathcal L(H_1,H_2)$ and $x\in H_1$ then $(T_n(x))$ is a Cauchy sequence in $H_2$.
