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Let $I=[0,1]$ and let $\displaystyle X:=\left\{f: I\times \mathbb R\to \mathbb R\colon \sup_{(t,x)}\frac{|f(t,x)|}{1+|x|}<\infty\right\}$. Prove that $X$, equipped with the norm $\displaystyle \|f\|:=\sup_{(t,x)}\frac{|f(t,x)|}{1+|x|}$ is a Banach space.

My first attempt was to use the characterization that $X$ is a Banach space if and only if every absolutely convergent series converges, but no success.

Then I've noticed that I can prove convergence on every Ball of arbitrarily large radius, however still i cannot conclude on the whole $I\times \mathbb R$. Maybe Ascoli Arzela, but, honestly, i don't know.

Hope you can help me. Thank you.

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But Arzelà-Ascoli is a theorem about $C(K)$ endowed with the sup norm. Here you have a different norm. So I'm not sure how to apply Arzelà-Ascoli in this case. –  Rudy the Reindeer Jul 28 '12 at 14:32

1 Answer 1

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  • The fact that $X$ is a normed space follows from the properties of supremum.
  • Take $\{f_k\}\subset X$ a Cauchy sequence. In particular, for each $(t,x)\in [0,1]\times \Bbb R$, the sequence $\{f_k(t,x)\}$ is Cauchy, hence converges to some $f(t,x)$. Fix $\varepsilon>0$, and take $N=n(\varepsilon)$ such that the norm of $f_k-f_j$ is $\leq \varepsilon$ whenever $j,k\geq N$. We have for all $(t,x)\in I\times \Bbb R$ and $k,j\geq N$ that $$|f_j(t,x)-f_k(t,x)|\leq (1+|x|)\varepsilon.$$ We can take the limit in the last equation to show that $f\in X$. Further more, we can show that this gives the convergence of $f_j$ to $f$.

An alternative way is the following. First show that if $X$ is a non-emptyset, the set of bounded real-valued functions defined on $X$, $B(X)$, endowed with the supremum norm is a Banach space. In this particular case, show that $X$ is a closed subspace of $B(X)$.

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