# $B$ Banach $\implies B^*$ Banach.

Since we only need to check that $$B^*$$ is complete, we should prove Cauchy sequence $$\lbrace l_n \rbrace$$ converges.

In general idea, $$|(l-l_n)(f)| \leq |(l-l_m)(f)| + |(l_m-l_n)(f)| \leq |(l-l_m)(f)| + \epsilon/2||f||$$ then let m be large enough. We have $$|(l-l_n)(f)| \leq \epsilon ||f||$$ And we finish the proof.

(This proof is in Stein's book Functional Analysis)

However, can I prove it in this way? I mean prove Cauchy sequence $$\lbrace l_n \rbrace$$ converges.

Can I use the definition of norm that $$||l-l_n||=\underset{||f||=1}{sup}|(l-l_n)(f)|$$ and due to $$|(l-l_n)(f)|$$ converges to $$0$$ when $$n$$ goes to infinity, we have $$l_n$$ converges to $$l$$.

It seems to be wrong, but I don't know why.

• The implication you're looking for is true but a bit strange to write down. If $X$ and $Y$ are normed vector spaces then the space of bounded, linear operators from $X$ to $Y$, $L(X,Y)$ is complete (and hence a Banach space) if and only if $Y$ is complete. $X^* = L(B, \mathbb{R})$ and so is a Banach space since $\mathbb{R}$ is complete. This doesn't require $X$ to be complete. – Rhys Steele Jan 14 '19 at 22:37
• Thank you for your reply! It's correct? But why does the book not use this simpler idea? I think the idea is probably wrong. – Busy Zhu Jan 14 '19 at 23:01
• If I'm honest I find your attempt confusing. Given a Cauchy sequence $\{l_n\}_{n \geq 0}$ in $B^*$ you don't a priori have a good candidate for its limit $l$, so the $l$ you write down isn't defined here. Further, writing a bound like $|(l-l_n)(f)| \leq |(l-l_m)(f)| + \frac{\varepsilon}{2} \|f\|$ isn't particularly helpful because you'll still be left to bound $|(l-l_m)(f)|$ in terms of $\|f\|$. For this proof to work you'd need to be able to pick $m$ independent of $f$ and if you could do that you'd just bound $|(l-l_n)(f)|$ for large enough $n$ straight away rather than doing this first. – Rhys Steele Jan 14 '19 at 23:05
• The usual technique is to check that for $f \in B$, $(l_n(f))$ is a Cauchy sequence in $\mathbb{R}$ and so converges to some number $L(f) \in \mathbb{R}$. You can then define a map $l: B \to \mathbb{R}$ by $l(f) = L(f)$. It's then not to hard to check that $l$ is linear by properties of limits. Then you check that $l$ in fact must be a bounded linear map and finally you show that $l_n \to l$ in $B^*$ by sending $m \to \infty$ in $|l_n(f) - l_m(f)| \leq \varepsilon \|f\|$ for suitably large $m,n$. – Rhys Steele Jan 14 '19 at 23:10

The space $$B^*$$ is the dual space of $$B$$, which is the space that contains all linear functionals such that $$f : B \to \mathbb R$$.
But, note that $$\mathbb R$$ is a complete space with respect to its metric (norm), the absolute value. That means that $$\mathbb R$$ is Banach.
• @BusyZhu You want to prove that $B^*$ is a Banach space. All you need to take into account is that $\mathbb R$ is complete, thus Banach. That means that if $x \in B^*$ and is Cauchy, then it converges and the proof is really straightforward. – Rebellos Jan 14 '19 at 22:38