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I have this problem to solve and I am not sure if I'm doing it right.

Let $X$ denote a real vector space consisting of all continuous functions $u \colon[0, \alpha] \to \mathbb R$ such that their norm

$$\|u\| := \sup_{x \in [0, \alpha]} \left| \frac{u(x)}{x} \right|$$

is finite. Is $X$ a Banach space?

I need to prove that $X$ is complete (meaning that all Cauchy sequences converge in $X$). Let $u_{n}$ be a Cauchy sequence in $X$. It means that:

$$( \forall \epsilon > 0) ( \exists N>0) (\forall m,n>N) (||u_{n}-u_{m}||<\epsilon)$$

From the last inequality I deduced that $$( \forall \epsilon > 0)( \exists N>0)(\forall m,n>N)(\forall x\in[0,\alpha])(\frac{|u_{n}(x)-u_{m}(x)|}{x}<\epsilon)$$

It means that for all $x \in \mathbb R$ the sequence ${u_{n}(x)}/{x}$ is Cauchy in $\mathbb{R}$ and therefore converges (due to completeness of $\mathbb R$). Let $\frac{u(x)}{x}$ denote the limit of $\frac{u_{n}(x)}{x}$ as $n$ approaches $\infty$. So:

$$(\forall \epsilon > 0)( \exists N>0)(\forall m,n>N)(\forall x\in[0,\alpha])(\frac{|u_{n}(x)-u(x)|}{x}<\epsilon).$$

It means that $\frac{u_{n}(x)}{x}$ converges uniformly to $\frac{u(x)}{x}$ and belongs itself to the space $\mathcal C [0, \alpha]$. I showed that all Cauchy sequences have a limit in $X$. It this proof correct?

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  • $\begingroup$ Sorry, but the “definition” of $X$ doesn't really define anything. $\endgroup$ – egreg Oct 25 '15 at 11:07
  • $\begingroup$ Isn't $X$ just a space of continuous functions on $[0,\alpha]$ with the norm above? $\endgroup$ – sarnow Oct 25 '15 at 11:16
  • $\begingroup$ I can't see any definition of a norm. Are you perhaps considering $X$ as the set of continuous functions $u\in C[0,\alpha]$ (this ambient space with the norm defined by $\|u\|=\sup\limits_{x\in[0,\alpha]}|u(x)|$) such that $\|u\|=\sup\limits_{x\in(0,\alpha]}\frac{u(x)}{x}$? $\endgroup$ – egreg Oct 25 '15 at 11:20
  • $\begingroup$ I edited my question, I didn't write it OK at first, sorry about it. $\endgroup$ – sarnow Oct 25 '15 at 11:27

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