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!
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.