2
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – user251257 Aug 18 '16 at 23:30
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.