I have a question concerning the proof of the result stated in the title.
To prove this theorem, we consider $(f_n)_{n\in\mathbb{N}}$ a Cauchy sequence in $L(E,F)$ :
$$
\forall\varepsilon>0, \exists N\in\mathbb{N} : m>n>N\implies \sup_{x\neq 0}\frac{\lVert f_m(x)-f_n(x)\rVert_{F}}{\lVert x\rVert_{E}}<\varepsilon
$$
which is equivalent to
$$
\forall\varepsilon>0, \exists N\in\mathbb{N} : m>n>N\implies\forall x\in E\setminus\{0\} : \frac{\lVert f_m(x)-f_n(x)\rVert_{F}}{\lVert x\rVert_{E}}<\varepsilon
$$
From there, we know that we have to consider the sequence $(f_m(x))_{m\in\mathbb{N}}$ with $x$ fixed to demonstrate its convergence in $F$ since $F$ is complete. Then, as is often the case when showing that a space is a Banach space, we let $m$ tend towards infinity in the inequality above, invoking the continuity of the norm to demonstrate the convergence of the initially considered sequence.
I wanted to proceed a bit differently this time and trying to prove this by using the formal definition of the convergence, but I am encountering difficulties in my proof, which is currently incorrect since I have a dependency of $N_1$ on $x$, which is not appropriate. I'm detailing the outline of the proof here.
Let $\varepsilon>0$
Since for $x$ fixed, the sequence $(f_m(x))_{m\in\mathbb{N}}$ converges in $F$ to $f(x)$. Consider $\varepsilon_1 = \frac{\varepsilon\lVert x\rVert_{E}}{2}$, there exists $N_1\in\mathbb{N}$ such that
$$
m>N_1\implies \lVert f_m(x) – f(x)\rVert_{F} < \varepsilon_1
$$
Since $(f_n)_{n\in\mathbb{N}}$ a Cauchy sequence in $L(E,F)$, consider $\epsilon_2 = \frac{\varepsilon}{2}$, then there exists $N_2\in\mathbb{N}$ such that
$$
m>n>N_2\implies \forall x\neq 0 : \lVert f_m(x) – f_n(x)\rVert_{F} < \epsilon_2\lVert x\rVert_{E}
$$
Thus we have for $m>n>max(N_1,N_2)$
$$
\lVert f_n(x) – f(x)\rVert_{F} \leq \lVert f_m(x) – f_n(x)\rVert_{F} + \lVert f_m(x) – f(x)\rVert_{F} < \varepsilon_1 + \varepsilon_2 = \varepsilon\lVert x\rVert_{E}
$$
So, I would like to have your assistance on this attempt and know how I could proceed to obtain something correct using the delta-epsilon method.
Thank you very much!
What I would like is to start from $\lVert f_n(x) – f(x)\rVert_{F}$ and find an upper bound by exploiting the convergence of the sequence $(f_m(x))$ to reach the final result. It is in this sense that I thought the triangle inequality could be useful.
Best Answer
$\newcommand{\L}{\mathscr{L}}$What matters most is the uniformity of the convergence that you get by very definition of the norm on $\L(E;F)$. You are quite right that the dependence of $N_1$ on $x$ is inappropriate, but it is easily fixed by leveraging the uniformity.
Fix normed linear spaces $E$ and $F$ with $F$ complete. Take a Cauchy sequence $(T_n)_{n\in\Bbb N}\subseteq\L(E;F)$. Denote by $B$ the closed unit ball of $E$.
Because for $x\in E$ and $n,m\in\Bbb N$ we have $\|T_n(x)-T_m(x)\|\le\|T_n-T_m\|\cdot\|x\|$ we infer $(T_n(x))_{n\in\Bbb N}$ is Cauchy in $F$ thus convergent to some point which we shall denote by $T(x)$. You have hopefully verified that $x\mapsto T(x)$ defines a linear mapping $E\to F$. There are three more things to do here.
(1) Verify that $x\mapsto T(x)$ defines a linear map $T:E\to F$.
(2) Check that $T\in\L(E;F)$ i.e. $T$ is continuous.
(3) Find "explicit" $\epsilon$-$N$ estimates to show that $\lim_{n\to\infty}T_n=T$ in $\L(E;F)$.
The points $(1),(2)$ are very much duplicates that I won't answer here. The third point might also be a duplicate, but I take it in good faith you're asking this question because other MSE posts about this theorem didn't satisfy you.
Fix $\epsilon>0$. There is, by definition, an $N\in\Bbb N$ such that if $n,m>N$ we have $\|T_n-T_m\|<\epsilon/3$.
I claim that if $n>N$ we have $\|T_n(x)-T(x)\|<\epsilon/2$ for all $x\in B$. If this is so, then we will have shown $\|T_n-T\|\le\epsilon/2<\epsilon$. Then we are done; this "$N$" is successful and we find $\lim_{n\to\infty}T_n=T$ in $\L(E;F)$.
Let's put this to the test. Fix $n>N$.
Take any $x\in B$. I know that there is an $m\in\Bbb N$, $m>N$, so large that $\|T_m(x)-T(x)\|<\epsilon/6$. This $m$ may depend on $x$. I can conclude: $$\begin{align}\|T_n(x)-T(x)\|&\le\|T_n(x)-T_m(x)\|+\|T_m(x)-T(x)\|\\&<\|T_n-T_m\|\cdot\|x\|+\frac{\epsilon}{6}\\&<\frac{\epsilon}{3}\cdot1+\frac{\epsilon}{6}\\&=\frac{\epsilon}{2}\end{align}$$As desired. This is true for any $x\in B$ as desired (the final bound is completely independent of $x$, we can always procure a particular such $m$ for any given $x$ but we do not actually care what $m$ is).