# Showing a set with a supremum norm is a Banach space

Problem Statement: Let $X$ be a set and $E$, a Banach space with norm $||\cdot||_{E}$. Let $B_{E}(X)$ denote the set of all bounded functions $f:X\rightarrow E$ with the supremum norm $||f||=\sup_{x\in X}||f(x)||_{E}$. Show that $B_{E}(X)$ is a Banach space (i.e. show that it is complete in this norm).

I am just now starting in my first course in Real Analysis, and we have just been given our first homework, and I am already quite lost in how to approach these types of proofs. (I have taken an Abstract Algebra sequence in which I became very comfortable with the material and styles of proofs, but Analysis already seems to be much more daunting.)

I have these definitions to help me:

A Banach space is a complete normed vector space.

A complete normed vector space is a normed vector space in which Cauchy sequences converge.

A sequence $(v_{n})_{n\geq 1}$ in a normed vector space $E$ is a Cauchy sequence if $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$ $(m,n\geq N)$ such that $||v_{n}-v_{m}||<\epsilon$.

I am assuming that the norm $||\cdot||_{E}$ in the problem is referring to the standard Euclidean norm. So basically I need to show that an arbitrary Cauchy sequence of functions in $B_{E}(X)$ converges?

My professor did a proof in class that $B_{\mathbb{R}}(X)$, the space of bounded functions from a set $X$ to $\mathbb{R}$, with the supremum norm, is complete. He began letting an arbitrary sequence $(f_{n})$ be Cauchy, and then showed that the $\lim f_{n} = f_{0}$ for some (I assume finite) $f_{0}$. But many of the steps were confusing to me, especially considering it was my first day in Analysis.

Any suggestions on how to approach this problem would be greatly appreciated!

• The space $E$ needs not be Euclidean, so $\|\cdot\|_E$ needs not be euclidean. You need to show 2 statements: (1) the supremum norm is indeed a norm. (2) completeness. Since you already have a proof for $E=\mathbb R$, you could just copy it and replace $\mathbb R$ by $E$. It is basically the same proof. – user251257 Aug 18 '16 at 23:30

Consider a sequence of objects (i.e. functions in this case) $(f_{n}) \in B_{E}(X)$. Let this sequence be cauchy in $B_{E}(X)$.

Now we need to find a candidate function in $B_{E}(X)$ that would satisfy our needs. So lets change the space to E and consider $[f_{n}(x)]$ where $x\in X$ (a sequence of objects in E). Now for a fixed $x \in X$. This is still a cauchy sequence in E and E is Banach. Hence the sequence would converge to an object in E. Hence as we change $x\in X$, we get a new converged object in E for every $x\in X$ . This can now be called a mapping from X to E. Lets represent it by $F:X\rightarrow E$. $F(x)$ is thus our candidate function.

Now you need to prove two more things

a. $f_{n}\rightarrow F$ (Why? Because $f_{n}(x) \in E$ while $f_{n} \in B_{E}(X)$. In other words, we have only proven pointwise convergence in E but now we have prove convergence in $B_{E}(X)$)

b. prove that $F$ is bounded implying $F\in B_{E}(X)$

Now lets prove (a). Remember $(f_{n}) \in B_{E}(X)$ is cauchy in $B_{E}(X)$. Hence, $\forall m,n \geq N \|f_{n}-f_{m}\|<\epsilon$. Thus, for an arbitrary $x\in X$ $\|f_{n}(x)-f_{m}(x)\|_{E}<\epsilon$. With $n$ fixed and $m \rightarrow \infty$ $\|f_{n}(x)-F(x)\|_{E}<\epsilon$. Since $x\in X$ was arbitrary, then its true $\forall x \in X$. Hence $\|f_{n}-F\| = \sup_{x\in X}\|f_{n}(x)-F(x)\|_{E}<\epsilon$ and $f_{n}\rightarrow F$ in $B_{E}(X)$.

You see its rather confusing to switch between spaces and using appropriate norms for them, otherwise these proofs are quite standard.

Use the proof for $E = \mathbb{R}$ to prove boundedness.