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

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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.

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Hint :

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.