Question on a proof that a space of linear map between normed spaces is a Banach space when the arrival space is a Banach space

Question

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!

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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.

Answer 1

$\newcommand{\L}{\mathscr{L}}$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.

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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).

Answer 2

I think FShrike's answer is working too hard here: there a couple of points worth emphasizing in how this theorem "works".

Firstly, the basic result is

Theorem 1: Let $M_1,M_2$ be metric spaces and let $\text{Map}_b(M_1,M_2)$ be the space of bounded functions between them, equipped with the supremum metric. Then $\text{Map}_b(M_1,M_2)$ is complete if $M_2$ is complete.

Theorem 2 If $M$ is a metric space and $Y$ is a Banach space, then the space $\text{Map}_b(M,Y)$ of bounded functions from $M$ to $Y$ is a Banach space.

The vector space structure actually makes Theorem 2 somewhat easier to prove, though the essential ideas in the proofs are the same.

Before discussing the proof of Theorem 2, I should say that the above theorems give templates for a range of similar theorems, which usually follow from Theorem 1 or 2 and the fact that various classes of functions embed as closed subspaces of $\text{Map}_b(M,Y)$. For example, the space of bounded continuous maps $\mathcal C_b(M_1,M_2)$ can be shown to be complete by this method (i.e. proving that the uniform limit of continuous functions is continuous).

Lastly before turning to the proof, I should also mention how $\mathcal L(X,Y)$ fits into this setting! A continuous linear map $T\colon X\to Y$ is said to be "bounded" if, for all $x \in X$, $\|T(x)\|\leq C.\|x\|$ for some $C>0$. This appears at first sight to be a different technical meaning of the term "bounded" to the one used in the definition of $\text{Map}_b(X,Y)$, where we say that $f\colon X\to Y$ is bounded if its image is a bounded subset of $Y$. In fact the usages can be reconciled using the closed unit ball $\mathbb B_X = \{x\in X: \|x\|\leq 1\}$: The key observations are:

1. Compatibility with scaling shows that, if $T\colon X \to Y$, then $T$ is completely determined by its restriction $T_{|\mathbb B_X}$ to the closed unit ball $\mathbb B_X$.
2. Moreover, noting that $\|T(x)\|\leq C.\|x\|$ holds for any $C$ when $x=0$, and that if $x \neq 0$, then the inequality is equivalent to $\|T(x/\|x\|)\|\leq C$, and hence $T$ is a bounded linear map if and only if $T_{|\mathbb B_X}$ is bounded.

Using 1. and 2. we see that $r\colon \mathcal L(X,Y) \to \text{Map}_b(\mathbb B_X,Y)$ given by $r(T) = T_{|\mathbb B_X}$ is an isometric embedding of $\mathcal L(X,Y)\to \text{Map}_b(\mathbb B_X,Y)$.

Now one can show that $\mathcal L(X,Y)$ is a Banach space by first showing that $\text{Map}_b(M,Y)$ is a Banach space for any metric space $M$, and then checking in the case $M=\mathbb B_X$, that the image of $r$ is closed in $\text{Map}_b(\mathbb B_X, Y)$. (This is essentially point 1. in the 3 steps FShrike's answer states).

And now, finally:

Proof of Theorem 2

There are two parts to the proof.

Part 1: first one shows that if $(f_n)$ is a Cauchy sequence in $\text{Map}_b(M,Y)$, then because $Y$ is complete, the sequence $(f_n)$ converges pointwise to a function $L\colon M\to Y$. This is where the completeness of $Y$ plays a role: since $(f_n)$ is Cauchy, the sequence $(f_n(x))_{n \geq 0}$ in $Y$ obtained by evaluating the $f_n$s at some $x \in M$ is Cauchy, and hence it converges, to $L(x)$ say, by the completeness of $Y$. This gives us a function $L\colon M \to Y$ which is our candidate for the limit of the sequence $(f_n)_{n \geq 0}$ in $\text{Map}_b(M,Y)$.

Part 2: It remains to establish that $L \in \text{Map}_b(M,Y)$ and that $f_n \to L$ as $n\to \infty$ in the normed vector space $\text{Map}_b(M,Y)$.

In fact, because $\text{Map}_b(M,Y)$ is a normed vector space, both of these assertions follow if one shows that $(L-f_n)_{n \geq 0}$ is a sequence in $\text{Map}_b(M,Y)$ which tends to $0$. To see that $(L-f_n)_{n \geq 0}$ is a sequence $\text{Map}_b(M,Y)$ which tends to zero, let $\epsilon>0$ be given. Then since $(f_n)_{n \geq 0}$ is Cauchy, there is an $N>0$ such that for all $n,m \geq N$ we have $\|f_m-f_n\|_{\infty} <\epsilon/2$. But then, for all $x \in M$, $$ \|L(x)-f_n(x)\| = \lim_{m\to \infty}\|f_m(x)-f_n(x)\| \leq \epsilon/2, $$ so that $\|L-f_n\|_{\infty} \leq \epsilon/2<\epsilon$ for all $n\geq N$, and hence $(L-f_n)_{n \geq 0}$ is a sequence in $\text{Map}_b(M,Y)$ converging to $0$ as required.

Addendum: Pointwise convergence is often brushed under the carpet when discussing things like uniform convergence, because the notion of pointwise convergence is not one which you can obtain by equipping the space of functions $Y^M$ from $M$ to $Y$ with a suitable metric. That said, it does arise from a perfectly natural topology on $Y^M$: if one forgets the metric space structure on $M$, then $Y^M$ can be viewed as $M$-many copies of $Y$, i.e. $Y^M = \prod_{m \in M} \{m\}\times Y$ is the Cartesian product sets $\{m\}\times Y$, and this can be give the standard product topology. Convergence in this topology is exactly pointwise convergence. Moreover, because $Y^M$ is a vector space, the notion of a Cauchy sequence makes sense even though we do not have a metric: indeed we define $(f_n)_{n \geq 0}$ to be Cauchy if, given any open neighbourhood $U$ of $0$, there is an $M>0$ such that for all $m,n\geq M$ we have $f_n-f_m \in U$.

Part 1 of the proof can then be rephrased as follows:

i) $Y^M$ is complete if $Y$ is complete.

ii) the inclusion $\text{Map}_b(M,Y) \hookrightarrow Y^M$ takes Cauchy sequences to Cauchy sequences.

(Indeed the second part of the proof can then be rephrased as the verification that $\text{Map}_b(M,Y)$ embeds as a closed subset of $Y^M$.)