# Proving Completeness of a $\alpha$-Holder space

In the linked question: Holder Continuous Functions on $$[0,1]$$ are complete + Banach space OP is trying to show an $$\alpha$$-Holder space ($$\Lambda_{\alpha}$$) is a Banach space. The first answer reads proceeds in three steps: producing a limit $$f$$, showing that $$f_n \rightarrow f$$ in norm, and finally showing $$f$$ is in the space.

To produce $$f$$ the poster assumes there's a Cauchy sequence in $$\Lambda_{\alpha}$$ and then states there exists a pointwise limit $$f$$ of this cauchy sequence.

My question: If we show $$f \in \Lambda_{\alpha}$$ aren't we done? -- we've assumed that a cauchy sequence in $$\Lambda_{\alpha}$$ and proved the limit lives in $$\Lambda_{\alpha}$$. If so, why does the poster bother checking $$f_n \rightarrow f$$ in norm separately in step 1? They should be able to just skip to step 2, right?

• Because before proving that, we don't know if $f$ is the limit in the norm sense, and not just a limit in the pointwise sense. The completeness is a statement about the convergence in a particular topology, which is here the norm topology. The fact that the pointwise limit is in the same normed space of all the $f_n$ doesn't imply that the convergence is in the norm sense. – Phil-W Nov 18 '18 at 22:30
• Indeed, consider the sequence of functions $f_n$ defined by $x \mapsto x^n$ in the space of bounded continuous functions on the halfopen interval $[0,1)$ equipped with the supremum norm. The pointwise limit is the zero function which certainly lives in this space, but the norm distance between $f_n$ and $0$ is $1$ for any $n$. – Thomas Bakx Nov 18 '18 at 22:40
• @Phil-W feel free to add your comment as an answer, I can accept it – yoshi Nov 18 '18 at 23:31